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  • compare two following values in numpy array

    - by Billy Mitchell
    What is the best way to touch two following values in an numpy array? example: npdata = np.array([13,15,20,25]) for i in range( len(npdata) ): print npdata[i] - npdata[i+1] this looks really messed up and additionally needs exception code for the last iteration of the loop. any ideas? Thanks!

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  • SQL column length query

    - by Mike
    i've been using the following query: select LEN(columnname) as columnmame from dbo.amu_datastaging This works, but is there a way to only return the greatest value instead of all the values? So if i return 1million records and the longest length is 400, the query would just return the value of 400?

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  • Unknown syntax error.

    - by matt1024
    Why do I get a syntax error running this code? If I remove the highlighted section (return cards[i]) I get the error highlighting the function call instead. Please help :) def dealcards(): for i in range(len(cards)): cards[i] = '' for j in range(8): cards[i] = cards[i].append(random.randint(0,9) return cards[i] print (dealcards())

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  • I'm writing a spellchecking program, how do I replace ch in a string?

    - by Ajay Hopkins
    What am I doing wrong/what can I do? import sys import string def remove(file): punctuation = string.punctuation for ch in file: if len(ch) > 1: print('error - ch is larger than 1 --| {0} |--'.format(ch)) if ch in punctuation: ch = ' ' return ch else: return ch ref = (open("ref.txt","r")) test_file = (open("test.txt", "r")) dictionary = ref.read().split() file = test_file.read().lower() file = remove(file) print(file) This is in Python 3.1.2

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  • Using fft2 with reshaping for an RGB filter

    - by Mahmoud Aladdin
    I want to apply a filter on an image, for example, blurring filter [[1/9.0, 1/9.0, 1/9.0], [1/9.0, 1/9.0, 1/9.0], [1/9.0, 1/9.0, 1/9.0]]. Also, I'd like to use the approach that convolution in Spatial domain is equivalent to multiplication in Frequency domain. So, my algorithm will be like. Load Image. Create Filter. convert both Filter & Image to Frequency domains. multiply both. reconvert the output to Spatial Domain and that should be the required output. The following is the basic code I use, the image is loaded and displayed as cv.cvmat object. Image is a class of my creation, it has a member image which is an object of scipy.matrix and toFrequencyDomain(size = None) uses spf.fftshift(spf.fft2(self.image, size)) where spf is scipy.fftpack and dotMultiply(img) uses scipy.multiply(self.image, image) f = Image.fromMatrix([[1/9.0, 1/9.0, 1/9.0], [1/9.0, 1/9.0, 1/9.0], [1/9.0, 1/9.0, 1/9.0]]) lena = Image.fromFile("Test/images/lena.jpg") print lena.image.shape lenaf = lena.toFrequencyDomain(lena.image.shape) ff = f.toFrequencyDomain(lena.image.shape) lenafm = lenaf.dotMultiplyImage(ff) lenaff = lenafm.toTimeDomain() lena.display() lenaff.display() So, the previous code works pretty well, if I told OpenCV to load the image via GRAY_SCALE. However, if I let the image to be loaded in color ... lena.image.shape will be (512, 512, 3) .. so, it gives me an error when using scipy.fttpack.ftt2 saying "When given, Shape and Axes should be of same length". What I tried next was converted my filter to 3-D .. as [[[1/9.0, 1/9.0, 1/9.0], [1/9.0, 1/9.0, 1/9.0], [1/9.0, 1/9.0, 1/9.0]], [[1/9.0, 1/9.0, 1/9.0], [1/9.0, 1/9.0, 1/9.0], [1/9.0, 1/9.0, 1/9.0]], [[1/9.0, 1/9.0, 1/9.0], [1/9.0, 1/9.0, 1/9.0], [1/9.0, 1/9.0, 1/9.0]]] And, not knowing what the axes argument do, I added it with random numbers as (-2, -1, -1), (-1, -1, -2), .. etc. until it gave me the correct filter output shape for the dotMultiply to work. But, of course it wasn't the correct value. Things were totally worse. My final trial, was using fft2 function on each of the components 2-D matrices, and then re-making the 3-D one, using the following code. # Spiltting the 3-D matrix to three 2-D matrices. for i, row in enumerate(self.image): r.append(list()) g.append(list()) b.append(list()) for pixel in row: r[i].append(pixel[0]) g[i].append(pixel[1]) b[i].append(pixel[2]) rfft = spf.fftshift(spf.fft2(r, size)) gfft = spf.fftshift(spf.fft2(g, size)) bfft = spf.fftshift(spf.fft2(b, size)) newImage.image = sp.asarray([[[rfft[i][j], gfft[i][j], bfft[i][j]] for j in xrange(len(rfft[i]))] for i in xrange(len(rfft))] ) return newImage Any help on what I made wrong, or how can I achieve that for both GreyScale and Coloured pictures.

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  • How to check for palindrome using Python logic

    - by DrOnline
    My background is only a 6 month college class in basic C/C++, and I'm trying to convert to Python. I may be talking nonsense, but it seems to me C, at least at my level, is very for-loop intensive. I solve most problems with these loops. And it seems to me the biggest mistake people do when going from C to Python is trying to implement C logic using Python, which makes things run slowly, and it's just not making the most of the language. I see on this website: http://hyperpolyglot.org/scripting (serach for "c-style for", that Python doesn't have C-style for loops. Might be outdated, but I interpret it to mean Python has its own methods for this. I've tried looking around, I can't find much up to date (Python 3) advice for this. How can I solve a palindrome challenge in Python, without using the for loop? I've done this in C in class, but I want to do it in Python, on a personal basis. The problem is from the Euler Project, great site btw. def isPalindrome(n): lst = [int(n) for n in str(n)] l=len(lst) if l==0 || l==1: return True elif len(lst)%2==0: for k in range (l) ##### else: while (k<=((l-1)/2)): if (list[]): ##### for i in range (999, 100, -1): for j in range (999,100, -1): if isPalindrome(i*j): print(i*j) break I'm missing a lot of code here. The five hashes are just reminders for myself. Concrete questions: 1) In C, I would make a for loop comparing index 0 to index max, and then index 0+1 with max-1, until something something. How to best do this in Python? 2) My for loop (in in range (999, 100, -1), is this a bad way to do it in Python? 3) Does anybody have any good advice, or good websites or resources for people in my position? I'm not a programmer, I don't aspire to be one, I just want to learn enough so that when I write my bachelor's degree thesis (electrical engineering), I don't have to simultaneously LEARN an applicable programming language while trying to obtain good results in the project. "How to go from basic C to great application of Python", that sort of thing. 4) Any specific bits of code to make a great solution to this problem would also be appreciated, I need to learn good algorithms.. I am envisioning 3 situations. If the value is zero or single digit, if it is of odd length, and if it is of even length. I was planning to write for loops... PS: The problem is: Find the highest value product of two 3 digit integers that is also a palindrome.

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  • Find last match with python regular expression

    - by SDD
    I wanto to match the last occurence of a simple pattern in a string, e.g. list = re.findall(r"\w+ AAAA \w+", "foo bar AAAA foo2 AAAA bar2) print "last match: ", list[len(list)-1] however, if the string is very long, a huge list of matches is generated. Is there a more direct way to match the second occurence of "AAAA" or should I use this workaround?

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  • Does this introduce security vulnerabilities?

    - by mcmt
    I don't think I'm missing anything. Then again I'm kind of a newbie. def GET(self, filename): name = urllib.unquote(filename) full = path.abspath(path.join(STATIC_PATH, filename)) #Make sure request is not tricksy and tries to get out of #the directory, e.g. filename = "../.ssh/id_rsa". GET OUTTA HERE assert full[:len(STATIC_PATH)] == STATIC_PATH, "bad path" return open(full).read()

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  • Java: Is there a way to obtain the bytecode for a class at runtime?

    - by Adam Paynter
    In Java, is there a way (at runtime) to obtain the bytecode which defined a particular class? Put another way, is there a way to obtain the byte[] array passed to ClassLoader.defineClass(String name, byte[] b, int off, int len) when a particular class was loaded? I see that this method is declared final, so creating a custom ClassLoader to intercept class definitions seems out of the question. In the past, I have used the class's ClassLoader to obtain the bytecode via the getResourceAsStream(String) method, but I would prefer a more canonical solution.

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  • Why is a .net generic dictionary so big

    - by thefroatgt
    I am serializing a generic dictionary in VB.net and I am very surprised that it is about 1.3kb with a single item. Am I doing something wrong, or is there something else I should be doing? I have a large number of dictionaries and it is killing me to send them all across the wire. The code I use for serialization is Dim dictionary As New Dictionary(Of Integer, Integer) Dim stream As New MemoryStream Dim bformatter As New BinaryFormatter() dictionary.Add(1, 1) bformatter.Serialize(stream, dictionary) Dim len As Long = stream.Length

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  • AppEngine: how do cursors work?

    - by victor a.k.a. python for ever
    hello, i have the following code def get(self): date = datetime.date.today() loc_query = Location.all() last_cursor = memcache.get('location_cursor') if last_cursor: loc_query.with_cursor(last_cursor) loc_result = loc_query.fetch(1) for loc in loc_result: self.record(loc, date) taskqueue.add( url='/task/query/simplegeo', params={'date':date, 'locid':loc.key().id()} ) if len(loc_result): memcache.add('location_cursor', loc_query.cursor()) taskqueue.add(url='/task/count/', method='GET') else: memcache.add('location_cursor', None) i don't know what i'm doing wrong, but i am getting the same cursor which is not the effect i wanted. why isn't the cursor moving?

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  • MS Access ADODB.recordset character limit is 2036!? Can this be increased?

    - by souper-dragon
    In the following AccessVBA code, I am trying to write a record to a memo field called "Recipient_Display": oRec1.Fields("RECIPIENT_DISPLAY") = Left(sRecipientDisplayNames, Len(sRecipientDisplayNames) - 2) When the string contains 2036 characters, the write completes. Above this number I get the following error: Run-time error'-2147217887(80040e21)': Could not update; currently locked by another session on this machine. What is the significance of this number 2036 and is there a property I can adjust that will allow the above update to take place?

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  • itertools.product eliminating repeated reversed tuples

    - by genclik27
    I asked a question yesterday and thanks to Tim Peters, it is solved. The question is here; itertools.product eliminating repeated elements The new question is further version of this. This time I will generate tuples inside of tuples. Here is an example; lis = [[(1,2), (3,4)], [(5,2), (1,2)], [(2,1), (1,2)]] When I use it in itertools.product function this is what I get, ((1, 2), (5, 2), (2, 1)) ((1, 2), (5, 2), (1, 2)) ((1, 2), (1, 2), (2, 1)) ((1, 2), (1, 2), (1, 2)) ((3, 4), (5, 2), (2, 1)) ((3, 4), (5, 2), (1, 2)) ((3, 4), (1, 2), (2, 1)) ((3, 4), (1, 2), (1, 2)) I want to change it in a way that if a sequence has (a,b) inside of it, then it can not have (b,a). In this example if you look at this sequence ((3, 4), (1, 2), (2, 1)) it has (1,2) and (2,1) inside of it. So, this sequence ((3, 4), (1, 2), (2, 1)) should not be considered in the results. As I said, I asked similar question before, in that case it was not considering duplicate elements. I try to adapt it to my problem. Here is modified code. Changed parts in old version are taken in comments. def reverse_seq(seq): s = [] for i in range(len(seq)): s.append(seq[-i-1]) return tuple(s) def uprod(*seqs): def inner(i): if i == n: yield tuple(result) return for elt in sets[i] - reverse: #seen.add(elt) rvrs = reverse_seq(elt) reverse.add(rvrs) result[i] = elt for t in inner(i+1): yield t #seen.remove(elt) reverse.remove(rvrs) sets = [set(seq) for seq in seqs] n = len(sets) #seen = set() reverse = set() result = [None] * n for t in inner(0): yield t In my opinion this code should work but I am getting error for the input lis = [[(1,2), (3,4)], [(5,2), (1,2)], [(2,1), (1,2)]]. I could not understand where I am wrong. for i in uprod(*lis): print i Output is, ((1, 2), (1, 2), (1, 2)) Traceback (most recent call last): File "D:\Users\SUUSER\workspace tree\sequence_covering _array\denemeler_buraya.py", line 39, in <module> for i in uprod(*lis): File "D:\Users\SUUSER\workspace tree\sequence_covering _array\denemeler_buraya.py", line 32, in uprod for t in inner(0): File "D:\Users\SUUSER\workspace tree\sequence_covering _array\denemeler_buraya.py", line 22, in inner for t in inner(i+1): File "D:\Users\SUUSER\workspace tree\sequence_covering _array\denemeler_buraya.py", line 25, in inner reverse.remove(rvrs) KeyError: (2, 1) Thanks,

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  • im writing a spellchecking program, how do i replace ch in a string..eg..

    - by Ajay Hopkins
    what am i doing wrong/what can i do?? import sys import string def remove(file): punctuation = string.punctuation for ch in file: if len(ch) > 1: print('error - ch is larger than 1 --| {0} |--'.format(ch)) if ch in punctuation: ch = ' ' return ch else: return ch ref = (open("ref.txt","r")) test_file = (open("test.txt", "r")) dictionary = ref.read().split() file = test_file.read().lower() file = remove(file) print(file) p.s, this is in Python 3.1.2

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  • Use Twisted's getPage as urlopen?

    - by RadiantHex
    Hi folks, I would like to use Twisted non-blocking getPage method within a webapp, but it feels quite complicated to use such function compared to urlopen. This is an example of what I'm trying to achive: def web_request(request): response = urllib.urlopen('http://www.example.org') return HttpResponse(len(response.read())) Is it so hard to have something similar with getPage?

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  • Most concise way to convert from date format: yyyy[3 digit day of year] to SQL datetime

    - by Seth Reno
    I'm working with an existing database where all dates are stored as integers in the following format: yyyy[3 digit day of year]. For example: 2010-01-01 == 2010001 2010-12-31 == 2010356 I'm using the following SQL to convert to a datetime: DATEADD(d, CAST(SUBSTRING( CAST(NEW_BIZ_OBS_DATE AS VARCHAR), 5, LEN(NEW_BIZ_OBS_DATE) - 4 ) AS INT) - 1, CAST('1/1/' + SUBSTRING(CAST(NEW_BIZ_OBS_DATE AS VARCHAR),1,4) AS DATETIME)) Does anyone have a more concise way to do this?

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  • A python code to convert a number from any base to the base of 10 giving errors . What is wrong with this code?

    - by mekasperasky
    import math def baseencode(number, base): ##Converting a number of any base to base10 if number == 0: return '0' for i in range(0,len(number)): if number[i]!= [A-Z]: num = num + number[i]*pow(i,base) else : num = num + (9 + ord(number[i])) *pow(i,base) return num a = baseencode('20',5) print a Errors I get are Traceback (most recent call last): File "doubtrob.py", line 19, in <module> a = baseencode('20',5) File "doubtrob.py", line 13, in baseencode if number[i]!= [A-Z]: NameError: global name 'A' is not defined

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  • Why won't haproxy capture my cookie?

    - by mike
    I'm having trouble getting frontend cookie capture to work in haproxy. I have this in my config: frontend frontend 0.0.0.0:9999 [snip] capture cookie foo len 10 Then I use nc to talk directly to the server and send it: GET / HTTP/1.1 Cookie: foo=bar I get a log line, but there's a "-" where the captured cookie should be.

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  • os.walk in python not running with cmd line parameter passed as path

    - by kartiku
    Hello, I needed to find the number of files in a folder on the system. This is what i used: file_count = sum((len(f) for _, _, f in os.walk('path'))) This works fine when we specify the path as a string in quotes, but when I enter a variable name that holds the path, type(file_count) is a generator object, and hence cannot be used as an integer. How to solve this and why does this happen?

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  • Python recursion with list returns None

    - by newman
    def foo(a): a.append(1) if len(a) > 10: print a return a else: foo(a) Why this recursive function returns None (see transcript below)? I can't quite understand what I am doing wrong. In [263]: x = [] In [264]: y = foo(x) [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1] In [265]: print y None

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  • How to get the width of a string in pixels?

    - by MA1
    I am using wxPython's HyperTreeList and I want to set the column width exactly equal to length of the largest string in it. To accomplish that, I'd like to to convert a python string size into pixels. For Example: If we have a string like str = "python" len(str) = 6 How could I convert the above string length/size into pixels? Is there another way?

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  • Detect if 2 HTML fragments have identical hierarchical structure

    - by sergzach
    An example of fragments that have identical hierarchical structure: (1) <div> <span>It's a message</span> </div> (2) <div> <span class='bold'>This is a new text</span> </div> An example of fragments that have different structure: (1) <div> <span><b>It's a message</b></span> </div> (2) <div> <span>This is a new text</span> </div> So, fragments with a similar structure correspond to one hierarchical tree (the same tag names, the same hierarchical structure). How can I detect if 2 elements (html fragments) have the same structure simply with lxml? I have a function that does not work properly for some more difficult case (than the example): def _is_equal( el1, el2 ): # input: 2 elements with possible equal structure and tag names # e.g. root = lxml.html.fromstring( buf ) # el1 = root[ 0 ] # el2 = root[ 1 ] # move from top to bottom, compare elements result = False if el1.tag == el2.tag: # has no children if len( el1 ) == len( el2 ): if len( el1 ) == 0: return True else: # iterate one of them, for example el1 i = 0 for child1 in el1: child2 = el2[ i ] is_equal2 = _is_equal( child1, child2 ) if not is_equal2: return False return True else: return False else: return False The code fails to detect that 2 divs with class='tovar2' have an identical structure: <body> <div class="tovar2"> <h2 class="new"> <a href="http://modnyedeti-krsk.ru/magazin/product/333193003"> ?????? ?/? </a> </h2> <ul class="art"> <li> ???????: <span>1759</span> </li> </ul> <div> <div class="wrap" style="width:180px;"> <div class="new"> <img src="shop_files/new-t.png" alt=""> </div> <a class="highslide" href="http://modnyedeti-krsk.ru/d/459730/d/820.jpg" onclick="return hs.expand(this)"> <img src="shop_files/fr_5.gif" style="background:url(/d/459730/d/548470803_5.jpg) 50% 50% no-repeat scroll;" alt="?????? ?/?" height="160" width="180"> </a> </div> </div> <form action="" onsubmit="return addProductForm(17094601,333193003,3150.00,this,false);"> <ul class="bott "> <li class="price">????:<br> <span> <b> 3 150 </b> ???. </span> </li> <li class="amount">???-??:<br><input class="number" onclick="this.select()" value="1" name="product_amount" type="text"> </li> <li class="buy"><input value="" type="submit"> </li> </ul> </form> </div> <div class="tovar2"> <h2 class="new"> <a href="http://modnyedeti-krsk.ru/magazin/product/333124803">?????? ?/?</a> </h2> <ul class="art"> <li> ???????: <span>1759</span> </li> </ul> <div> <div class="wrap" style="width:180px;"> <div class="new"> <img src="shop_files/new-t.png" alt=""> </div> <a class="highslide" href="http://modnyedeti-krsk.ru/d/459730/d/820.jpg" onclick="return hs.expand(this)"> <img src="shop_files/fr_5.gif" style="background:url(/d/459730/d/548470803_5.jpg) 50% 50% no-repeat scroll;" alt="?????? ?/?" height="160" width="180"> </a> </div> </div> <form action="" onsubmit="return addProductForm(17094601,333124803,3150.00,this,false);"> <ul class="bott "> <li class="price">????:<br> <span> <b>3 150</b> ???. </span> </li> <li class="amount">???-??:<br><input class="number" onclick="this.select()" value="1" name="product_amount" type="text"> </li> <li class="buy"> <input value="" type="submit"> </li> </ul> </form> </div> </body>

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