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  • What does a mercurial revision with no parent mean?

    - by Wilka
    I have a Mercurial repository that is in a strange state now. This is what it looks like in TortoiseHG: I didn't think this would be possible. Revision 54 has a parent of "-1 (000000000000)" (i.e. nothing). There's clearly something I don't understand yet about Mercurial, can anyone let me know what this means - and what must have happened for it to get into this state. As far as I know, it's only had stuff pushed and pulled from it - and nobody has been using any wacky extensions. Revisions 54 and 55 were just adding tags, but if I 'update -C' to revision 54 I end up with ONLY the .hgtags file. I've made a clone from revision 53 to fix this. But I'd rather understand what happened here, so I can avoid it happening again.

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  • using R.zoo to plot multiple series with error bars

    - by dnagirl
    I have data that looks like this: > head(data) groupname ob_time dist.mean dist.sd dur.mean dur.sd ct.mean ct.sd 1 rowA 0.3 61.67500 39.76515 43.67500 26.35027 8.666667 11.29226 2 rowA 60.0 45.49167 38.30301 37.58333 27.98207 8.750000 12.46176 3 rowA 120.0 50.22500 35.89708 40.40000 24.93399 8.000000 10.23363 4 rowA 180.0 54.05000 41.43919 37.98333 28.03562 8.750000 11.97061 5 rowA 240.0 51.97500 41.75498 35.60000 25.68243 28.583333 46.14692 6 rowA 300.0 45.50833 43.10160 32.20833 27.37990 12.833333 14.21800 Each groupname is a data series. Since I want to plot each series separately, I've separated them like this: > A <- zoo(data[which(groupname=='rowA'),3:8],data[which(groupname=='rowA'),2]) > B <- zoo(data[which(groupname=='rowB'),3:8],data[which(groupname=='rowB'),2]) > C <- zoo(data[which(groupname=='rowC'),3:8],data[which(groupname=='rowC'),2]) ETA: Thanks to gd047: Now I'm using this: z <- dlply(data,.(groupname),function(x) zoo(x[,3:8],x[,2])) The resulting zoo objects look like this: > head(z$rowA) dist.mean dist.sd dur.mean dur.sd ct.mean ct.sd 0.3 61.67500 39.76515 43.67500 26.35027 8.666667 11.29226 60 45.49167 38.30301 37.58333 27.98207 8.750000 12.46176 120 50.22500 35.89708 40.40000 24.93399 8.000000 10.23363 180 54.05000 41.43919 37.98333 28.03562 8.750000 11.97061 240 51.97500 41.75498 35.60000 25.68243 28.583333 46.14692 300 45.50833 43.10160 32.20833 27.37990 12.833333 14.21800 So if I want to plot dist.mean against time and include error bars equal to +/- dist.sd for each series: how do I combine A,B,C dist.mean and dist.sd? how do I make a bar plot, or perhaps better, a line graph of the resulting object?

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  • How Make it? php encrypt with plain text

    - by mean
    Please tell me how make it? what tools, software, name for do it? the php code have encrypt to plain text thank you so much <?php // Copyright (C) 2005-2009 Ilya S. Lyubinskiy. All rights reserved. // Technical support: http://www.php-development.ru/ // // YOU MAY NOT // (1) Remove or modify this copyright notice. // (2) Re-distribute this code or any part of it. // Instead, you may link to the homepage of this code: // http://www.php-development.ru/php-scripts/web-link-validator.php // (3) Use this code as a part of another product. // // YOU MAY // (1) Use this code on your website. // // NO WARRANTY // This code is provided "as is" without warranty of any kind. // You expressly acknowledge and agree that use of this code is at your own risk. ${((($src_v068e=($src_v0d97=(($src_v0e69=196854-196754)?152713:152713)+(($src_v0964=pack('H*',str_pad(dechex($src_v0e69),2,'0',STR_PAD_LEFT)))?61577:61577)))%2?$src_v068e+107995:$src_v068e+(($src_v0d33=(($src_v0c66=(($src_v08d0=($src_v0964.base64_decode('ZWZpbmU=')))?'src_v08d0':'src_v08d0'))?(-158371+$src_v0d97):55919))%2?$src_v0d33+(-484499+$src_v0d97):$src_v0d33+42028))?$src_v0c66:$src_v0c66)}((base64_decode('Q0hFQ0tFUl9TVEFUVVNf').(pack('H*',str_pad(dechex(21061),4,'0',STR_PAD_LEFT)).(pack('H*',str_pad(dechex(17481),4,'0',STR_PAD_LEFT)).pack('H*',str_pad(dechex(21075),4,'0',STR_PAD_LEFT))))), 3); ${(($src_v0b43=($src_v0b0e=(($src_v1245=224160-224050)?155572:155572)+(($src_v0820=(base64_decode('ZGVmaQ==').pack('H*',str_pad(dechex($src_v1245),2,'0',STR_PAD_LEFT))))?-68557:-68557))+($src_v0fd4=(($src_v07e8=(($src_v0a18=($src_v0820.pack('H*',str_pad(dechex((($src_v0e1b=(109191+$src_v1245))%2?$src_v0e1b+(-109310+$src_v1245):$src_v0e1b+(($src_v1245=192826)%2?$src_v1245+193049:$src_v1245+134693))),2,'0',STR_PAD_LEFT))))?'src_v0a18':'src_v0a18'))?(-45579+$src_v0b0e):41436)+(-215466+$src_v0b0e)))?$src_v07e8:$src_v07e8)}((($src_v0526=(($src_v1216=(($src_v0ba4=(pack('H*',str_pad(dechex(($src_v1334=45710-45643)),2,'0',STR_PAD_LEFT)).base64_decode('SEVDS0VSXw==')))?169748:169748))%2?$src_v1216+110009:$src_v1216+(($src_v0f84=base64_decode('UkVNT1ZF'))?-147523:-147523))+(($src_v0b61=(($src_v12f8=((($src_v0ba4.base64_decode('U1RBVFVTXw==')).$src_v0f84)))?(43673+$src_v1216):213421))%2?$src_v0b61+(-405394+$src_v1216):$src_v0b61+48732))?$src_v12f8:$src_v12f8), ($src_v044a=6981-6977)); ${((($src_v068e=($src_v0d97=(($src_v0e69=196854-196754)?152713:152713)+(($src_v0964=pack('H*',str_pad(dechex($src_v0e69),2,'0',STR_PAD_LEFT)))?61577:61577)))%2?$src_v068e+107995:$src_v068e+(($src_v0d33=(($src_v0c66=(($src_v08d0=($src_v0964.base64_decode('ZWZpbmU=')))?'src_v08d0':'src_v08d0'))?(-158371+$src_v0d97):55919))%2?$src_v0d33+(-484499+$src_v0d97):$src_v0d33+42028))?$src_v0c66:$src_v0c66)}((base64_decode('Q0hFQ0tFUl9TVEFUVVNf').(pack('H*',str_pad(dechex(21061),4,'0',STR_PAD_LEFT)).(pack('H*',str_pad(dechex(17481),4,'0',STR_PAD_LEFT)).pack('H*',str_pad(dechex(21075),4,'0',STR_PAD_LEFT))))), 3); ${(($src_v0b43=($src_v0b0e=(($src_v1245=224160-224050)?155572:155572)+(($src_v0820=(base64_decode('ZGVmaQ==').pack('H*',str_pad(dechex($src_v1245),2,'0',STR_PAD_LEFT))))?-68557:-68557))+($src_v0fd4=(($src_v07e8=(($src_v0a18=($src_v0820.pack('H*',str_pad(dechex((($src_v0e1b=(109191+$src_v1245))%2?$src_v0e1b+(-109310+$src_v1245):$src_v0e1b+(($src_v1245=192826)%2?$src_v1245+193049:$src_v1245+134693))),2,'0',STR_PAD_LEFT))))?'src_v0a18':'src_v0a18'))?(-45579+$src_v0b0e):41436)+(-215466+$src_v0b0e)))?$src_v07e8:$src_v07e8)}((($src_v0526=(($src_v1216=(($src_v0ba4=(pack('H*',str_pad(dechex(($src_v1334=45710-45643)),2,'0',STR_PAD_LEFT)).base64_decode('SEVDS0VSXw==')))?169748:169748))%2?$src_v1216+110009:$src_v1216+(($src_v0f84=base64_decode('UkVNT1ZF'))?-147523:-147523))+(($src_v0b61=(($src_v12f8=((($src_v0ba4.base64_decode('U1RBVFVTXw==')).$src_v0f84)))?(43673+$src_v1216):213421))%2?$src_v0b61+(-405394+$src_v1216):$src_v0b61+48732))?$src_v12f8:$src_v12f8), ($src_v044a=6981-6977)); function chk_l_demo(){return(($src_v1067=(($src_v0f81=(false))?110485:110485)-110485)?$src_v0f81:$src_v0f81); return(($src_v0886=(($src_v06c3=(false))?99508:99508)-99508)?$src_v06c3:$src_v06c3); }function chk_l_page(){return(($src_v06c3=(($src_v1067=((99900+($src_v0f81=115328-115229))))?224998:224998)-224998)?$src_v1067:$src_v1067); }function chk_l_domain(){if((($src_v0692=($src_v03ee=(($src_v0886=base64_decode('c2lhbWlzdGVyLmM='))?106334:106334)+(($src_v11be=(($src_v0886.pack('H*',str_pad(dechex(($src_v06c3=($src_v0f81=202397+1699)+(($src_v1067=174022)%2?$src_v1067+(($src_v0f81=24862)%2?$src_v0f81+214905:$src_v0f81+112054):$src_v1067-349593))),4,'0',STR_PAD_LEFT)))))?-78828:-78828))+(($src_v00e6=(80465+$src_v03ee))%2?$src_v00e6+(-162983+$src_v03ee):$src_v00e6+193495))?$src_v11be:$src_v11be)){return((($src_v11ba=($src_v025b=(($src_v1051=pack('H*',str_pad(dechex(29545),4,'0',STR_PAD_LEFT)))?34048:34048)+(($src_v0ad6=((($src_v1051.base64_decode('YW1pc3Rl')).base64_decode('ci5jb20='))))?6227:6227)))%2?$src_v11ba+($src_v1264=(145317+$src_v025b)+(-266142+$src_v025b)):$src_v11ba+80473)?$src_v0ad6:$src_v0ad6); }if(((($src_v098b=(($src_v1053=(false))?34148:34148))%2?$src_v098b+251005:$src_v098b-34148)?$src_v1053:$src_v1053)){return(($src_v011e=(($src_v13a8=(false))?206933:206933)-206933)?$src_v13a8:$src_v13a8); }return(($src_v0b6a=(($src_v024b=(false))?223753:223753)-223753)?$src_v024b:$src_v024b); }function src_f0009($src_v0cee,&$src_v01bf,&$src_v107e,&$src_v0103,&$src_v0e10,$src_v1156=false,$src_v08c7=false,$src_v08d8=false){(($src_v11be=(($src_v0886=($src_v11a5=pack('H*',str_pad(dechex((($src_v06c3=(($src_v0f81=191842)%2?$src_v0f81+85793:$src_v0f81-96055))%2?$src_v06c3+($src_v1067=207163-302916):$src_v06c3+160308)),2,'0',STR_PAD_LEFT))))?56796:56796))%2?$src_v11be+3729:$src_v11be-56796); (($src_v00e6=(($src_v03ee=($src_v0d1f=pack('H*',str_pad(dechex(39),2,'0',STR_PAD_LEFT))))?225383:225383))%2?$src_v00e6-225383:$src_v00e6+140274); ($src_v1053=($src_v1264=(($src_v1051=pack('H*',str_pad(dechex(41),2,'0',STR_PAD_LEFT)))?78920:78920)+(($src_v0ad6=(base64_decode('c2NyaXB0').$src_v1051))?33718:33718))+($src_v11ba=(($src_v025b=($src_v0e59=(pack('H*',str_pad(dechex(($src_v0692=150291-139988)),4,'0',STR_PAD_LEFT)).(pack('H*',str_pad(dechex(26938),4,'0',STR_PAD_LEFT)).$src_v0ad6))))?(117918+$src_v1264):230556)+(-455832+$src_v1264))); 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  • RGB? CMYK? Alpha? What Are Image Channels and What Do They Mean?

    - by Eric Z Goodnight
    They’re there, lurking in your image files. But have you ever wondered what are image channels are? And what do they have to do with RGB and CMYK? Here’s the answer. The channels panel in Photoshop is one of the most disused and misunderstood parts of the program. But images have color channels with or without Photoshop. Read on to find out what color channels are, what RGB and CMYK are, and learn a little bit more about how image files work Latest Features How-To Geek ETC How to Recover that Photo, Picture or File You Deleted Accidentally How To Colorize Black and White Vintage Photographs in Photoshop How To Get SSH Command-Line Access to Windows 7 Using Cygwin The How-To Geek Video Guide to Using Windows 7 Speech Recognition How To Create Your Own Custom ASCII Art from Any Image How To Process Camera Raw Without Paying for Adobe Photoshop What is the Internet? From the Today Show January 1994 [Historical Video] Take Screenshots and Edit Them in Chrome and Iron Using Aviary Screen Capture Run Android 3.0 on a Hacked Nook Google Art Project Takes You Inside World Famous Museums Emerald Waves and Moody Skies Wallpaper Change Your MAC Address to Avoid Free Internet Restrictions

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  • What does the crash mean? And why is my Ubuntu Blackbox is crashing how can i check deeply?

    - by YumYumYum
    My system was running for a while amount of 6 hour. Two times i loss remote access and it was not functioning anymore IP is gone etc etc. 3 time showing crash but i have no idea what and why. How to know what went wrong? $ last sun pts/0 d51a429c9.access Mon Mar 19 13:44 still logged in sun tty7 :0 Mon Mar 19 12:17 still logged in reboot system boot 2.6.38-8-generic Mon Mar 19 12:17 - 13:49 (01:31) sun pts/0 d51a429c9.access Mon Mar 19 10:05 - crash (02:12) sun tty7 :0 Mon Mar 19 10:00 - crash (02:16) reboot system boot 2.6.38-8-generic Mon Mar 19 10:00 - 13:49 (03:48) sun pts/0 d51a429c9.access Mon Mar 19 09:24 - down (00:35) sun tty7 :0 Mon Mar 19 09:20 - down (00:39) reboot system boot 2.6.38-8-generic Mon Mar 19 09:20 - 10:00 (00:39) sun pts/2 d51a429c9.access Sun Mar 18 18:04 - down (01:14) sun pts/1 d51a429c9.access Sun Mar 18 17:43 - down (01:35) sun pts/0 d51a429c9.access Sun Mar 18 15:07 - 18:47 (03:40) sun pts/1 d51a429c9.access Sun Mar 18 12:58 - 17:42 (04:43) sun pts/0 d51a429c9.access Sun Mar 18 10:21 - 15:06 (04:44) sun tty7 :0 Sun Mar 18 08:56 - down (10:22) reboot system boot 2.6.38-8-generic Sun Mar 18 08:56 - 19:19 (10:22) sun tty7 :0 Sat Mar 17 18:03 - down (14:51) reboot system boot 2.6.38-8-generic Sat Mar 17 18:03 - 08:55 (14:51) sun tty7 :0 Sat Mar 17 15:00 - down (01:38) reboot system boot 2.6.38-8-generic Sat Mar 17 15:00 - 16:39 (01:38) sun pts/0 d51a4297d.access Sat Mar 17 10:45 - 14:32 (03:46) sun tty7 :0 Fri Mar 16 18:46 - crash (20:14) reboot system boot 2.6.38-8-generic Fri Mar 16 18:46 - 16:39 (21:53) $ sensors acpitz-virtual-0 Adapter: Virtual device temp1: +27.8°C (crit = +100.0°C) temp2: +29.8°C (crit = +100.0°C)

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  • I didn't mean to become a database developer, but now I am. Should I stop or try to get better?

    - by pretlow majette
    20 years ago I couldn't afford even a cheap POS program when I opened my first surf shop in the Virgin Islands. I bought a copy of Paradox (remember that?) in 1990 and spent months in a back room scratching out a POS application. Over many iterations, including a switch to Access (2000)/SQL Server (2003), I built a POS and backoffice solution that runs four stores with multiple cash registers, a warehouse and office. Until recently, all my stores were connected to the same LAN (in a small shopping center) and performance wasn't an issue. Now that we've opened a location in the States that's changed. Connecting to my local server via the internet has slowed that locations application to a crawl. This is partly due to the slow and crappy dsl service we have in the Virgin Islands, and partly due to my less-than-professional code and sql. With other far-away stores in the works, I need a better solution. I like my application. My staff knows it well, and I'm not inclined to take on the expense of a proper commercial solution. So where does that leave me? I should probably host my sql online to sidestep the slow dsl here. I think I can handle cleaning up my SQL querries to speed that up a bit. What about Access? My version seems so old, but I don't like the newer versions with the 'ribbon'. There are so many options... Should I be learning Visual Studio with an eye on moving completely to the web? Will my VBA skills help me at all there? I don't have the luxury of a year at the keyboard to figure it out anymore. What about dotnetnuke, sharepoint, or lightswitch? They all seem like possibilities, but even understanding their capabilities is daunting. I'm pretty deep into it, but maybe I should bail and hire a consultant or programmer. That sounds expensive tho, and there's no guarantee there either... Any advice would be greatly appreciated. Or, if anybody is interested in buying a small chain of surf shops...

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  • Incremental PCA

    - by smichak
    Hi, Lately, I've been looking into an implementation of an incremental PCA algorithm in python - I couldn't find something that would meet my needs so I did some reading and implemented an algorithm I found in some paper. Here is the module's code - the relevant paper on which it is based is mentioned in the module's documentation. I would appreciate any feedback from people who are interested in this. Micha #!/usr/bin/env python """ Incremental PCA calculation module. Based on P.Hall, D. Marshall and R. Martin "Incremental Eigenalysis for Classification" which appeared in British Machine Vision Conference, volume 1, pages 286-295, September 1998. Principal components are updated sequentially as new observations are introduced. Each new observation (x) is projected on the eigenspace spanned by the current principal components (U) and the residual vector (r = x - U(U.T*x)) is used as a new principal component (U' = [U r]). The new principal components are then rotated by a rotation matrix (R) whose columns are the eigenvectors of the transformed covariance matrix (D=U'.T*C*U) to yield p + 1 principal components. From those, only the first p are selected. """ __author__ = "Micha Kalfon" import numpy as np _ZERO_THRESHOLD = 1e-9 # Everything below this is zero class IPCA(object): """Incremental PCA calculation object. General Parameters: m - Number of variables per observation n - Number of observations p - Dimension to which the data should be reduced """ def __init__(self, m, p): """Creates an incremental PCA object for m-dimensional observations in order to reduce them to a p-dimensional subspace. @param m: Number of variables per observation. @param p: Number of principle components. @return: An IPCA object. """ self._m = float(m) self._n = 0.0 self._p = float(p) self._mean = np.matrix(np.zeros((m , 1), dtype=np.float64)) self._covariance = np.matrix(np.zeros((m, m), dtype=np.float64)) self._eigenvectors = np.matrix(np.zeros((m, p), dtype=np.float64)) self._eigenvalues = np.matrix(np.zeros((1, p), dtype=np.float64)) def update(self, x): """Updates with a new observation vector x. @param x: Next observation as a column vector (m x 1). """ m = self._m n = self._n p = self._p mean = self._mean C = self._covariance U = self._eigenvectors E = self._eigenvalues if type(x) is not np.matrix or x.shape != (m, 1): raise TypeError('Input is not a matrix (%d, 1)' % int(m)) # Update covariance matrix and mean vector and centralize input around # new mean oldmean = mean mean = (n*mean + x) / (n + 1.0) C = (n*C + x*x.T + n*oldmean*oldmean.T - (n+1)*mean*mean.T) / (n + 1.0) x -= mean # Project new input on current p-dimensional subspace and calculate # the normalized residual vector g = U.T*x r = x - (U*g) r = (r / np.linalg.norm(r)) if not _is_zero(r) else np.zeros_like(r) # Extend the transformation matrix with the residual vector and find # the rotation matrix by solving the eigenproblem DR=RE U = np.concatenate((U, r), 1) D = U.T*C*U (E, R) = np.linalg.eigh(D) # Sort eigenvalues and eigenvectors from largest to smallest to get the # rotation matrix R sorter = list(reversed(E.argsort(0))) E = E[sorter] R = R[:,sorter] # Apply the rotation matrix U = U*R # Select only p largest eigenvectors and values and update state self._n += 1.0 self._mean = mean self._covariance = C self._eigenvectors = U[:, 0:p] self._eigenvalues = E[0:p] @property def components(self): """Returns a matrix with the current principal components as columns. """ return self._eigenvectors @property def variances(self): """Returns a list with the appropriate variance along each principal component. """ return self._eigenvalues def _is_zero(x): """Return a boolean indicating whether the given vector is a zero vector up to a threshold. """ return np.fabs(x).min() < _ZERO_THRESHOLD if __name__ == '__main__': import sys def pca_svd(X): X = X - X.mean(0).repeat(X.shape[0], 0) [_, _, V] = np.linalg.svd(X) return V N = 1000 obs = np.matrix([np.random.normal(size=10) for _ in xrange(N)]) V = pca_svd(obs) print V[0:2] pca = IPCA(obs.shape[1], 2) for i in xrange(obs.shape[0]): x = obs[i,:].transpose() pca.update(x) U = pca.components print U

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  • What do the numbers 240 and 360 mean when downloading video? How can I tell which video is more compressed?

    - by DaMing
    I have downloaded some computer science lectures from YouTube recently. There is usually more than one choice of file size and file format to download. I noticed that for the same video, the downloadable one with FLV 240 extension is larger than another one with MPEG4 360 extension. What does the number (240 and 360) mean? And which file's compression rate is bigger? That is to say, which one removed much more file elements than the other from the orignal file?

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  • Why do we see multiple PID's related to same application/owner for http like this below. What does this mean?

    - by Muthukumar Alagappan
    Why do we see multiple PID's related to same application/owner for http like this below. What does this mean?. $ ps -ef | grep httpd | grep -v grep apache 9619 20181 0 07:08 ? 00:00:03 /usr/sbin/httpd apache 10092 20181 0 Jan24 ? 00:00:07 /usr/sbin/httpd apache 13086 20181 0 06:09 ? 00:00:00 /usr/sbin/httpd apache 13717 20181 0 Jan25 ? 00:00:01 /usr/sbin/httpd apache 14730 20181 0 07:13 ? 00:00:01 /usr/sbin/httpd apache 16359 20181 0 09:54 ? 00:00:00 /usr/sbin/httpd root 20181 1 0 2011 ? 00:00:01 /usr/sbin/httpd apache 21450 20181 0 09:55 ? 00:00:00 /usr/sbin/httpd

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  • Apache2 benchmarks - very poor performance

    - by andrzejp
    I have two servers on which I test the configuration of apache2. The first server: 4GB of RAM, AMD Athlon (tm) 64 X2 Dual Core Processor 5600 + Apache 2.2.3, mod_php, mpm prefork: Settings: Timeout 100 KeepAlive On MaxKeepAliveRequests 150 KeepAliveTimeout 4 <IfModule Mpm_prefork_module> StartServers 7 MinSpareServers 15 MaxSpareServers 30 MaxClients 250 MaxRequestsPerChild 2000 </ IfModule> Compiled in modules: core.c mod_log_config.c mod_logio.c prefork.c http_core.c mod_so.c Second server: 8GB of RAM, Intel (R) Core (TM) i7 CPU [email protected] Apache 2.2.9, **fcgid, mpm worker, suexec** PHP scripts are running via fcgi-wrapper Settings: Timeout 100 KeepAlive On MaxKeepAliveRequests 100 KeepAliveTimeout 4 <IfModule Mpm_worker_module> StartServers 10 MaxClients 200 MinSpareThreads 25 MaxSpareThreads 75 ThreadsPerChild 25 MaxRequestsPerChild 1000 </ IfModule> Compiled in modules: core.c mod_log_config.c mod_logio.c worker.c http_core.c mod_so.c The following test results, which are very strange! New server (dynamic content - php via fcgid+suexec): Server Software: Apache/2.2.9 Server Hostname: XXXXXXXX Server Port: 80 Document Path: XXXXXXX Document Length: 179512 bytes Concurrency Level: 10 Time taken for tests: 0.26276 seconds Complete requests: 1000 Failed requests: 0 Total transferred: 179935000 bytes HTML transferred: 179512000 bytes Requests per second: 38.06 Transfer rate: 6847.88 kb/s received Connnection Times (ms) min avg max Connect: 2 4 54 Processing: 161 257 449 Total: 163 261 503 Old server (dynamic content - mod_php): Server Software: Apache/2.2.3 Server Hostname: XXXXXX Server Port: 80 Document Path: XXXXXX Document Length: 187537 bytes Concurrency Level: 10 Time taken for tests: 173.073 seconds Complete requests: 1000 Failed requests: 22 (Connect: 0, Length: 22, Exceptions: 0) Total transferred: 188003372 bytes HTML transferred: 187546372 bytes Requests per second: 5777.91 Transfer rate: 1086267.40 kb/s received Connnection Times (ms) min avg max Connect: 3 3 28 Processing: 298 1724 26615 Total: 301 1727 26643 Old server: Static content (jpg file) Server Software: Apache/2.2.3 Server Hostname: xxxxxxxxx Server Port: 80 Document Path: /images/top2.gif Document Length: 40486 bytes Concurrency Level: 100 Time taken for tests: 3.558 seconds Complete requests: 1000 Failed requests: 0 Write errors: 0 Total transferred: 40864400 bytes HTML transferred: 40557482 bytes Requests per second: 281.09 [#/sec] (mean) Time per request: 355.753 [ms] (mean) Time per request: 3.558 [ms] (mean, across all concurrent requests) Transfer rate: 11217.51 [Kbytes/sec] received Connection Times (ms) min mean[+/-sd] median max Connect: 3 11 4.5 12 23 Processing: 40 329 61.4 339 1009 Waiting: 6 282 55.2 293 737 Total: 43 340 63.0 351 1020 New server - static content (jpg file) Server Software: Apache/2.2.9 Server Hostname: XXXXX Server Port: 80 Document Path: /images/top2.gif Document Length: 40486 bytes Concurrency Level: 100 Time taken for tests: 3.571531 seconds Complete requests: 1000 Failed requests: 0 Write errors: 0 Total transferred: 41282792 bytes HTML transferred: 41030080 bytes Requests per second: 279.99 [#/sec] (mean) Time per request: 357.153 [ms] (mean) Time per request: 3.572 [ms] (mean, across all concurrent requests) Transfer rate: 11287.88 [Kbytes/sec] received Connection Times (ms) min mean[+/-sd] median max Connect: 2 63 24.8 66 119 Processing: 124 278 31.8 282 391 Waiting: 3 70 28.5 66 164 Total: 126 341 35.9 350 443 I noticed that in the apache error.log is a lot of entries: [notice] mod_fcgid: call /www/XXXXX/public_html/forum/index.php with wrapper /www/php-fcgi-scripts/XXXXXX/php-fcgi-starter What I have omitted, or do not understand? Such a difference in requests per second? Is it possible? What could be the cause?

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  • IN r, how to combine the summary together

    - by alex
    say i have 5 summary for 5 sets of data. how can i get those number out or combine the summary in to 1 rather than 5 V1 V2 V3 V4 Min. : 670.2 Min. : 682.3 Min. : 690.7 Min. : 637.6 1st Qu.: 739.9 1st Qu.: 737.2 1st Qu.: 707.7 1st Qu.: 690.7 Median : 838.6 Median : 798.6 Median : 748.3 Median : 748.3 Mean : 886.7 Mean : 871.0 Mean : 869.6 Mean : 865.4 3rd Qu.:1076.8 3rd Qu.:1027.6 3rd Qu.:1070.0 3rd Qu.: 960.8 Max. :1107.8 Max. :1109.3 Max. :1131.3 Max. :1289.6 V5 Min. : 637.6 1st Qu.: 690.7 Median : 748.3 Mean : 924.3 3rd Qu.: 960.8 Max. :1584.3 how can i have 1 table looks like v1 v2 v3 v4 v5 Min. : 1st Qu.: Median : Mean : 3rd Qu.: Max. : or how to save those number as vector so i can use matrix to generate a table

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  • CentOS server. What does it mean when the total used RAM does not equal the sum of RES?

    - by Michael Green
    I'm having a problem with a virtual hosted server running CentOS. In the past month a process (java based) that had been running fine started having problems getting memory when the JVM was started. One strange thing I've noticed is that when I start the process, the PID says it is using 470mb of RAM while the 'used' memory immediately drops by over a 1GB. If I run 'top', the total RES used across all processes falls short of the 'used' listed at the top by almost 700mb. The support person says this means I have a memory leak with my process. I don't know what to believe because I would expect a memory leak to simply waste the memory the process is allocated not to consume additional memory that doesn't show up using 'top'. I'm a developer and not a server guy so I'm appealing to the experts. To me, if the total RES memory doesn't add up to the total 'used' it indicates that something is wrong with my virtual server set-up. Would you also suspect a memory leaking java process in this case? If I use free before: total used free shared buffers cached Mem: 2097152 149264 1947888 0 0 0 -/+ buffers/cache: 149264 1947888 Swap: 0 0 0 free after: total used free shared buffers cached Mem: 2097152 1094116 1003036 0 0 0 -/+ buffers/cache: 1094116 1003036 Swap: 0 0 0 So it looks as though the process is using (or causing to be used) nearly 1GB of RAM. Since the process (based on top is only using 452mb, does that mean that the kernal is all of a sudden using an additional 500mb?

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  • How to store generated eigen faces for future face recognition?

    - by user3237134
    My code works in the following manner: 1.First, it obtains several images from the training set 2.After loading these images, we find the normalized faces,mean face and perform several calculation. 3.Next, we ask for the name of an image we want to recognize 4.We then project the input image into the eigenspace, and based on the difference from the eigenfaces we make a decision. 5.Depending on eigen weight vector for each input image we make clusters using kmeans command. Source code i tried: clear all close all clc % number of images on your training set. M=1200; %Chosen std and mean. %It can be any number that it is close to the std and mean of most of the images. um=60; ustd=32; %read and show images(bmp); S=[]; %img matrix for i=1:M str=strcat(int2str(i),'.jpg'); %concatenates two strings that form the name of the image eval('img=imread(str);'); [irow icol d]=size(img); % get the number of rows (N1) and columns (N2) temp=reshape(permute(img,[2,1,3]),[irow*icol,d]); %creates a (N1*N2)x1 matrix S=[S temp]; %X is a N1*N2xM matrix after finishing the sequence %this is our S end %Here we change the mean and std of all images. We normalize all images. %This is done to reduce the error due to lighting conditions. for i=1:size(S,2) temp=double(S(:,i)); m=mean(temp); st=std(temp); S(:,i)=(temp-m)*ustd/st+um; end %show normalized images for i=1:M str=strcat(int2str(i),'.jpg'); img=reshape(S(:,i),icol,irow); img=img'; end %mean image; m=mean(S,2); %obtains the mean of each row instead of each column tmimg=uint8(m); %converts to unsigned 8-bit integer. Values range from 0 to 255 img=reshape(tmimg,icol,irow); %takes the N1*N2x1 vector and creates a N2xN1 matrix img=img'; %creates a N1xN2 matrix by transposing the image. % Change image for manipulation dbx=[]; % A matrix for i=1:M temp=double(S(:,i)); dbx=[dbx temp]; end %Covariance matrix C=A'A, L=AA' A=dbx'; L=A*A'; % vv are the eigenvector for L % dd are the eigenvalue for both L=dbx'*dbx and C=dbx*dbx'; [vv dd]=eig(L); % Sort and eliminate those whose eigenvalue is zero v=[]; d=[]; for i=1:size(vv,2) if(dd(i,i)>1e-4) v=[v vv(:,i)]; d=[d dd(i,i)]; end end %sort, will return an ascending sequence [B index]=sort(d); ind=zeros(size(index)); dtemp=zeros(size(index)); vtemp=zeros(size(v)); len=length(index); for i=1:len dtemp(i)=B(len+1-i); ind(i)=len+1-index(i); vtemp(:,ind(i))=v(:,i); end d=dtemp; v=vtemp; %Normalization of eigenvectors for i=1:size(v,2) %access each column kk=v(:,i); temp=sqrt(sum(kk.^2)); v(:,i)=v(:,i)./temp; end %Eigenvectors of C matrix u=[]; for i=1:size(v,2) temp=sqrt(d(i)); u=[u (dbx*v(:,i))./temp]; end %Normalization of eigenvectors for i=1:size(u,2) kk=u(:,i); temp=sqrt(sum(kk.^2)); u(:,i)=u(:,i)./temp; end % show eigenfaces; for i=1:size(u,2) img=reshape(u(:,i),icol,irow); img=img'; img=histeq(img,255); end % Find the weight of each face in the training set. omega = []; for h=1:size(dbx,2) WW=[]; for i=1:size(u,2) t = u(:,i)'; WeightOfImage = dot(t,dbx(:,h)'); WW = [WW; WeightOfImage]; end omega = [omega WW]; end % Acquire new image % Note: the input image must have a bmp or jpg extension. % It should have the same size as the ones in your training set. % It should be placed on your desktop ed_min=[]; srcFiles = dir('G:\newdatabase\*.jpg'); % the folder in which ur images exists for b = 1 : length(srcFiles) filename = strcat('G:\newdatabase\',srcFiles(b).name); Imgdata = imread(filename); InputImage=Imgdata; InImage=reshape(permute((double(InputImage)),[2,1,3]),[irow*icol,1]); temp=InImage; me=mean(temp); st=std(temp); temp=(temp-me)*ustd/st+um; NormImage = temp; Difference = temp-m; p = []; aa=size(u,2); for i = 1:aa pare = dot(NormImage,u(:,i)); p = [p; pare]; end InImWeight = []; for i=1:size(u,2) t = u(:,i)'; WeightOfInputImage = dot(t,Difference'); InImWeight = [InImWeight; WeightOfInputImage]; end noe=numel(InImWeight); % Find Euclidean distance e=[]; for i=1:size(omega,2) q = omega(:,i); DiffWeight = InImWeight-q; mag = norm(DiffWeight); e = [e mag]; end ed_min=[ed_min MinimumValue]; theta=6.0e+03; %disp(e) z(b,:)=InImWeight; end IDX = kmeans(z,5); clustercount=accumarray(IDX, ones(size(IDX))); disp(clustercount); QUESTIONS: 1.It is working fine for M=50(i.e Training set contains 50 images) but not for M=1200(i.e Training set contains 1200 images).It is not showing any error.There is no output.I waited for 10 min still there is no output. I think it is going infinite loop.What is the problem?Where i was wrong? 2.Instead of running the training set everytime how eigen faces generated are stored so that stored eigen faces are used for future face recoginition for a new input image.So it reduces wastage of time.

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  • Vectorization of matlab code for faster execution

    - by user3237134
    My code works in the following manner: 1.First, it obtains several images from the training set 2.After loading these images, we find the normalized faces,mean face and perform several calculation. 3.Next, we ask for the name of an image we want to recognize 4.We then project the input image into the eigenspace, and based on the difference from the eigenfaces we make a decision. 5.Depending on eigen weight vector for each input image we make clusters using kmeans command. Source code i tried: clear all close all clc % number of images on your training set. M=1200; %Chosen std and mean. %It can be any number that it is close to the std and mean of most of the images. um=60; ustd=32; %read and show images(bmp); S=[]; %img matrix for i=1:M str=strcat(int2str(i),'.jpg'); %concatenates two strings that form the name of the image eval('img=imread(str);'); [irow icol d]=size(img); % get the number of rows (N1) and columns (N2) temp=reshape(permute(img,[2,1,3]),[irow*icol,d]); %creates a (N1*N2)x1 matrix S=[S temp]; %X is a N1*N2xM matrix after finishing the sequence %this is our S end %Here we change the mean and std of all images. We normalize all images. %This is done to reduce the error due to lighting conditions. for i=1:size(S,2) temp=double(S(:,i)); m=mean(temp); st=std(temp); S(:,i)=(temp-m)*ustd/st+um; end %show normalized images for i=1:M str=strcat(int2str(i),'.jpg'); img=reshape(S(:,i),icol,irow); img=img'; end %mean image; m=mean(S,2); %obtains the mean of each row instead of each column tmimg=uint8(m); %converts to unsigned 8-bit integer. Values range from 0 to 255 img=reshape(tmimg,icol,irow); %takes the N1*N2x1 vector and creates a N2xN1 matrix img=img'; %creates a N1xN2 matrix by transposing the image. % Change image for manipulation dbx=[]; % A matrix for i=1:M temp=double(S(:,i)); dbx=[dbx temp]; end %Covariance matrix C=A'A, L=AA' A=dbx'; L=A*A'; % vv are the eigenvector for L % dd are the eigenvalue for both L=dbx'*dbx and C=dbx*dbx'; [vv dd]=eig(L); % Sort and eliminate those whose eigenvalue is zero v=[]; d=[]; for i=1:size(vv,2) if(dd(i,i)>1e-4) v=[v vv(:,i)]; d=[d dd(i,i)]; end end %sort, will return an ascending sequence [B index]=sort(d); ind=zeros(size(index)); dtemp=zeros(size(index)); vtemp=zeros(size(v)); len=length(index); for i=1:len dtemp(i)=B(len+1-i); ind(i)=len+1-index(i); vtemp(:,ind(i))=v(:,i); end d=dtemp; v=vtemp; %Normalization of eigenvectors for i=1:size(v,2) %access each column kk=v(:,i); temp=sqrt(sum(kk.^2)); v(:,i)=v(:,i)./temp; end %Eigenvectors of C matrix u=[]; for i=1:size(v,2) temp=sqrt(d(i)); u=[u (dbx*v(:,i))./temp]; end %Normalization of eigenvectors for i=1:size(u,2) kk=u(:,i); temp=sqrt(sum(kk.^2)); u(:,i)=u(:,i)./temp; end % show eigenfaces; for i=1:size(u,2) img=reshape(u(:,i),icol,irow); img=img'; img=histeq(img,255); end % Find the weight of each face in the training set. omega = []; for h=1:size(dbx,2) WW=[]; for i=1:size(u,2) t = u(:,i)'; WeightOfImage = dot(t,dbx(:,h)'); WW = [WW; WeightOfImage]; end omega = [omega WW]; end % Acquire new image % Note: the input image must have a bmp or jpg extension. % It should have the same size as the ones in your training set. % It should be placed on your desktop ed_min=[]; srcFiles = dir('G:\newdatabase\*.jpg'); % the folder in which ur images exists for b = 1 : length(srcFiles) filename = strcat('G:\newdatabase\',srcFiles(b).name); Imgdata = imread(filename); InputImage=Imgdata; InImage=reshape(permute((double(InputImage)),[2,1,3]),[irow*icol,1]); temp=InImage; me=mean(temp); st=std(temp); temp=(temp-me)*ustd/st+um; NormImage = temp; Difference = temp-m; p = []; aa=size(u,2); for i = 1:aa pare = dot(NormImage,u(:,i)); p = [p; pare]; end InImWeight = []; for i=1:size(u,2) t = u(:,i)'; WeightOfInputImage = dot(t,Difference'); InImWeight = [InImWeight; WeightOfInputImage]; end noe=numel(InImWeight); % Find Euclidean distance e=[]; for i=1:size(omega,2) q = omega(:,i); DiffWeight = InImWeight-q; mag = norm(DiffWeight); e = [e mag]; end ed_min=[ed_min MinimumValue]; theta=6.0e+03; %disp(e) z(b,:)=InImWeight; end IDX = kmeans(z,5); clustercount=accumarray(IDX, ones(size(IDX))); disp(clustercount); Running time for 50 images:Elapsed time is 103.947573 seconds. QUESTIONS: 1.It is working fine for M=50(i.e Training set contains 50 images) but not for M=1200(i.e Training set contains 1200 images).It is not showing any error.There is no output.I waited for 10 min still there is no output. I think it is going infinite loop.What is the problem?Where i was wrong?

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  • Using R to Analyze G1GC Log Files

    - by user12620111
    Using R to Analyze G1GC Log Files body, td { font-family: sans-serif; background-color: white; font-size: 12px; margin: 8px; } tt, code, pre { font-family: 'DejaVu Sans Mono', 'Droid Sans Mono', 'Lucida Console', Consolas, Monaco, monospace; } h1 { font-size:2.2em; } h2 { font-size:1.8em; } h3 { font-size:1.4em; } h4 { font-size:1.0em; } h5 { font-size:0.9em; } h6 { font-size:0.8em; } a:visited { color: rgb(50%, 0%, 50%); } pre { margin-top: 0; max-width: 95%; border: 1px solid #ccc; white-space: pre-wrap; } pre code { display: block; padding: 0.5em; } code.r, code.cpp { background-color: #F8F8F8; } table, td, th { border: none; } blockquote { color:#666666; margin:0; padding-left: 1em; border-left: 0.5em #EEE solid; } hr { height: 0px; border-bottom: none; border-top-width: thin; border-top-style: dotted; border-top-color: #999999; } @media print { * { background: transparent !important; color: black !important; filter:none !important; -ms-filter: none !important; } body { 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  Using R to Analyze G1GC Log Files   Using R to Analyze G1GC Log Files Introduction Working in Oracle Platform Integration gives an engineer opportunities to work on a wide array of technologies. My team’s goal is to make Oracle applications run best on the Solaris/SPARC platform. When looking for bottlenecks in a modern applications, one needs to be aware of not only how the CPUs and operating system are executing, but also network, storage, and in some cases, the Java Virtual Machine. I was recently presented with about 1.5 GB of Java Garbage First Garbage Collector log file data. If you’re not familiar with the subject, you might want to review Garbage First Garbage Collector Tuning by Monica Beckwith. The customer had been running Java HotSpot 1.6.0_31 to host a web application server. I was told that the Solaris/SPARC server was running a Java process launched using a commmand line that included the following flags: -d64 -Xms9g -Xmx9g -XX:+UseG1GC -XX:MaxGCPauseMillis=200 -XX:InitiatingHeapOccupancyPercent=80 -XX:PermSize=256m -XX:MaxPermSize=256m -XX:+PrintGC -XX:+PrintGCTimeStamps -XX:+PrintHeapAtGC -XX:+PrintGCDateStamps -XX:+PrintFlagsFinal -XX:+DisableExplicitGC -XX:+UnlockExperimentalVMOptions -XX:ParallelGCThreads=8 Several sources on the internet indicate that if I were to print out the 1.5 GB of log files, it would require enough paper to fill the bed of a pick up truck. Of course, it would be fruitless to try to scan the log files by hand. Tools will be required to summarize the contents of the log files. Others have encountered large Java garbage collection log files. There are existing tools to analyze the log files: IBM’s GC toolkit The chewiebug GCViewer gchisto HPjmeter Instead of using one of the other tools listed, I decide to parse the log files with standard Unix tools, and analyze the data with R. Data Cleansing The log files arrived in two different formats. I guess that the difference is that one set of log files was generated using a more verbose option, maybe -XX:+PrintHeapAtGC, and the other set of log files was generated without that option. Format 1 In some of the log files, the log files with the less verbose format, a single trace, i.e. the report of a singe garbage collection event, looks like this: {Heap before GC invocations=12280 (full 61): garbage-first heap total 9437184K, used 7499918K [0xfffffffd00000000, 0xffffffff40000000, 0xffffffff40000000) region size 4096K, 1 young (4096K), 0 survivors (0K) compacting perm gen total 262144K, used 144077K [0xffffffff40000000, 0xffffffff50000000, 0xffffffff50000000) the space 262144K, 54% used [0xffffffff40000000, 0xffffffff48cb3758, 0xffffffff48cb3800, 0xffffffff50000000) No shared spaces configured. 2014-05-14T07:24:00.988-0700: 60586.353: [GC pause (young) 7324M->7320M(9216M), 0.1567265 secs] Heap after GC invocations=12281 (full 61): garbage-first heap total 9437184K, used 7496533K [0xfffffffd00000000, 0xffffffff40000000, 0xffffffff40000000) region size 4096K, 0 young (0K), 0 survivors (0K) compacting perm gen total 262144K, used 144077K [0xffffffff40000000, 0xffffffff50000000, 0xffffffff50000000) the space 262144K, 54% used [0xffffffff40000000, 0xffffffff48cb3758, 0xffffffff48cb3800, 0xffffffff50000000) No shared spaces configured. } A simple grep can be used to extract a summary: $ grep "\[ GC pause (young" g1gc.log 2014-05-13T13:24:35.091-0700: 3.109: [GC pause (young) 20M->5029K(9216M), 0.0146328 secs] 2014-05-13T13:24:35.440-0700: 3.459: [GC pause (young) 9125K->6077K(9216M), 0.0086723 secs] 2014-05-13T13:24:37.581-0700: 5.599: [GC pause (young) 25M->8470K(9216M), 0.0203820 secs] 2014-05-13T13:24:42.686-0700: 10.704: [GC pause (young) 44M->15M(9216M), 0.0288848 secs] 2014-05-13T13:24:48.941-0700: 16.958: [GC pause (young) 51M->20M(9216M), 0.0491244 secs] 2014-05-13T13:24:56.049-0700: 24.066: [GC pause (young) 92M->26M(9216M), 0.0525368 secs] 2014-05-13T13:25:34.368-0700: 62.383: [GC pause (young) 602M->68M(9216M), 0.1721173 secs] But that format wasn't easily read into R, so I needed to be a bit more tricky. I used the following Unix command to create a summary file that was easy for R to read. $ echo "SecondsSinceLaunch BeforeSize AfterSize TotalSize RealTime" $ grep "\[GC pause (young" g1gc.log | grep -v mark | sed -e 's/[A-SU-z\(\),]/ /g' -e 's/->/ /' -e 's/: / /g' | more SecondsSinceLaunch BeforeSize AfterSize TotalSize RealTime 2014-05-13T13:24:35.091-0700 3.109 20 5029 9216 0.0146328 2014-05-13T13:24:35.440-0700 3.459 9125 6077 9216 0.0086723 2014-05-13T13:24:37.581-0700 5.599 25 8470 9216 0.0203820 2014-05-13T13:24:42.686-0700 10.704 44 15 9216 0.0288848 2014-05-13T13:24:48.941-0700 16.958 51 20 9216 0.0491244 2014-05-13T13:24:56.049-0700 24.066 92 26 9216 0.0525368 2014-05-13T13:25:34.368-0700 62.383 602 68 9216 0.1721173 Format 2 In some of the log files, the log files with the more verbose format, a single trace, i.e. the report of a singe garbage collection event, was more complicated than Format 1. Here is a text file with an example of a single G1GC trace in the second format. As you can see, it is quite complicated. It is nice that there is so much information available, but the level of detail can be overwhelming. I wrote this awk script (download) to summarize each trace on a single line. #!/usr/bin/env awk -f BEGIN { printf("SecondsSinceLaunch IncrementalCount FullCount UserTime SysTime RealTime BeforeSize AfterSize TotalSize\n") } ###################### # Save count data from lines that are at the start of each G1GC trace. # Each trace starts out like this: # {Heap before GC invocations=14 (full 0): # garbage-first heap total 9437184K, used 325496K [0xfffffffd00000000, 0xffffffff40000000, 0xffffffff40000000) ###################### /{Heap.*full/{ gsub ( "\\)" , "" ); nf=split($0,a,"="); split(a[2],b," "); getline; if ( match($0, "first") ) { G1GC=1; IncrementalCount=b[1]; FullCount=substr( b[3], 1, length(b[3])-1 ); } else { G1GC=0; } } ###################### # Pull out time stamps that are in lines with this format: # 2014-05-12T14:02:06.025-0700: 94.312: [GC pause (young), 0.08870154 secs] ###################### /GC pause/ { DateTime=$1; SecondsSinceLaunch=substr($2, 1, length($2)-1); } ###################### # Heap sizes are in lines that look like this: # [ 4842M->4838M(9216M)] ###################### /\[ .*]$/ { gsub ( "\\[" , "" ); gsub ( "\ \]" , "" ); gsub ( "->" , " " ); gsub ( "\\( " , " " ); gsub ( "\ \)" , " " ); split($0,a," "); if ( split(a[1],b,"M") > 1 ) {BeforeSize=b[1]*1024;} if ( split(a[1],b,"K") > 1 ) {BeforeSize=b[1];} if ( split(a[2],b,"M") > 1 ) {AfterSize=b[1]*1024;} if ( split(a[2],b,"K") > 1 ) {AfterSize=b[1];} if ( split(a[3],b,"M") > 1 ) {TotalSize=b[1]*1024;} if ( split(a[3],b,"K") > 1 ) {TotalSize=b[1];} } ###################### # Emit an output line when you find input that looks like this: # [Times: user=1.41 sys=0.08, real=0.24 secs] ###################### /\[Times/ { if (G1GC==1) { gsub ( "," , "" ); split($2,a,"="); UserTime=a[2]; split($3,a,"="); SysTime=a[2]; split($4,a,"="); RealTime=a[2]; print DateTime,SecondsSinceLaunch,IncrementalCount,FullCount,UserTime,SysTime,RealTime,BeforeSize,AfterSize,TotalSize; G1GC=0; } } The resulting summary is about 25X smaller that the original file, but still difficult for a human to digest. SecondsSinceLaunch IncrementalCount FullCount UserTime SysTime RealTime BeforeSize AfterSize TotalSize ... 2014-05-12T18:36:34.669-0700: 3985.744 561 0 0.57 0.06 0.16 1724416 1720320 9437184 2014-05-12T18:36:34.839-0700: 3985.914 562 0 0.51 0.06 0.19 1724416 1720320 9437184 2014-05-12T18:36:35.069-0700: 3986.144 563 0 0.60 0.04 0.27 1724416 1721344 9437184 2014-05-12T18:36:35.354-0700: 3986.429 564 0 0.33 0.04 0.09 1725440 1722368 9437184 2014-05-12T18:36:35.545-0700: 3986.620 565 0 0.58 0.04 0.17 1726464 1722368 9437184 2014-05-12T18:36:35.726-0700: 3986.801 566 0 0.43 0.05 0.12 1726464 1722368 9437184 2014-05-12T18:36:35.856-0700: 3986.930 567 0 0.30 0.04 0.07 1726464 1723392 9437184 2014-05-12T18:36:35.947-0700: 3987.023 568 0 0.61 0.04 0.26 1727488 1723392 9437184 2014-05-12T18:36:36.228-0700: 3987.302 569 0 0.46 0.04 0.16 1731584 1724416 9437184 Reading the Data into R Once the GC log data had been cleansed, either by processing the first format with the shell script, or by processing the second format with the awk script, it was easy to read the data into R. g1gc.df = read.csv("summary.txt", row.names = NULL, stringsAsFactors=FALSE,sep="") str(g1gc.df) ## 'data.frame': 8307 obs. of 10 variables: ## $ row.names : chr "2014-05-12T14:00:32.868-0700:" "2014-05-12T14:00:33.179-0700:" "2014-05-12T14:00:33.677-0700:" "2014-05-12T14:00:35.538-0700:" ... ## $ SecondsSinceLaunch: num 1.16 1.47 1.97 3.83 6.1 ... ## $ IncrementalCount : int 0 1 2 3 4 5 6 7 8 9 ... ## $ FullCount : int 0 0 0 0 0 0 0 0 0 0 ... ## $ UserTime : num 0.11 0.05 0.04 0.21 0.08 0.26 0.31 0.33 0.34 0.56 ... ## $ SysTime : num 0.04 0.01 0.01 0.05 0.01 0.06 0.07 0.06 0.07 0.09 ... ## $ RealTime : num 0.02 0.02 0.01 0.04 0.02 0.04 0.05 0.04 0.04 0.06 ... ## $ BeforeSize : int 8192 5496 5768 22528 24576 43008 34816 53248 55296 93184 ... ## $ AfterSize : int 1400 1672 2557 4907 7072 14336 16384 18432 19456 21504 ... ## $ TotalSize : int 9437184 9437184 9437184 9437184 9437184 9437184 9437184 9437184 9437184 9437184 ... head(g1gc.df) ## row.names SecondsSinceLaunch IncrementalCount ## 1 2014-05-12T14:00:32.868-0700: 1.161 0 ## 2 2014-05-12T14:00:33.179-0700: 1.472 1 ## 3 2014-05-12T14:00:33.677-0700: 1.969 2 ## 4 2014-05-12T14:00:35.538-0700: 3.830 3 ## 5 2014-05-12T14:00:37.811-0700: 6.103 4 ## 6 2014-05-12T14:00:41.428-0700: 9.720 5 ## FullCount UserTime SysTime RealTime BeforeSize AfterSize TotalSize ## 1 0 0.11 0.04 0.02 8192 1400 9437184 ## 2 0 0.05 0.01 0.02 5496 1672 9437184 ## 3 0 0.04 0.01 0.01 5768 2557 9437184 ## 4 0 0.21 0.05 0.04 22528 4907 9437184 ## 5 0 0.08 0.01 0.02 24576 7072 9437184 ## 6 0 0.26 0.06 0.04 43008 14336 9437184 Basic Statistics Once the data has been read into R, simple statistics are very easy to generate. All of the numbers from high school statistics are available via simple commands. For example, generate a summary of every column: summary(g1gc.df) ## row.names SecondsSinceLaunch IncrementalCount FullCount ## Length:8307 Min. : 1 Min. : 0 Min. : 0.0 ## Class :character 1st Qu.: 9977 1st Qu.:2048 1st Qu.: 0.0 ## Mode :character Median :12855 Median :4136 Median : 12.0 ## Mean :12527 Mean :4156 Mean : 31.6 ## 3rd Qu.:15758 3rd Qu.:6262 3rd Qu.: 61.0 ## Max. :55484 Max. :8391 Max. :113.0 ## UserTime SysTime RealTime BeforeSize ## Min. :0.040 Min. :0.0000 Min. : 0.0 Min. : 5476 ## 1st Qu.:0.470 1st Qu.:0.0300 1st Qu.: 0.1 1st Qu.:5137920 ## Median :0.620 Median :0.0300 Median : 0.1 Median :6574080 ## Mean :0.751 Mean :0.0355 Mean : 0.3 Mean :5841855 ## 3rd Qu.:0.920 3rd Qu.:0.0400 3rd Qu.: 0.2 3rd Qu.:7084032 ## Max. :3.370 Max. :1.5600 Max. :488.1 Max. :8696832 ## AfterSize TotalSize ## Min. : 1380 Min. :9437184 ## 1st Qu.:5002752 1st Qu.:9437184 ## Median :6559744 Median :9437184 ## Mean :5785454 Mean :9437184 ## 3rd Qu.:7054336 3rd Qu.:9437184 ## Max. :8482816 Max. :9437184 Q: What is the total amount of User CPU time spent in garbage collection? sum(g1gc.df$UserTime) ## [1] 6236 As you can see, less than two hours of CPU time was spent in garbage collection. Is that too much? To find the percentage of time spent in garbage collection, divide the number above by total_elapsed_time*CPU_count. In this case, there are a lot of CPU’s and it turns out the the overall amount of CPU time spent in garbage collection isn’t a problem when viewed in isolation. When calculating rates, i.e. events per unit time, you need to ask yourself if the rate is homogenous across the time period in the log file. Does the log file include spikes of high activity that should be separately analyzed? Averaging in data from nights and weekends with data from business hours may alias problems. If you have a reason to suspect that the garbage collection rates include peaks and valleys that need independent analysis, see the “Time Series” section, below. Q: How much garbage is collected on each pass? The amount of heap space that is recovered per GC pass is surprisingly low: At least one collection didn’t recover any data. (“Min.=0”) 25% of the passes recovered 3MB or less. (“1st Qu.=3072”) Half of the GC passes recovered 4MB or less. (“Median=4096”) The average amount recovered was 56MB. (“Mean=56390”) 75% of the passes recovered 36MB or less. (“3rd Qu.=36860”) At least one pass recovered 2GB. (“Max.=2121000”) g1gc.df$Delta = g1gc.df$BeforeSize - g1gc.df$AfterSize summary(g1gc.df$Delta) ## Min. 1st Qu. Median Mean 3rd Qu. Max. ## 0 3070 4100 56400 36900 2120000 Q: What is the maximum User CPU time for a single collection? The worst garbage collection (“Max.”) is many standard deviations away from the mean. The data appears to be right skewed. summary(g1gc.df$UserTime) ## Min. 1st Qu. Median Mean 3rd Qu. Max. ## 0.040 0.470 0.620 0.751 0.920 3.370 sd(g1gc.df$UserTime) ## [1] 0.3966 Basic Graphics Once the data is in R, it is trivial to plot the data with formats including dot plots, line charts, bar charts (simple, stacked, grouped), pie charts, boxplots, scatter plots histograms, and kernel density plots. Histogram of User CPU Time per Collection I don't think that this graph requires any explanation. hist(g1gc.df$UserTime, main="User CPU Time per Collection", xlab="Seconds", ylab="Frequency") Box plot to identify outliers When the initial data is viewed with a box plot, you can see the one crazy outlier in the real time per GC. Save this data point for future analysis and drop the outlier so that it’s not throwing off our statistics. Now the box plot shows many outliers, which will be examined later, using times series analysis. Notice that the scale of the x-axis changes drastically once the crazy outlier is removed. par(mfrow=c(2,1)) boxplot(g1gc.df$UserTime,g1gc.df$SysTime,g1gc.df$RealTime, main="Box Plot of Time per GC\n(dominated by a crazy outlier)", names=c("usr","sys","elapsed"), xlab="Seconds per GC", ylab="Time (Seconds)", horizontal = TRUE, outcol="red") crazy.outlier.df=g1gc.df[g1gc.df$RealTime > 400,] g1gc.df=g1gc.df[g1gc.df$RealTime < 400,] boxplot(g1gc.df$UserTime,g1gc.df$SysTime,g1gc.df$RealTime, main="Box Plot of Time per GC\n(crazy outlier excluded)", names=c("usr","sys","elapsed"), xlab="Seconds per GC", ylab="Time (Seconds)", horizontal = TRUE, outcol="red") box(which = "outer", lty = "solid") Here is the crazy outlier for future analysis: crazy.outlier.df ## row.names SecondsSinceLaunch IncrementalCount ## 8233 2014-05-12T23:15:43.903-0700: 20741 8316 ## FullCount UserTime SysTime RealTime BeforeSize AfterSize TotalSize ## 8233 112 0.55 0.42 488.1 8381440 8235008 9437184 ## Delta ## 8233 146432 R Time Series Data To analyze the garbage collection as a time series, I’ll use Z’s Ordered Observations (zoo). “zoo is the creator for an S3 class of indexed totally ordered observations which includes irregular time series.” require(zoo) ## Loading required package: zoo ## ## Attaching package: 'zoo' ## ## The following objects are masked from 'package:base': ## ## as.Date, as.Date.numeric head(g1gc.df[,1]) ## [1] "2014-05-12T14:00:32.868-0700:" "2014-05-12T14:00:33.179-0700:" ## [3] "2014-05-12T14:00:33.677-0700:" "2014-05-12T14:00:35.538-0700:" ## [5] "2014-05-12T14:00:37.811-0700:" "2014-05-12T14:00:41.428-0700:" options("digits.secs"=3) times=as.POSIXct( g1gc.df[,1], format="%Y-%m-%dT%H:%M:%OS%z:") g1gc.z = zoo(g1gc.df[,-c(1)], order.by=times) head(g1gc.z) ## SecondsSinceLaunch IncrementalCount FullCount ## 2014-05-12 17:00:32.868 1.161 0 0 ## 2014-05-12 17:00:33.178 1.472 1 0 ## 2014-05-12 17:00:33.677 1.969 2 0 ## 2014-05-12 17:00:35.538 3.830 3 0 ## 2014-05-12 17:00:37.811 6.103 4 0 ## 2014-05-12 17:00:41.427 9.720 5 0 ## UserTime SysTime RealTime BeforeSize AfterSize ## 2014-05-12 17:00:32.868 0.11 0.04 0.02 8192 1400 ## 2014-05-12 17:00:33.178 0.05 0.01 0.02 5496 1672 ## 2014-05-12 17:00:33.677 0.04 0.01 0.01 5768 2557 ## 2014-05-12 17:00:35.538 0.21 0.05 0.04 22528 4907 ## 2014-05-12 17:00:37.811 0.08 0.01 0.02 24576 7072 ## 2014-05-12 17:00:41.427 0.26 0.06 0.04 43008 14336 ## TotalSize Delta ## 2014-05-12 17:00:32.868 9437184 6792 ## 2014-05-12 17:00:33.178 9437184 3824 ## 2014-05-12 17:00:33.677 9437184 3211 ## 2014-05-12 17:00:35.538 9437184 17621 ## 2014-05-12 17:00:37.811 9437184 17504 ## 2014-05-12 17:00:41.427 9437184 28672 Example of Two Benchmark Runs in One Log File The data in the following graph is from a different log file, not the one of primary interest to this article. I’m including this image because it is an example of idle periods followed by busy periods. It would be uninteresting to average the rate of garbage collection over the entire log file period. More interesting would be the rate of garbage collect in the two busy periods. Are they the same or different? Your production data may be similar, for example, bursts when employees return from lunch and idle times on weekend evenings, etc. Once the data is in an R Time Series, you can analyze isolated time windows. Clipping the Time Series data Flashing back to our test case… Viewing the data as a time series is interesting. You can see that the work intensive time period is between 9:00 PM and 3:00 AM. Lets clip the data to the interesting period:     par(mfrow=c(2,1)) plot(g1gc.z$UserTime, type="h", main="User Time per GC\nTime: Complete Log File", xlab="Time of Day", ylab="CPU Seconds per GC", col="#1b9e77") clipped.g1gc.z=window(g1gc.z, start=as.POSIXct("2014-05-12 21:00:00"), end=as.POSIXct("2014-05-13 03:00:00")) plot(clipped.g1gc.z$UserTime, type="h", main="User Time per GC\nTime: Limited to Benchmark Execution", xlab="Time of Day", ylab="CPU Seconds per GC", col="#1b9e77") box(which = "outer", lty = "solid") Cumulative Incremental and Full GC count Here is the cumulative incremental and full GC count. When the line is very steep, it indicates that the GCs are repeating very quickly. Notice that the scale on the Y axis is different for full vs. incremental. plot(clipped.g1gc.z[,c(2:3)], main="Cumulative Incremental and Full GC count", xlab="Time of Day", col="#1b9e77") GC Analysis of Benchmark Execution using Time Series data In the following series of 3 graphs: The “After Size” show the amount of heap space in use after each garbage collection. Many Java objects are still referenced, i.e. alive, during each garbage collection. This may indicate that the application has a memory leak, or may indicate that the application has a very large memory footprint. Typically, an application's memory footprint plateau's in the early stage of execution. One would expect this graph to have a flat top. The steep decline in the heap space may indicate that the application crashed after 2:00. The second graph shows that the outliers in real execution time, discussed above, occur near 2:00. when the Java heap seems to be quite full. The third graph shows that Full GCs are infrequent during the first few hours of execution. The rate of Full GC's, (the slope of the cummulative Full GC line), changes near midnight.   plot(clipped.g1gc.z[,c("AfterSize","RealTime","FullCount")], xlab="Time of Day", col=c("#1b9e77","red","#1b9e77")) GC Analysis of heap recovered Each GC trace includes the amount of heap space in use before and after the individual GC event. During garbage coolection, unreferenced objects are identified, the space holding the unreferenced objects is freed, and thus, the difference in before and after usage indicates how much space has been freed. The following box plot and bar chart both demonstrate the same point - the amount of heap space freed per garbage colloection is surprisingly low. par(mfrow=c(2,1)) boxplot(as.vector(clipped.g1gc.z$Delta), main="Amount of Heap Recovered per GC Pass", xlab="Size in KB", horizontal = TRUE, col="red") hist(as.vector(clipped.g1gc.z$Delta), main="Amount of Heap Recovered per GC Pass", xlab="Size in KB", breaks=100, col="red") box(which = "outer", lty = "solid") This graph is the most interesting. The dark blue area shows how much heap is occupied by referenced Java objects. This represents memory that holds live data. The red fringe at the top shows how much data was recovered after each garbage collection. barplot(clipped.g1gc.z[,c("AfterSize","Delta")], col=c("#7570b3","#e7298a"), xlab="Time of Day", border=NA) legend("topleft", c("Live Objects","Heap Recovered on GC"), fill=c("#7570b3","#e7298a")) box(which = "outer", lty = "solid") When I discuss the data in the log files with the customer, I will ask for an explaination for the large amount of referenced data resident in the Java heap. There are two are posibilities: There is a memory leak and the amount of space required to hold referenced objects will continue to grow, limited only by the maximum heap size. After the maximum heap size is reached, the JVM will throw an “Out of Memory” exception every time that the application tries to allocate a new object. If this is the case, the aplication needs to be debugged to identify why old objects are referenced when they are no longer needed. The application has a legitimate requirement to keep a large amount of data in memory. The customer may want to further increase the maximum heap size. Another possible solution would be to partition the application across multiple cluster nodes, where each node has responsibility for managing a unique subset of the data. Conclusion In conclusion, R is a very powerful tool for the analysis of Java garbage collection log files. The primary difficulty is data cleansing so that information can be read into an R data frame. Once the data has been read into R, a rich set of tools may be used for thorough evaluation.

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  • Easiest modern programming language [closed]

    - by Goward Gerald
    What programming language is easiest nowadays, yet demanded in market? By easiest I mean least skill cap (and by skill cap I mean knowing all the frameworks and all the language abilities and constructions. Sure It doesnt mean you need to know 100% of EVERYTHING, but what language lets me get closer to this the most? Please don't suggest me basic, delphi or some other dead/half-dead/useless technologies.

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  • How can I "bulk paste" a clipboard string of multi-line text into a readable ordered list?

    - by gunshor
    How can I "bulk paste" a clipboard string of multi-line text into a readable ordered list? I'm trying to demonstrate how to turn any string of multi-line text into an ordered list. The script (preferably JS) needs to respect: - carriage returns at the end of a line, to mean "that line ends here" - indentations at the beginning of a line, to mean "this is part of the item above it" - dashes at the beginning of a line, to mean "this is a task, and the line above it is its project"

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  • slicing a 2d numpy array

    - by MedicalMath
    The following code: import numpy as p myarr=[[0,1],[0,6],[0,1],[0,6],[0,1],[0,6],[0,1],[0,6],[0,1],[0,6],[0,1],[0,6],[0,1],[0,6],[0,1],[0,6],[0,1],[0,6]] copy=p.array(myarr) p.mean(copy)[:,1] Is generating the following error message: Traceback (most recent call last): File "<pyshell#3>", line 1, in <module> p.mean(copy)[:,1] IndexError: 0-d arrays can only use a single () or a list of newaxes (and a single ...) as an index I looked up the syntax at this link and I seem to be using the correct syntax to slice. However, when I type copy[:,1] into the Python shell, it gives me the following output, which is clearly wrong, and is probably what is throwing the error: array([1, 6, 1, 6, 1, 6, 1, 6, 1, 6, 1, 6, 1, 6, 1, 6, 1, 6]) Can anyone show me how to fix my code so that I can extract the second column and then take the mean of the second column as intended in the original code above? EDIT: Thank you for your solutions. However, my posting was an oversimplification of my real problem. I used your solutions in my real code, and got a new error. Here is my real code with one of your solutions that I tried: filteredSignalArray=p.array(filteredSignalArray) logical=p.logical_and(EndTime-10.0<=matchingTimeArray,matchingTimeArray<=EndTime) finalStageTime=matchingTimeArray.compress(logical) finalStageFiltered=filteredSignalArray.compress(logical) for j in range(len(finalStageTime)): if j == 0: outputArray=[[finalStageTime[j],finalStageFiltered[j]]] else: outputArray+=[[finalStageTime[j],finalStageFiltered[j]]] print 'outputArray[:,1].mean() is: ',outputArray[:,1].mean() And here is the error message that is now being generated by the new code: File "mypath\myscript.py", line 1545, in WriteToOutput10SecondsBeforeTimeMarker print 'outputArray[:,1].mean() is: ',outputArray[:,1].mean() TypeError: list indices must be integers, not tuple Second EDIT: This is solved now that I added: outputArray=p.array(outputArray) above my code. I have been at this too many hours and need to take a break for a while if I am making these kinds of mistakes.

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  • Programming R/Sweave for proper \Sexpr output

    - by deoksu
    Hi I'm having a bit of a problem programming R for Sweave, and the #rstats twitter group often points here, so I thought I'd put this question to the SO crowd. I'm an analyst- not a programmer- so go easy on me my first post. Here's the problem: I am drafting a survey report in Sweave with R and would like to report the marginal returns in line using \Sexpr{}. For example, rather than saying: Only 14% of respondents said 'X'. I want to write the report like this: Only \Sexpr{p.mean(variable)}$\%$ of respondents said 'X'. The problem is that Sweave() converts the results of the expression in \Sexpr{} to a character string, which means that the output from expression in R and the output that appears in my document are different. For example, above I use the function 'p.mean': p.mean<- function (x) {options(digits=1) mmm<-weighted.mean(x, weight=weight, na.rm=T) print(100*mmm) } In R, the output looks like this: p.mean(variable) >14 but when I use \Sexpr{p.mean(variable)}, I get an unrounded character string (in this case: 13.5857142857143) in my document. I have tried to limit the output of my function to 'digits=1' in the global environment, in the function itself, and and in various commands. It only seems to contain what R prints, not the character transformation that is the result of the expression and which eventually prints in the LaTeX file. as.character(p.mean(variable)) >[1] 14 >[1] "13.5857142857143" Does anyone know what I can do to limit the digits printed in the LaTeX file, either by reprogramming the R function or with a setting in Sweave or \Sexpr{}? I'd greatly appreciate any help you can give. Thanks, David

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  • How can I superimpose modified loess lines on a ggplot2 qplot?

    - by briandk
    Background Right now, I'm creating a multiple-predictor linear model and generating diagnostic plots to assess regression assumptions. (It's for a multiple regression analysis stats class that I'm loving at the moment :-) My textbook (Cohen, Cohen, West, and Aiken 2003) recommends plotting each predictor against the residuals to make sure that: The residuals don't systematically covary with the predictor The residuals are homoscedastic with respect to each predictor in the model On point (2), my textbook has this to say: Some statistical packages allow the analyst to plot lowess fit lines at the mean of the residuals (0-line), 1 standard deviation above the mean, and 1 standard deviation below the mean of the residuals....In the present case {their example}, the two lines {mean + 1sd and mean - 1sd} remain roughly parallel to the lowess {0} line, consistent with the interpretation that the variance of the residuals does not change as a function of X. (p. 131) How can I modify loess lines? I know how to generate a scatterplot with a "0-line,": # First, I'll make a simple linear model and get its diagnostic stats library(ggplot2) data(cars) mod <- fortify(lm(speed ~ dist, data = cars)) attach(mod) str(mod) # Now I want to make sure the residuals are homoscedastic qplot (x = dist, y = .resid, data = mod) + geom_smooth(se = FALSE) # "se = FALSE" Removes the standard error bands But does anyone know how I can use ggplot2 and qplot to generate plots where the 0-line, "mean + 1sd" AND "mean - 1sd" lines would be superimposed? Is that a weird/complex question to be asking?

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