Search Results

Search found 1466 results on 59 pages for 'matlab toolbox'.

Page 36/59 | < Previous Page | 32 33 34 35 36 37 38 39 40 41 42 43  | Next Page >

  • read angles in radian and convert them in degrees/minutes/seconds

    - by Amadou
    n=0; disp('This program performs an angle conversion'); disp('input data set to a straight line. Enter the name'); disp('of the file containing the input Lambda in radian: '); filename = input(' ','s'); [fid,msg] = fopen(filename,'rt'); if fid < 0 disp(msg); else A=textscan(fid, '%g',1); while ~feof(fid) Lambda = A(1); n = n + 1; A = textscan(fid, '%f',1); end fclose(fid); end Alpha=Lambda*180/pi; fprintf('Angle converted from radian to degree/minutes/seconds:\n'); fprintf('Alpha =%12d\n',Alpha); fprintf('No of angles =%12d\n',n);

    Read the article

  • I need help on this data file to be edited in SOM_PAK format

    - by Mola
    Hi Experts, I am working on Self Organizing Map (SOM) Implementation and i have a microarray dataset which i am trying to read in using some_read_data function, but i keep having an errors when i edit it to have it in SOM_PAK form which is recognise by SOM for reading such as ??? Error using == somtoolbox\som_read_data.m Only 69 vector components on input file data line 1 (dimension is 70) Error in == SomMainFunction at 3 sD = som_read_data('B_r2.txt'); but when i try to read the data without editing which is the original file as shown here: http://rapidshare.com/files/376239367/DLBCL.txt.html It indicates "Data read OK", but i have the following error ??? Error using == unknown Out of memory. Type HELP MEMORY for your options. Error in == somtoolbox\som_bmus.m at 189 Bmus = zeros(dlen,length(which_bmus)); Error in == somvis\somvis_p_matrix.m at 41 [dummy dists] = som_bmus (dat, dat, 2:datlen); Error in == SomMainFunction at 16 [pheight rad_real perc] = somvis_p_matrix(sM,sD); You can get the datafile from here:http://rapidshare.com/files/376239367/DLBCL.txt.html I need someone to help me correct this data for me and put it in SOM_PAK format. I have tried getting it in SOM_PAK format, but it still giving me errors:

    Read the article

  • Solving for the coefficent of linear equations with one known coefficent

    - by CppLearner
    clc; clear all; syms y a2 a3 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % [ 0.5 0.25 0.125 ] [ a2 ] [ y ] % [ 1 1 1 ] [ a3 ] = [ 3 ] % [ 2 4 8 ] [ 6 ] [ 2 ] %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% M = [0.5 0.25 0.125; 1 1 1; 2 4 8]; t = [a2 a3 6]; r = [y 3 2]; sol = M * t' s1 = solve(sol(1), a2) % solve for a2 s2 = solve(sol(2), a3) % solve for a3 This is what I have so far. These are my output sol = conj(a2)/2 + conj(a3)/4 + 3/4 conj(a2) + conj(a3) + 6 2*conj(a2) + 4*conj(a3) + 48 s1 = - conj(a3)/2 - 3/2 - Im(a3)*i s2 = - conj(a2) - 6 - 2*Im(a2)*i sol looks like what we would have if we put them back into equation form: 0.5 * a2 + 0.25 * a3 + 0.125 * a4 a2 + a3 + a4 = 3 2*a2 + 4*a3 + 8*a4 = 2 where a4 is known == 6. My problem is, I am stuck with how to use solve to actually solve these equations to get the values of a2 and a3. s2 solve for a3 but it doesn't match what we have on paper (not quite). a2 + a3 + 6 = 3 should yield a3 = -3 - a2. because of the imaginary. Somehow I need to equate the vector solution sol to the values [y 3 2] for each row.

    Read the article

  • Error in Ordinary Differential Equation representation

    - by Priya M
    UPDATE I am trying to find the Lyapunov Exponents given in link LE. I am trying to figure it out and understand it by taking the following eqs for my case. These are a set of ordinary differential equations (these are just for testing how to work with cos and sin as ODE) f(1)=ALPHA*(y-x); f(2)=x*(R-z)-y; f(3) = 10*cos(x); and x=X(1); y=X(2); cos(y)=X(3); f1 means dx/dt;f2 dy/dt and f3 in this case would be -10sinx. However,when expressing as x=X(1);y=X(2);i am unsure how to express for cos.This is just a trial example i was doing so as to know how to work with equations where we have a cos,sin etc terms as a function of another variable. When using ode45 to solve these Eqs [T,Res]=sol(3,@test_eq,@ode45,0,0.01,20,[7 2 100 ],10); it throws the following error ??? Attempted to access (2); index must be a positive integer or logical. Error in ==> Eq at 19 x=X(1); y=X(2); cos(x)=X(3); Is my representation x=X(1); y=X(2); cos(y)=X(3); alright? How to resolve the error? Thank you

    Read the article

  • xml2struct cannot access element(s)

    - by csetzkorn
    I try to use this: xml2struct when I use this xml: <XMLname attrib1="Some value"> <Element>Some text</Element> <DifferentElement attrib2="2">Some more text</DifferentElement> <DifferentElement attrib3="2" attrib4="1">Even more text</DifferentElement> </XMLname> I can create a struct: test = xml2struct('C:\bla\bla.xml'); (tested it with class(test)) It looks like this: test = Name: 'XMLname' Attributes: [1x1 struct] Data: '' Children: [1x7 struct] But I cannot access: test.XMLname.Element.Text I get: ??? Reference to non-existent field 'XMLname'. Any ideas?

    Read the article

  • HOw to Extract image pixels and perform computation on it

    - by gavishna
    What i want to do is extract the pixels into array A(i,j).Let A(i,j) represent the pixel in the ith row & jth colm of the image. Let B(i,j) be the transformed array containing bits such that B(i,j) = 1 if (A(i,j)/2^t) mod 2 =1, where t=0,1,2,3,4,5,6,7 = 0 otherwise Then the array B(i,j) will be mutliplied by another array say C(i,j) representing another image. Then i want to plot the result of B*C as an image. How do i take in the pixels for the above conditions for both gray and RGB color format.Also how do i solve the above problem.Any helo will be usefull because i am not an expert in image processing.

    Read the article

  • Unpacking varargin to individual variables

    - by Richie Cotton
    I'm writing a wrapper to a function that takes varargin as its inputs. I want to preserve the function signature in the wrapper, but nesting varargin causes all the variable to be lumped together. function inner(varargin) %#ok<VANUS> % An existing function disp(nargin) end function outer(varargin) % My wrapper inner(varargin); end outer('foo', 1:3, {}) % Uh-oh, this is 1 I need a way to unpack varargin in the outer function, so that I have a list of individual variables. There is a really nasty way to do this by constructing a string of the names of the variables to pass the inner, and calling eval. function outer2(varargin) %#ok<VANUS> % My wrapper, second attempt inputstr = ''; for i = 1:nargin inputstr = [inputstr 'varargin{' num2str(i) '}']; %#ok<AGROW> if i < nargin inputstr = [inputstr ', ']; %#ok<AGROW> end end eval(['inner(' inputstr ')']); end outer2('foo', 1:3, {}) % 3, as it should be Can anyone think of a less hideous way of doing things, please?

    Read the article

  • A think on sum(X, 1)

    - by Runner
    >> X = [0 1 2 3 4 5] >> sum(X, 1) ans = 3 5 7 sum(X, 1) should sum along the 1st dimension(row) as per the document says: S = SUM(X,DIM) sums along the dimension DIM. But why does it actually sums along the 2nd dimension(column)?

    Read the article

  • why arrayfun does NOT improve my struct array operation performance

    - by HaveF
    here is the input data: % @param Landmarks: % Landmarks should be 1*m struct. % m is the number of training set. % Landmark(i).data is a n*2 matrix old function: function Landmarks=CenterOfGravity(Landmarks) % align center of gravity for i=1 : length(Landmarks) Landmarks(i).data=Landmarks(i).data - ones(size(Landmarks(i).data,1),1)... *mean(Landmarks(i).data); end end new function which use arrayfun: function [Landmarks] = center_to_gravity(Landmarks) Landmarks = arrayfun(@(struct_data)... struct('data', struct_data.data - repmat(mean(struct_data.data), [size(struct_data.data, 1), 1]))... ,Landmarks); end %function center_to_gravity when using profiler, I find the usage of time is NOT what I expected: Function Total Time Self Time* CenterOfGravity 0.011s 0.004 s center_to_gravity 0.029s 0.001 s Can someone tell me why? BTW...I can't add "arrayfun" as a new tag for my reputation.

    Read the article

  • calling function in radiobutton group

    - by vijisai
    thank you very much. with your help, i am now able to call the function for each radio button. however, i get a error message Reference to non-existent field 'ics_si' ics_si is my function, which has the following code, i do not know where i am making a mistake i have created the edit box for user to input the values for bore and stroke. and vdisp is calculated and the result is displayed in the third edit box. function ics_si_Callback(hObject, eventdata, handles) b = str2double(get(handles.bore,'String')); s = str2double(get(handles.stroke,'String')); vdisp = (pi * b * b * s*10^(-3))/4; set(handles.vdisp,'String',vdisp); this code must be called when i press the first or second radio button. i.e. when the radio button is pressed, it should call the function ics_si, calculate it and display the result. how to get this.

    Read the article

  • ??? Error using ==> dlmread at 55 Filename must be a string.

    - by Tim
    [file_input, pathname] = uigetfile( ... {'*.txt', 'Text (*.txt)'; ... '*.xls', 'Excel (*.xls)'; ... '*.*', 'All Files (*.*)'}, ... 'Select files'); D = uiimport(file_input); M = dlmread(D); X = freed(M); Getting errors with dlmread......"??? Error using == dlmread at 55 Filename must be a string."..need to get the data from dlmread to "freed"

    Read the article

  • How to unblend two images from a blend image

    - by gavishna
    I have blended/merged 2 images img1 and img2 with this code which is woking fine.What i want to know is how to obtain the original two images img1 and img2.The code for blending is as under img1=imread('C:\MATLAB7\Picture5.jpg'); img2=imread('C:\MATLAB7\Picture6.jpg'); for i=1:size(img1,1) for j=1:size(img1,2) for k=1:size(img1,3) output(i,j,k)=(img1(i,j,k)+img2(i,j,k))/2; end end end imshow(output,[0 255]);

    Read the article

  • Triangulation & Direct linear transform

    - by srand
    Following Hartley/Zisserman's Multiview Geometery, Algorithm 12: The optimal triangulation method (p318), I got the corresponding image points xhat1 and xhat2 (step 10). In step 11, one needs to compute the 3D point Xhat. One such method is Direct Linear Transform (DLT), mentioned in 12.2 (p312) and 4.1 (p88). The homogenous method (DLT), p312-313, states that it finds a solution as the unit singular vector corresponding to the smallest singular value of A, thus, A = [xhat1(1) * P1(3,:)' - P1(1,:)' ; xhat1(2) * P1(3,:)' - P1(2,:)' ; xhat2(1) * P2(3,:)' - P2(1,:)' ; xhat2(2) * P2(3,:)' - P2(2,:)' ]; [Ua Ea Va] = svd(A); Xhat = Va(:,end); plot3(Xhat(1),Xhat(2),Xhat(3), 'r.'); However, A is a 16x1 matrix, resulting in a Va that is 1x1. What am I doing wrong (and a fix) in getting the 3D point? For what its worth sample data: xhat1 = 1.0e+009 * 4.9973 -0.2024 0.0027 xhat2 = 1.0e+011 * 2.0729 2.6624 0.0098 P1 = 699.6674 0 392.1170 0 0 701.6136 304.0275 0 0 0 1.0000 0 P2 = 1.0e+003 * -0.7845 0.0508 -0.1592 1.8619 -0.1379 0.7338 0.1649 0.6825 -0.0006 0.0001 0.0008 0.0010 A = <- my computation 1.0e+011 * -0.0000 0 0.0500 0 0 -0.0000 -0.0020 0 -1.3369 0.2563 1.5634 2.0729 -1.7170 0.3292 2.0079 2.6624

    Read the article

  • measuring uncertainty in matlabs svmclassify

    - by Mark
    I'm doing contextual object recognition and I need a prior for my observations. e.g. this space was labeled "dog", what's the probability that it was labeled correctly? Do you know if matlabs svmclassify has an argument to return this level of certainty with it's classification? If not, matlabs svm has the following structures in it: SVM = SupportVectors: [11x124 single] Alpha: [11x1 double] Bias: 0.0915 KernelFunction: @linear_kernel KernelFunctionArgs: {} GroupNames: {11x1 cell} SupportVectorIndices: [11x1 double] ScaleData: [1x1 struct] FigureHandles: [] Can you think of any ways to compute a good measure of uncertainty from these? (Which support vector to use?) Papers/articles explaining uncertainty in SVMs welcome. More in depth explanations of matlabs SVM are also welcome. If you can't do it this way, can you think of any other libraries with SVMs that have this measure of uncertainty?

    Read the article

  • Problem with xor operation

    - by gavishna
    Kindly tell me why the original image is not coming with this code. The resulting image receive is yellowish in color,instead of being similar to the image Img_new Img=imread(‘lena_color.tif’); Img_new=rgb2gray(img); Send=zeroes(size(Img_new); Receive= zeroes(size(Img_new); Mask= rand(size(Img_new); for i=1 :256 for j=1:256 Send(i,j)=xor( Img_new(i,j),mask(i,j)); End End image(send); imshow(send); for i=1 :256 for j=1:256 receive(i,j)=xor( send(i,j),mask(i,j)); End End image(receive); imshow(receive); plz help

    Read the article

  • Better way to compare neighboring cells in matrix

    - by HyperCube
    Suppose I have a matrix of size 100x100 and I would like to compare each pixel to its direct neighbor (left, upper, right, lower) and then do some operations on the current matrix or a new one of the same size. A sample code in Python/Numpy could look like the following: (the comparison 0.5 has no meaning, I just want to give a working example for some operation while comparing the neighbors) import numpy as np my_matrix = np.random.rand(100,100) new_matrix = np.array((100,100)) my_range = np.arange(1,99) for i in my_range: for j in my_range: if my_matrix[i,j+1] > 0.5: new_matrix[i,j+1] = 1 if my_matrix[i,j-1] > 0.5: new_matrix[i,j-1] = 1 if my_matrix[i+1,j] > 0.5: new_matrix[i+1,j] = 1 if my_matrix[i-1,j] > 0.5: new_matrix[i-1,j] = 1 if my_matrix[i+1,j+1] > 0.5: new_matrix[i+1,j+1] = 1 if my_matrix[i+1,j-1] > 0.5: new_matrix[i+1,j-1] = 1 if my_matrix[i-1,j+1] > 0.5: new_matrix[i-1,j+1] = 1 This can get really nasty if I want to step into one neighboring cell and start from it to do a similar task... Do you have some suggestions how this can be done in a more efficient manner? Is this even possible?

    Read the article

  • How to calculate the current index?

    - by niko
    Hi, I have written an algorithm which iteratively solves the problem. The first iteration consists of 6 steps and all the following iterations consist of 5 steps (first step is skipped). What I want to calculate is the current (local) step in the iteration from current global step. For example if there are 41 steps in total which means there are 8 iterations: indices from 1 to 6 belong to 1st iteration indices from 7 to 11 belong to second iteration ... For calculating the current iteration I have written the following code: if(currentStep <= 6) iteration = 1; else iteration = floor((currentStep - 7)/5) + 2; end The problem remains in calculating local steps. in first iteration the performed steps are: 1, 2, 3, 4, 5, 6 in all the following iterations the performing steps are 2, 3, 4, 5, 6 So what has to be done is to transform the array of global steps [1 2 3 4 5 6 7 8 9 10 11 12 13 ... 41] into array of local steps [1 2 3 4 5 6 2 3 4 5 6 2 3 ... 6]. I would appreciate if anyone could help in finding the solution to a given problem. Thank you!

    Read the article

  • How to convert a one column integer data file into a mask for image

    - by gavishna
    I have a data file which contains integers say in range 0-255 containing about 1000 integers which are random in nature.I want to use that as a mask or to multiply an image which is in RGb and another image which is in gray format. HOw do i go about this, how do i convert/represent this data file in matrix format of image dimension ?Kindly suggest. also is it possible to obtain a 3D histogram?

    Read the article

  • divide the image into 3*3 blocks

    - by Jayanth Silesh
    I have a matrix that does not happen to have dimensions that are multiples of 3 or it might. How can we divide the entire image into blocks of 3*3 matrices. (Can ignore the last ones which does not come under the 3*3 multiples. Also, the 3*3 matrices can be be saved in arrays. a=3; b=3; %window size x=size(f,1)/a; y=size(f,2)/b; %f is the original image m=a*ones(1,x); n=b*ones(1,y); I=mat2cell(f,m,n);

    Read the article

< Previous Page | 32 33 34 35 36 37 38 39 40 41 42 43  | Next Page >