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• problems in trying ieee 802.15.4 working from msk

  - by asel

  Hi, i took a msk code from dsplog.com and tried to modify it to test the ieee 802.15.4. There are several links on that site for ieee 802.15.4. Currently I am getting simulated ber results all approximately same for all the cases of Eb_No values. Can you help me to find why? thanks in advance! clear PN = [ 1 1 0 1 1 0 0 1 1 1 0 0 0 0 1 1 0 1 0 1 0 0 1 0 0 0 1 0 1 1 1 0; 1 1 1 0 1 1 0 1 1 0 0 1 1 1 0 0 0 0 1 1 0 1 0 1 0 0 1 0 0 0 1 0; 0 0 1 0 1 1 1 0 1 1 0 1 1 0 0 1 1 1 0 0 0 0 1 1 0 1 0 1 0 0 1 0; 0 0 1 0 0 0 1 0 1 1 1 0 1 1 0 1 1 0 0 1 1 1 0 0 0 0 1 1 0 1 0 1; 0 1 0 1 0 0 1 0 0 0 1 0 1 1 1 0 1 1 0 1 1 0 0 1 1 1 0 0 0 0 1 1; 0 0 1 1 0 1 0 1 0 0 1 0 0 0 1 0 1 1 1 0 1 1 0 1 1 0 0 1 1 1 0 0; 1 1 0 0 0 0 1 1 0 1 0 1 0 0 1 0 0 0 1 0 1 1 1 0 1 1 0 1 1 0 0 1; 1 0 0 1 1 1 0 0 0 0 1 1 0 1 0 1 0 0 1 0 0 0 1 0 1 1 1 0 1 1 0 1; 1 0 0 0 1 1 0 0 1 0 0 1 0 1 1 0 0 0 0 0 0 1 1 1 0 1 1 1 1 0 1 1; 1 0 1 1 1 0 0 0 1 1 0 0 1 0 0 1 0 1 1 0 0 0 0 0 0 1 1 1 0 1 1 1; 0 1 1 1 1 0 1 1 1 0 0 0 1 1 0 0 1 0 0 1 0 1 1 0 0 0 0 0 0 1 1 1; 0 1 1 1 0 1 1 1 1 0 1 1 1 0 0 0 1 1 0 0 1 0 0 1 0 1 1 0 0 0 0 0; 0 0 0 0 0 1 1 1 0 1 1 1 1 0 1 1 1 0 0 0 1 1 0 0 1 0 0 1 0 1 1 0; 0 1 1 0 0 0 0 0 0 1 1 1 0 1 1 1 1 0 1 1 1 0 0 0 1 1 0 0 1 0 0 1; 1 0 0 1 0 1 1 0 0 0 0 0 0 1 1 1 0 1 1 1 1 0 1 1 1 0 0 0 1 1 0 0; 1 1 0 0 1 0 0 1 0 1 1 0 0 0 0 0 0 1 1 1 0 1 1 1 1 0 1 1 1 0 0 0; ]; N = 5*10^5; % number of bits or symbols fsHz = 1; % sampling period T = 4; % symbol duration Eb_N0_dB = [0:10]; % multiple Eb/N0 values ct = cos(pi*[-T:N*T-1]/(2*T)); st = sin(pi*[-T:N*T-1]/(2*T)); for ii = 1:length(Eb_N0_dB) tx = []; % MSK Transmitter ipBit = round(rand(1,N/32)*15); for k=1:length(ipBit) sym = ipBit(k); tx = [tx PN((sym+1),1:end)]; end ipMod = 2*tx - 1; % BPSK modulation 0 -> -1, 1 -> 1 ai = kron(ipMod(1:2:end),ones(1,2*T)); % even bits aq = kron(ipMod(2:2:end),ones(1,2*T)); % odd bits ai = [ai zeros(1,T) ]; % padding with zero to make the matrix dimension match aq = [zeros(1,T) aq ]; % adding delay of T for Q-arm % MSK transmit waveform xt = 1/sqrt(T)*[ai.*ct + j*aq.*st]; % Additive White Gaussian Noise nt = 1/sqrt(2)*[randn(1,N*T+T) + j*randn(1,N*T+T)]; % white gaussian noise, 0dB variance % Noise addition yt = xt + 10^(-Eb_N0_dB(ii)/20)*nt; % additive white gaussian noise % MSK receiver % multiplying with cosine and sine waveforms xE = conv(real(yt).*ct,ones(1,2*T)); xO = conv(imag(yt).*st,ones(1,2*T)); bHat = zeros(1,N); bHat(1:2:end) = xE(2*T+1:2*T:end-2*T); % even bits bHat(2:2:end) = xO(3*T+1:2*T:end-T); % odd bits result=zeros(16,1); chiplen=32; seqstart=1; recovered = []; while(seqstart<length(bHat)) A = bHat(seqstart:seqstart+(chiplen-1)); for j=1:16 B = PN(j,1:end); result(j)=sum(A.*B); end [value,index] = max(result); recovered = [recovered (index-1)]; seqstart = seqstart+chiplen; end; %# create binary string - the 4 forces at least 4 bits bstr1 = dec2bin(ipBit,4); bstr2 = dec2bin(recovered,4); %# convert back to numbers (reshape so that zeros are preserved) out1 = str2num(reshape(bstr1',[],1))'; out2 = str2num(reshape(bstr2',[],1))'; % counting the errors nErr(ii) = size(find([out1 - out2]),2); end nErr/(length(ipBit)*4) % simulated ber theoryBer = 0.5*erfc(sqrt(10.^(Eb_N0_dB/10))) % theoretical ber

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• Drawing random smooth lines contained in a square [migrated]

  - by Doug Mercer

  I'm trying to write a matlab function that creates random, smooth trajectories in a square of finite side length. Here is my current attempt at such a procedure: function [] = drawroutes( SideLength, v, t) %DRAWROUTES Summary of this function goes here % Detailed explanation goes here %Some parameters intended to help help keep the particles in the box RandAccel=.01; ConservAccel=0; speedlimit=.1; G=10^(-8); % %Initialize Matrices Ax=zeros(v,10*t); Ay=Ax; vx=Ax; vy=Ax; x=Ax; y=Ax; sx=zeros(v,1); sy=zeros(v,1); % %Define initial position in square x(:,1)=SideLength*.15*ones(v,1)+(SideLength*.7)*rand(v,1); y(:,1)=SideLength*.15*ones(v,1)+(SideLength*.7)*rand(v,1); % for i=2:10*t %Measure minimum particle distance component wise from boundary %for each vehicle BorderGravX=[abs(SideLength*ones(v,1)-x(:,i-1)),abs(x(:,i-1))]'; BorderGravY=[abs(SideLength*ones(v,1)-y(:,i-1)),abs(y(:,i-1))]'; rx=min(BorderGravX)'; ry=min(BorderGravY)'; % %Set the sign of the repulsive force for k=1:v if x(k,i)<.5*SideLength sx(k)=1; else sx(k)=-1; end if y(k,i)<.5*SideLength sy(k)=1; else sy(k)=-1; end end % %Calculate Acceleration w/ random "nudge" and repulive force Ax(:,i)=ConservAccel*Ax(:,i-1)+RandAccel*(rand(v,1)-.5*ones(v,1))+sx*G./rx.^2; Ay(:,i)=ConservAccel*Ay(:,i-1)+RandAccel*(rand(v,1)-.5*ones(v,1))+sy*G./ry.^2; % %Ad hoc method of trying to slow down particles from jumping outside of %feasible region for h=1:v if abs(vx(h,i-1)+Ax(h,i))<speedlimit vx(h,i)=vx(h,i-1)+Ax(h,i); elseif (vx(h,i-1)+Ax(h,i))<-speedlimit vx(h,i)=-speedlimit; else vx(h,i)=speedlimit; end end for h=1:v if abs(vy(h,i-1)+Ay(h,i))<speedlimit vy(h,i)=vy(h,i-1)+Ay(h,i); elseif (vy(h,i-1)+Ay(h,i))<-speedlimit vy(h,i)=-speedlimit; else vy(h,i)=speedlimit; end end % %Update position x(:,i)=x(:,i-1)+(vx(:,i-1)+vx(:,i))/2; y(:,i)=y(:,i-1)+(vy(:,i-1)+vy(:,1))/2; % end %Plot position clf; hold on; axis([-100,SideLength+100,-100,SideLength+100]); cc=hsv(v); for j=1:v plot(x(j,1),y(j,1),'ko') plot(x(j,:),y(j,:),'color',cc(j,:)) end hold off; % end My original plan was to place particles within a square, and move them around by allowing their acceleration in the x and y direction to be governed by a uniformly distributed random variable. To keep the particles within the square, I tried to create a repulsive force that would push the particles away from the boundaries of the square. In practice, the particles tend to leave the desired "feasible" region after a relatively small number of time steps (say, 1000)." I'd love to hear your suggestions on either modifying my existing code or considering the problem from another perspective. When reading the code, please don't feel the need to get hung up on any of the ad hoc parameters at the very beginning of the script. They seem to help, but I don't believe any beside the "G" constant should truly be necessary to make this system work. Here is an example of the current output: Many of the vehicles have found their way outside of the desired square region, [0,400] X [0,400].

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• ASP.NET MVC : strange POST behavior

  - by user93422

  ASP.NET MVC 2 app I have two actions on my controller (Toons): [GET] List [POST] Add App is running on IIS7 integration mode, so /Toons/List works fine. But when I do POST (that redirects to /Toons/List internally) it redirects (with 302 Object Moved) back to /Toons/Add. The problem goes away if I use .aspx hack (that works in IIS6/IIS7 classic mode). But without .aspx - GET work fine, but POST redirects me onto itself but with GET. What am I missing? I'm hosting with webhost4life.com and they did change IIS7 to integrated mode already. EDIT: The code works as expected using UltiDev Cassini server. EDIT: It turned out to be trailing-slash-in-URL issue. Somehow IIS7 doesn't route request properly if there is no slash at the end. EDET: Explanation of the behavior What happens is when I request (POST) /Toons/List (without trailing slash), IIS doesn't find the handler (I do not have knowledge to understand how exactly IIS does URL-to-handler mapping) and redirects the request (using 302 code) to /Toons/List/ (notice trailing slash). A browser, according to the HTTP specification, must redirect the request using same method (POST in this case), but instead it handles 302 as if it is 303 and issues GET request for the new URL. This is incorrect, but known behavior of most browsers. The solution is either to use .aspx-hack to make it unambiguous for IIS how to map requests to ASP.NET handler, or configure IIS to handle everything in the virtual directory using ASP.NET handler. Q: what is a better way to handle this?

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• function taking in an input image and different kernel size

  - by drifterOcean19

  I have this filtering function that takes an input image, performs convolution using a given kernel, and returns the resulting image. However, I can't seem to work it out how to make it takes different kernel sizes.For example instead of pre-defined 3x3 kernel as below in the code, it could instead take 5x5 or 7x7. and then the user could input the type of kernel/filter they want(Depending on the intended effect). I can't seem to put my head around it. i'm quite new to matlab. function [newImg] = kernelFunc(imgB) img=imread(imgB); figure,imshow(img); img2=zeros(size(img)+2); newImg=zeros(size(img)); for rgb=1:3 for x=1:size(img,1) for y=1:size(img,2) img2(x+1,y+1,rgb)=img(x,y,rgb); end end end for rgb=1:3 for i= 1:size(img2,1)-2 for j=1:size(img2,2)-2 window=zeros(9,1); inc=1; for x=1:3 for y=1:3 window(inc)=img2(i+x-1,j+y-1,rgb); inc=inc+1; end end kernel=[1;2;1;2;4;2;1;2;1]/16; med=window.*kernel; disp(med); med=sum(med); med=floor(med); newImg(i,j,rgb)=med; end end end newImg=uint8(newImg); figure,imshow(newImg); end

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• Please help with my twodimensional array source code

  - by Baiba

  Here is what i have done but i have some questions: class masivins { public static void main (String args[]) { int mas[][] = {{0, 2, 7, 0, 8, 5, 3}, {0, 4, 0, 6, 0, 0, 0}, {0, 0, 0, 0, 3, 0, 0}, {7, 0, 0, 9, 1, 0, 7}, {5, 0, 4, 0, 0, 2, 0}}; int nulmas[] = new int [7]; int nul=0; for(int j=0; j<7; j++) { nul=0; for(int i=0; i<5; i++) { if(mas[i][j]==0) { nul++; } } nulmas[j]=nul; } for(int i=0; i<5; i++) { for(int j=0; j<7; j++) { System.out.println(mas[i][j]); } System.out.println(); } System.out.println(); for(int i=0; i<5; i++) { System.out.println("Zeros in each array column: " + nulmas[i]); } System.out.println(); } } so my questions are: 1) why after running project there are only 5 "Zeros in each array column....." shown? 2) what and where i need to change in this code to get out the number of column in which zeros are least?

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• python matrices - list index out of range

  - by user1888493

  I am writing a function, that takes a matrix as input, such as the one below. Then the it returns the matrix' inverse, where all the 1s are changed to 0s and all the 0s changed to 1s, while keeping the diagonal from top left to bottom right 0s. An example input: g1 = [[0, 1, 1, 0], [1, 0, 0, 1], [1, 0, 0, 1], [0, 1, 1, 0]] the function should output this: g1 = [[0, 0, 0, 1], [0, 0, 1, 0], [0, 1, 0, 0], [1, 0, 0, 0]] When I run the program, it raises a list index out of range error. I'm sure this happens, because the loops I have set up are trying to access values that do not exist. But how do I allow an input of unknown row and column size? I only know how to do this with a single list, but a list of lists? Following you see the transforming function, but not the test function that calls it: def inverse_graph(graph): # take in graph # change all zeros to ones and ones to zeros r, c = 0, 0 # row, column equal zero while (graph[r][c] == 0 or graph[r][c] == 1): # while the current row has a value. while (graph[r][c] == 0 or graph[r][c] == 1): # while the current column has a value if (graph[r][c] == 0): graph[r][c] = 1 elif (graph[r][c] == 1): graph[r][c] = 0 c+=1 c=0 r+=1 c=0 r=0 # sets diagonal to zeros while (g1[r][c] == 0 or g1[r][c] == 1): g1[r][c]=0 c+=1 r+=1 return graph

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• Show a number with specified number of significant digits

  - by dreeves

  I use the following function to convert a number to a string for display purposes (don't use scientific notation, don't use a trailing dot, round as specified): (* Show Number. Convert to string w/ no trailing dot. Round to the nearest r. *) Unprotect[Round]; Round[x_,0] := x; Protect[Round]; shn[x_, r_:0] := StringReplace[ ToString@NumberForm[Round[N@x,r], ExponentFunction->(Null&)], re@"\\.$"->""] (Note that re is an alias for RegularExpression.) That's been serving me well for years. But sometimes I don't want to specify the number of digits to round to, rather I want to specify a number of significant figures. For example, 123.456 should display as 123.5 but 0.00123456 should display as 0.001235. To get really fancy, I might want to specify significant digits both before and after the decimal point. For example, I might want .789 to display as 0.8 but 789.0 to display as 789 rather than 800. Do you have a handy utility function for this sort of thing, or suggestions for generalizing my function above? Related: Suppressing a trailing "." in numerical output from Mathematica

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• Have a trouble with the function roots

  - by user3707462

  Hey guys I have multiple problems with using function 'roots'. I Have to find zeros of 's^1000 + 1'. I made Y = zeros(1,1000) then manually changed the 1000th matrice to '1'. but then 'root' function does not work with it ! Another problem is that I am having trouble with matrix multiplication. The question is finding zeros(roots) of (s^6 + 6*s^5 + 15*s^4 + 20*s^3 + 15*s^2 + 6*s +1)*(s^6 + 6s^5 + 15*s^4 +15*s^2 +6*s +1) so i did: a = [1 6 15 20 15 6 1] b = [1 6 15 0 15 6 1] y = a.*b; roots(y) but this gives me -27.9355 + 0.0000i -8.2158 + 0.0000i 0.1544 + 0.9880i 0.1544 - 0.9880i -0.1217 + 0.0000i -0.0358 + 0.0000i where I calculate the original equation with wolfram then I have made matrix as : p = [1 12 66 200 375 492 524 492 375 200 66 12 1] roots(p) and this gives me : -3.1629 + 2.5046i -3.1629 - 2.5046i 0.3572 + 0.9340i 0.3572 - 0.9340i -1.0051 + 0.0000i -1.0025 + 0.0044i -1.0025 - 0.0044i -0.9975 + 0.0044i -0.9975 - 0.0044i -0.9949 + 0.0000i -0.1943 + 0.1539i -0.1943 - 0.1539i and I think the second solution is right (that is what wolfram alpha gave me) How would you answer these two questions through matlab guys?

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• Parallelizing for loop

  - by vman049

  I have MATLAB code which I'm trying to parallelize with a simple change from "for" to "parfor." I'm unable to do so because of an error I'm receiving on the variable "votes" which states: Valid indices for 'votes' are restricted in PARFOR loops. Explanation: For MATLAB to execute parfor loops efficiently, the amount of data sent to the MATLAB workers must be minimal. One of the ways MATLAB achieves this is by restricting the way variables can be indexed in parfor iterations. The indicated variable is indexed in a way that is incompatible with parfor. Suggested Action: Fix the indexing. For a description of the indexing restrictions, see “Sliced Variables” in the Parallel Computing Toolbox documentation. Below is my code: votes = zeros(num_layers, size(spikes, 1), size(SVMs_layer1, 1)); predDir = zeros(size(spikes, 1), 1); chronProb = zeros([num_layers, size(chronDists)]); for i = 1:num_layers switch i case 1 B = B1; k_elem_temp = k_elem1; rest_elem_temp = rest_elem1; case 2 B = B2; k_elem_temp = k_elem2; rest_elem_temp = rest_elem2; case 3 B = B3; k_elem_temp = k_elem3; rest_elem_temp = rest_elem3; end for j = 1:length(chronPred) if chronDists(i, j, :) ~= 0 parfor k = 1:8 chronProb(i, j, k) = logistic(B{k}(1) + chronDists(i, j, k).*(B{k}(2))); votes(i, j, k_elem_temp(k, :)) = votes(i, j, k_elem_temp(k, :)) + chronProb(i, j, k)/num_k(i)/num_layers; votes(i, j, rest_elem_temp(k, :)) = votes(i, j, rest_elem_temp(k, :)) + (1 - chronProb(i, j, k))/num_rest(i)/num_layers; end end end end Do you have any suggestions as to how I could adjust my code so that it runs in parallel? Thank you!

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• Multi Svn Repositories in Apache

  - by fampinheiro

  I have a set up with apache and subversion In the apache configuration i have <Location /svn> DAV svn SVNParentPath c:/svn </Location> Now i have multiple repositories a a_b a_c a_b_c a_b_d b and i want to map them as a/svn a/b/svn a/c/svn a/b/c/svn a/b/d/svn b/svn to do this without adding directives and restarting apache i tought of making this rules RewriteEngine On RewriteCond $1 !=svn RewriteCond $2 !=svn RewriteRule ^/([^/]+)/(.*?)/svn/(.*)$ /$1_$2/svn/$3 [N] RewriteRule ^/([^/]+)/svn/(.*)$ /svn/$1/$2 [L,PT] this way i rewrite them to /svn/a /svn/a_b /svn/a_c /svn/a_b_c /svn/a_b_d /svn/b The objective is that the client don't have the notion of this happening when a acess is made to a folder without trailing slash the mod dav return a redirect to the folder with the trailing slash exposing my internal url. can i rewrite the outgoing url ?!

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• How do I convert a Linux disk image into a sparse file?

  - by endolith

  I have a bunch of disk images, made with ddrescue, on an EXT partition, and I want to reduce their size without losing data, while still being mountable. How can I fill the empty space in the image's filesystem with zeros, and then convert the file into a sparse file so this empty space is not actually stored on disk? For example: > du -s --si --apparent-size Jimage.image 120G Jimage.image > du -s --si Jimage.image 121G Jimage.image This actually only has 50G of real data on it, though, so the second measurement should be much smaller. This supposedly will fill empty space with zeros: cat /dev/zero > zero.file rm zero.file But if sparse files are handled transparently, it might actually create a sparse file without writing anything to the virtual disk, ironically preventing me from turning the virtual disk image into a sparse file itself. :) Does it? Note: For some reason, sudo dd if=/dev/zero of=./zero.file works when cat does not on a mounted disk image.

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• mod_rewite Rule: root/? root/app/views/home/home.php

  - by Jonathon David Oates

  I am shocking at mod_rewite, here's the scenario: I need a rule that rewrites mydomain.com to mydomain.com/app/views/home/home.php. The rule, or set of rules rather, must also rewite mydomain.com/signin to mydomain.com/app/views/signin/signin.php, and work in a similar fashion for any subdirectory, for example: mydomain.com/subdir must redirect to mydomain.com/app/views/subdir/subdir.php. The rules must also work with or without the trailing slash, for example: ….com or ….com/. Thank you all, your help is much appreciated! If you could outline how and why your solution works or direct me to a good resource that explains it, I'd be exceptionally grateful! Edit: I have got a simple .htaccess file with this: Options +FollowSymLinks RewriteEngine On RewriteRule ^$ http://mydomain.local/~Jay/some_awesome_app/app/views/home/home.php This does the redirect but changes the URL in the address bar too! I've not got a trailing [R] flag so why would this be?

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• Zipcodes in CSV Generation

  - by BRADINO

  When exporting to CSV format, then opening in a spreadsheet program like Excel zipcodes that start with a zero or zeroes have the preceding zeros stripped off. Obviously it is because the spreadsheet sees that column as integers and preceding zeros in integers are useless. A quick and dirty trick to force Excel (hopefully you are using OpenOffice) to display the full zipcode, we wrap it in double quotes and put an equal sign in front of it, to force it to be a string like this: $zipcode = 00123; $data = '="' . $zipcode . '"' ; So if you are doing the straight query to CSV export, using the fputcsv function it would look something like this. Basically just overwrite the value in the row and then continue along. while ($row = mysql_fetch_assoc($query)){         $row['zipcode'] = '="'.$row['zipcode'].'"';     fputcsv($output, $row); } php csv zipcode csv number csv force string

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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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• Preffered lambda syntax?

  - by Roger Alsing

  I'm playing around a bit with my own C like DSL grammar and would like some oppinions. I've reserved the use of "(...)" for invocations. eg: foo(1,2); My grammar supports "trailing closures" , pretty much like Ruby's blocks that can be passed as the last argument of an invocation. Currently my grammar support trailing closures like this: foo(1,2) { //parameterless closure passed as the last argument to foo } or foo(1,2) [x] { //closure with one argument (x) passed as the last argument to foo print (x); } The reason why I use [args] instead of (args) is that (args) is ambigious: foo(1,2) (x) { } There is no way in this case to tell if foo expects 3 arguments (int,int,closure(x)) or if foo expects 2 arguments and returns a closure with one argument(int,int) - closure(x) So thats pretty much the reason why I use [] as for now. I could change this to something like: foo(1,2) : (x) { } or foo(1,2) (x) -> { } So the actual question is, what do you think looks best? [...] is somewhat wrist unfriendly. let x = [a,b] { } Ideas?

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• Preferred lambda syntax?

  - by Roger Alsing

  I'm playing around a bit with my own C like DSL grammar and would like some oppinions. I've reserved the use of "(...)" for invocations. eg: foo(1,2); My grammar supports "trailing closures" , pretty much like Ruby's blocks that can be passed as the last argument of an invocation. Currently my grammar support trailing closures like this: foo(1,2) { //parameterless closure passed as the last argument to foo } or foo(1,2) [x] { //closure with one argument (x) passed as the last argument to foo print (x); } The reason why I use [args] instead of (args) is that (args) is ambigious: foo(1,2) (x) { } There is no way in this case to tell if foo expects 3 arguments (int,int,closure(x)) or if foo expects 2 arguments and returns a closure with one argument(int,int) - closure(x) So thats pretty much the reason why I use [] as for now. I could change this to something like: foo(1,2) : (x) { } or foo(1,2) (x) -> { } So the actual question is, what do you think looks best? [...] is somewhat wrist unfriendly. let x = [a,b] { } Ideas?

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• async handler deleted by the wrong thread in django

  - by user3480706

  I'm run this algorithm in my django application.when i run several time from my GUI django local server will stopped and i got this error Exception RuntimeError: RuntimeError('main thread is not in main loop',) in ignored Tcl_AsyncDelete: async handler deleted by the wrong thread Aborted (core dumped) code print "Learning the sin function" network =MLP.MLP(2,10,1) samples = np.zeros(2000, dtype=[('x', float, 1), ('y', float, 1)]) samples['x'] = np.linspace(-5,5,2000) samples['y'] = np.sin(samples['x']) #samples['y'] = np.linspace(-4,4,2500) for i in range(100000): n = np.random.randint(samples.size) network.propagate_forward(samples['x'][n]) network.propagate_backward(samples['y'][n]) plt.figure(figsize=(10,5)) # Draw real function x = samples['x'] y = samples['y'] #x=np.linspace(-6.0,7.0,50) plt.plot(x,y,color='b',lw=1) samples1 = np.zeros(2000, dtype=[('x1', float, 1), ('y1', float, 1)]) samples1['x1'] = np.linspace(-4,4,2000) samples1['y1'] = np.sin(samples1['x1']) # Draw network approximated function for i in range(samples1.size): samples1['y1'][i] = network.propagate_forward(samples1['x1'][i]) plt.plot(samples1['x1'],samples1['y1'],color='r',lw=3) plt.axis([-2,2,-2,2]) plt.show() plt.close() return HttpResponseRedirect('/charts/charts') how can i fix this error ?need a quick help

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• Mat matrix multiplication, openCV?

  - by facebook-1593205594

  I initialized two Mat images as: Mat ft=Mat::zeros(src.rows,src.cols,CV_32FC1),h=Mat::zeros(src.rows,src.cols,CV_32FC1); and then i have some calculations: ft has fourier transform stored for an image, and h has matrix for Laplacian filtering in fourier domain.......they both have same dimensions, and then i did multiplication of them using both h*ft and gemm(h,ft,1,NULL,0,temp); function call but while executing it shows some problems..... it reads like this: opencv error assertion failed (some long code and at last says something about gemm in ....matmul.cpp)......termination called after throwing exception of 'cv::exception'

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• Runtime error of TASM language help!

  - by dominoos

  .model small .stack 400h .data message db "hello. ", 0ah, 0dh, "$" firstdigit db ? seconddigit db ? thirddigit db ? number dw ? newnumber db ? anumber dw 0d bnumber dw 0d Firstn db 0ah, 0dh, "Enter first 3 digit number: ","$" secondn db 0ah, 0dh, "Enter second 3 digit number: ","$" messageB db 0ah, 0dh, "HCF of two number is: ","$" linebreaker db 0ah, 0dh, ' ', 0ah, 0dh, '$' .code Start: mov ax, @data ; establish access to the data segment mov ds, ax ; mov number, 0d mov dx, offset message ; print the string "yob choi 0648293" mov ah, 9h int 21h num: mov dx, offset Firstn ; print the string "put 1st 3 digit" mov ah, 9h int 21h ;run JMP FirstFirst ; jump to FirstFirst FirstFirst: ;first digit mov ah, 1d ;bios code for read a keystroke int 21h ;call bios, it is understood that the ascii code will be returned in al mov firstdigit, al ;may as well save a copy sub al, 30h ;Convert code to an actual integer cbw ;CONVERT BYTE TO WORD. This takes whatever number is in al and ;extends it to ax, doubling its size from 8 bits to 16 bits ;The first digit now occupies all of ax as an integer mov cx, 100d ;This is so we can calculate 100*1st digit +10*2nd digit + 3rd digit mul cx ;start to accumulate the 3 digit number in the variable imul cx ;it is understood that the other operand is ax ; the result will use both dx::ax ;dx will contain only leading zeros add anumber, ax ;save ;Second Digit mov ah, 1d ;bios code for read a keystroke int 21h ;call bios, it is understood that the ascii code will be returned in al mov seconddigit, al ;may as well save a copy sub al, 30h ;Convert code to an actual integer cbw ;CONVERT BYTE TO WORD. This takes whatever number is in al and ;extends it to ax, boubling its size from 8 bits to 16 bits ;The first digit now occupies all of ax as an integer mov cx, 10d ;continue to accumulate the 3 digit number in the variable mul cx ;it is understood that the other operand is ax, containing first digit ;the result will use both dx::ax ;dx will contain only leading zeros. add anumber, ax ;save ;third Digit mov ah, 1d ;samething as above int 21h ; mov thirddigit, al ; sub al, 30h ; cbw ; add anumber, ax ; jmp num2 ;go to checks Num2: mov dx, offset secondn ; print the string "put 2nd 3 digits" mov ah, 9h int 21h ;run JMP SecondSecond SecondSecond: ;first digit mov ah, 1d ;bios code for read a keystroke int 21h ;call bios, it is understood that the ascii code will be returned in al mov firstdigit, al ;may as well save a copy sub al, 30h ;Convert code to an actual integer cbw ;CONVERT BYTE TO WORD. This takes whatever number is in al and ;extends it to ax, doubling its size from 8 bits to 16 bits ;The first digit now occupies all of ax as an integer mov cx, 100d ;This is so we can calculate 100*1st digit +10*2nd digit + 3rd digit mul cx ;start to accumulate the 3 digit number in the variable imul cx ;it is understood that the other operand is ax ; the result will use both dx::ax ;dx will contain only leading zeros add bnumber, ax ;save ;Second Digit mov ah, 1d ;bios code for read a keystroke int 21h ;call bios, it is understood that the ascii code will be returned in al mov seconddigit, al ;may as well save a copy sub al, 30h ;Convert code to an actual integer cbw ;CONVERT BYTE TO WORD. This takes whatever number is in al and ;extends it to ax, boubling its size from 8 bits to 16 bits ;The first digit now occupies all of ax as an integer mov cx, 10d ;continue to accumulate the 3 digit number in the variable mul cx ;it is understood that the other operand is ax, containing first digit ;the result will use both dx::ax ;dx will contain only leading zeros. add bnumber, ax ;save ;third Digit mov ah, 1d ;samething as above int 21h ; mov thirddigit, al ; sub al, 30h ; cbw ; add bnumber, ax ; jmp compare ;go to compare compare: CMP ax, anumber ;comparing numbB and Number JA comp1 ;go to comp1 if anumber is bigger CMP ax, anumber ; JB comp2 ;go to comp2 if anumber is smaller CMP ax, anumber ; JE equal ;go to equal if two numbers are the same JMP compare ;go to compare (avioding error) comp1: SUB ax, anumber; subtract smaller number from bigger number JMP compare ; comp2: SUB anumber, ax; subtract smaller number from bigger number JMP compare ; equal: mov ah, 9d ;make linkbreak after the 2nd 3 digit number mov dx, offset linebreaker int 21h mov ah, 9d ;print "HCF of two number is:" mov dx, offset messageB int 21h mov ax,anumber ;copying 2nd number into ax add al,30h ; converting to ascii mov newnumber,al ; copying from low part of register into newnumb mov ah, 2d ;bios code for print a character mov dl, newnumber ;we had saved the ascii code here int 21h ;call to bios JMP exit; exit: mov ah, 4ch int 21h ;exit the program End hi, this is a program that finds highest common factor of 2 different 3digit number. if i put 200, 235,312 (low numbers) it works fine. but if i put 500, 550, 654(bigger number) the program crashes after the 2nd 3digit number is entered. can you help me to find out what problem is?

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• SAX Parse Exception

  - by Anand

  Hi, I am using JMS Messaging in my java program. My messages are coming from IBM Main Frame and the messages are xml files. Mainframe pushwa messages that is xml files to the queue But when the java program reads the messages from the queue an additional character "?" gets added in front of the xml file. For example a file like this: <?xml version="1.0" encoding="UTF-8"?> becomes ?<?xml version="1.0" encoding="UTF-8"?> when I read the message from the queue. And when I try to parse this I catch the following exception SAX Exception org.xml.sax.SAXParseException: Content is not allowed in trailing section. Content is not allowed in trailing section What could be the reason for this additional character getting added ?

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• Regex question: Why isn't this matching?

  - by AllenG

  I have the following regex: (?<=\.\d+?)0+(?=\D|$) I'm running it against a string which contains the following: SVC~NU^0270~313.3~329.18~~10~~6.00: When it runs, it matches the 6.00 (correctly) which my logic then trims by one zero to turn into 6.0. The regex then runs again (or should) but fails to pick up the 6.0. I'm by no means an expert on Regex, but my understanding of my expression is that it's looking for a decimal with 1 or more optional (so, really zero or more) digits prior to one or more zeros which are then followed by any non-digit character or the line break. Assuming that interpretation is correct, I can't see why it wouldn't match on the second pass. For that matter, I'm not sure why my Regex.Replace isn't matching the full 6.00 on the first pass and removing both of the trailing zeros... Any suggestions?

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• Do I need to use decimal places when using floats? Is the "f" suffix necessary?

  - by Paulo Fierro

  I've seen several examples in books and around the web where they sometimes use decimal places when declaring float values even if they are whole numbers, and sometimes using an "f" suffix. Is this necessary? For example: [UIColor colorWithRed:0.8 green:0.914 blue:0.9 alpha:1.00]; How is this different from: [UIColor colorWithRed:0.8f green:0.914f blue:0.9f alpha:1.00f]; Does the trailing "f" mean anything special? Getting rid of the trailing zeros for the alpha value works too, so it becomes: [UIColor colorWithRed:0.8 green:0.914 blue:0.9 alpha:1]; So are the decimal zeros just there to remind myself and others that the value is a float? Just one of those things that has puzzled me so any clarification is welcome :)

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• RijndaelManaged Padding when data matches block size

  - by trampster

  If I use PKCS7 padding in RijndaelManaged with 16 bytes of data then I get 32 bytes of data output. It appears that for PKCS7 when the data size matches the block size it adds a whole extra block of data. If I use Zeros padding for 16 bytes of data I get out 16 bytes of data. So for Zeros padding if the data matches the block size then it doesn't pad. I have searched through the documentation and it says nothing about this difference in padding behavior. Can someone please point me to some kind of documentation which specifies what the padding behavior should be for the different padding modes when the data size matches the block size.

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• Using a Filter to serve a specific page?

  - by user246114

  Hi, I am using a class which implements Filter for my jsp stuff. It looks like this: public class MyFilter implements Filter { public void doFilter(ServletRequest request, ServletResponse response, FilterChain chain) throws IOException, ServletException { request.getRequestDispatcher("mypage.jsp").forward(request, response); } } So the target, "mypage.jsp", is just sitting in my top-level directory. The filter works fine if I'm entering urls like: http://www.mysite.com/foo http://www.mysite.com/boo but if I enter a trailing slash, I'll get a 404: http://www.mysite.com/foo/ http://www.mysite.com/boo/ HTTP ERROR: 404 /foo/mypage.jsp RequestURI=/foo/mypage.jsp it seems if I enter the trailing slash, then the filter thinks I want it to look for mypage.jsp in subfolder foo or boo, but I really always want it to just find it at: http://www.mysite.com/mypage.jsp how can I do that? Thank you

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• Matlab code help. Backward Euler method.

  - by m0s

  Here is the matlab/freemat code I got to solve ODE numerically using backward Euler method. However the results are inconsistent with my textbook results, and sometimes even ridiculously inconsistent. Please point out what is wrong with the code. I have already asked this question on mathoverflow.com no help there, hope someone here can help. function [x,y]=backEuler(f,xinit,yinit,xfinal,h) %f - this is your y prime %xinit - initial X %yinit - initial Y %xfinal - final X %h - step size n=(xfinal-xinit)/h; %Calculate steps %Inititialize arrays... %1st elements take xinit and yinit corespondigly, the rest fill with 0s x=[xinit zeros(1,n)]; y=[yinit zeros(1,n)]; %Numeric routine for i=1:n x(i+1)=x(i)+h; ynew=y(i)+h*(f(x(i),y(i))); y(i+1)=y(i)+h*f(x(i+1),ynew); end end

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