explanation for matlab code - matlab

Am new to matlab.
Can someone explain me the following code. this code is used for training the neural network
N = xlsread('data.xls','Sheet1');
N = N(1:150,:);
UN = xlsread('data.xls','Sheet2');
UN = UN(1:150,:);
traindata = [N ; UN];
save('traindata.mat','traindata');
label = [];
for i = 1 : size(N,1)*2
if( i <= size(N,1))
% label = [label ;sum(traindata(i,:))/size(traindata(i,:),2)];
label = [label ;sum(traindata(i,:))/10];
else
% label = [label ;sum(traindata(i,:))/size(traindata(i,:),2)];
label = [label ;sum(traindata(i,:))/10];
end
end
weightMat = BpTrainingProcess(4,0.0001,0.1,0.9,15,[size(traindata,1) 1],traindata,label);

I cannot find a Neural Network toolbox built-in that corresponds to BpTrainingProcess(), so this must be a file you have access to locally (or you need to obtain from the person who gave you this code). It likely strings together several function calls to Neural Network toolbox functions, or perhaps is an original implementation of a back-propagation training method.
Otherwise, the code has some drawbacks. For one, it doesn't appear that the interior if-else statement actually does anything. Even the lines that are commented out would leave a totally useless if-else setup. It looks like the if-else is intended to let you do different label normalization for the data loaded from Sheet1 of the Excel file vs. data loaded from Sheet2. Maybe that is important for you, but it's currently not happening in the program.
Lastly, the code uses an empty array for label and the proceeds to append rows to the empty array. This is unneeded because you already know how many rows there will be (it will total up to size(N,1)*2 = 150*2 = 300 rows. You could just as easily set label=zeros(300,1) and then use usual indexing at each iteration of the for-loop: label(i) = .... This saves time and space, but arguably won't matter much for a 300-row data set (assuming that the length of each row is not too large).
I put documentation next to the code below.
% The functionn 'xlsread()' reads data from an Excel file.
% Here it is storing the values from Sheet 1 of the file 'data.xls'
% into the variable N, and then using the syntax N = N(1:150,:) to
% change N from being all of the data into being only the first
% 150 rows of the data
N = xlsread('data.xls','Sheet1');
N = N(1:150,:);
% Now do the same thing for Sheet 2 from the Excel file.
UN = xlsread('data.xls','Sheet2');
UN = UN(1:150,:);
% This concatenates the two different data arrays together, making
% one large array where N is the top half and UN is the bottom half.
% This is basically just stacking N on top of UN into one array.
traindata = [N ; UN];
% This saves a copy of the newly stacked array into the Matlab data file
% 'traindata.mat'. From now on, you should be able to load the data from
% this file, without needing to read it from the Excel sheet above.
save('traindata.mat','traindata');
% This makes an empty array which will have new things appended to it below.
label = [];
% Because UN and N have the same number of rows, then the training data
% has twice as many rows. So this sets up a for loop that will traverse
% all of these rows of the training data. The 'size()' function can be
% used to get the different dimensions of an array.
for i = 1 : size(N,1)*2
% Here, an if statement is used to check if the current row number, i,
% is less than or equal to than the number of rows in N. This implies
% that this part of the if-statement is only for handling the top half
% of 'trainingdata', that is, the stuff coming from the variable N.
if( i <= size(N,1))
% The line below was already commented out. Maybe it had an old use
% but is no longer needed?
% label = [label ;sum(traindata(i,:))/size(traindata(i,:),2)];
% This syntax will append new rows to the variable 'label', which
% started out as an empty array. This is usually bad practice, memory-wise
% and also for readability.
% Here, the sum of the training data is being computed, and divided by 10
% in every case, and then appended as a new row in 'label'. Hopefully,
% if you are familiar with the data, you will know why the data in 'N'
% always needs to be divided by 10.
label = [label ;sum(traindata(i,:))/10];
% Otherwise, if i > # of rows then handle the data differently.
% Really this means the code below treats only data from the variable UN.
else
% The line below was already commented out. Maybe it had an old use
% but is no longer needed?
% label = [label ;sum(traindata(i,:))/size(traindata(i,:),2)];
% Just like above, the data is being divided by 10. Given that there
% is nothing different about the code here, and how it modifies 'label'
% there is no need for the if-else statements, and they only waste time.
label = [label ;sum(traindata(i,:))/10];
% This is needed to show the end of the if-else block.
end
% This is needed to show the end of the for-loop.
end
% This appears to be a Back-Propagation Neural Network training function.
% This doesn't match any built-in Matlab function I can find, but you might
% check in the Neural Network toolbox to see if the local function
% BpTrainingProcess is a wrapper for a collection of built-in training functions.
weightMat = BpTrainingProcess(4, 0.0001, 0.1, 0.9, 15,
[size(traindata,1) 1], traindata,label);
Here is a link to an example Matlab Neural Network toolbox function for back-propagation training. You might want to look around the documentation there to see if any of it resembles the interior of BpTrainingProcess().

Related

Matlab interp1 gives last row as NaN

I have a problem similar to here. However, it doesn't seem that there is a resolution.
My problem is as such: I need to import some files, for example, 5. There are 20 columns in each file, but the number of lines are varied. Column 1 is time in terms of crank-angle degrees, and the rest are data.
So my code first imports all of the files, finds the file with the most number of rows, then creates a multidimensional array with that many rows. The timing is in engine cycles so, I would then remove lines from the imported file that go beyond a whole engine cycle. This way, I always have data in terms of X whole engine cycles. Then I would just interpolate the data to the pre-allocated array to have a giant multi-dimensional array for the 5 data files.
However, this seems to always result in the last row of every column of every page being filled with NaNs. Please have a look at the code below. I can't see where I'm doing wrong. Oh, and by the way, as I have been screwed over before, this is NOT homework.
maxlines = 0;
maxcycle = 999;
for i = 1:1
filename = sprintf('C:\\Directory\\%s\\file.out',folder{i});
file = filelines(filename); % Import file clean
lines = size(file,1); % Find number of lines of file
if lines > maxlines
maxlines = lines; % If number of lines in this file is the most, save it
end
lastCAD = file(end,1); % Add simstart to shift the start of the cycle to 0 CAD
lastcycle = fix((lastCAD-simstart)./cycle); % Find number of whole engine cycles
if lastcycle < maxcycle
maxcycle = lastcycle; % Find lowest number of whole engine cycles amongst all designs
end
cols = size(file,2); % Find number of columns in files
end
lastcycleCAD = maxcycle.*cycle+simstart; % Define last CAD of whole cycle that can be used for analysis
% Import files
thermo = zeros(maxlines,cols,designs); % Initialize array to proper size
xq = linspace(simstart,lastcycleCAD,maxlines); % Define the CAD degrees
for i = 1:designs
filename = sprintf('C:\\Directory\\%s\\file.out',folder{i});
file = importthermo(filename, 6, inf); % Import the file clean
[~,lastcycleindex] = min(abs(file(:,1)-lastcycleCAD)); % Find index of end of last whole cycle
file = file(1:lastcycleindex,:); % Remove all CAD after that
thermo(:,1,i) = xq;
for j = 2:17
thermo(:,j,i) = interp1(file(:,1),file(:,j),xq);
end
sprintf('file from folder %s imported OK',folder{i})
end
thermo(end,:,:) = []; % Remove NaN row
Thank you very much for your help!
Are you sampling out of the range? if so, you need to tell interp1 that you want extrapolation
interp1(file(:,1),file(:,j),xq,'linear','extrap');

How to convert cell variable to classified variable inside of parfor loop in matlab? [duplicate]

I have this (quite long) Matlab code with nested loops where I want to parallelize the main time-consuming iteration. The only variable that (apparently) gives me problems is DMax, where I get the error:
Error: The variable DMax in a `parfor` cannot be classified.
See Parallel for Loops in MATLAB, "Overview".
This is a draft of my code:
t0=matrix (Maxiter,1); % This is a big matrix whose dimensions are reported in brachets
Maxiter = 1E6;
DMax = zeros(Maxiter,40);
% Other Stuff
for j=1:269
% Do more stuff
for soil=1:4
parfor i =1:Maxiter
k(i,soil) = a %k is a real number
a(i,soil) = b %similar to k
% Do a lot of stuff
for t= (floor(t0(i,soil))+1):40
DMax(i,t) = k(i,soil)*((t-t0(i,soil))^a(i,soil));
% Do some more stuff
end
end
end
end
for time=1:40
% Do the final stuff
end
I guess the problem is in the way I defined DMax, but I do not know what it could be more precisely. I already looked on the web but with not very satisfying results.
It is very clearly described in the documentation that each variable inside parfor must be classified into one of several types. Your DMax variable should be a sliced variable (arrays whose segments are operated on by different iterations of the loop), but in order to be classified as such, all the following conditions must hold:
Type of First-Level Indexing — The first level of indexing is either parentheses, (), or braces, {}.
Fixed Index Listing — Within the first-level parenthesis or braces, the list of indices is the same for all occurrences of a
given variable.
Form of Indexing — Within the list of indices for the variable, exactly one index involves the loop variable.
Shape of Array — The array maintains a constant shape. In assigning to a sliced variable, the right-hand side of the assignment cannot be [] or '', because these operators attempt to
delete elements.
Clearly, Fixed Index Listing property does not hold since you reference it as DMax(i,t) where t changes its values. There's an identical example described in the documentation, please pay attention. So one workaround would be to use a temporary variable inside the inner loop, and then assign the whole row back to DMax.
Also note that variable a cannot be classified into any category either. That's not to mention that it's not defined in your example at all. Please read the guide carefully and make sure it can be classified into one of the categories. Rewrite the code if needed, e.g. introducing new temporary variables.
Here's the code where DMax usage is corrected:
Maxiter = 1E6;
t0 = randn(Maxiter,1); % This is a big matrix whose dimensions are reported in brachets
DMax = zeros(Maxiter,40);
% Other Stuff
for j = 1:269
% Do more stuff
for soil = 1:4
parfor i = 1:Maxiter
k(i,soil) = a %k is a real number
a(i,soil) = b %similar to k
% Do a lot of stuff
tmp = zeros(1,40);
for t = (floor(t0(i,soil))+1):40
tmp(t) = k(i,soil)*((t-t0(i,soil))^a(i,soil));
% Do some more stuff
end
DMax(i,:) = tmp;
end
end
end
for time = 1:40
% Do the final stuff
end

Vectorization of matlab code for faster execution

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 100 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.What is the problem?Where i was wrong?
To answer your second question, you can simply 'save' any generated variable as a .mat file in your working directory (Current folder) which can be accessed later. So in your code, if the 'training eigenfaces' is given by the variable 'u', you can use the following:
save('eigenface.mat','u')
This creates a .mat file with the name eigenface.mat which contains the variable 'u', the eigenfaces. Note that this variable is saved in your Current Folder.
In a later instance when you are trying out with your test data, you can simply 'load' this variable:
load('eigenface.mat')
This automatically loads 'u' into your workspace.
You can also save additional variables in the same .mat file if necessary
save('eigenface.mat','u','v'...)
The answer to the first question is that the code is simply not done running yet. Vectorizing the code (instead of using a for loop), as suggested in the comment section above can improve the speed significantly.
[EDIT]
Since the images are not very big, the code is not significantly slowed down by the first for loop. You can improve the performance of the rest of the code by vectorizing the code. Vectorized operations are faster than for-loops. This link might be useful in understanding vectorization:
http://www.mathworks.com/help/matlab/matlab_prog/vectorization.html
For example, the second for-loop can be replaced by the following vectorized form, as suggested in the comments
tempS = double(S);
meanS = mean(S,1);
stdS = std(S,0,1);
S = (tempS - meanS) ./ stdS;
Use MATLAB's timer functions tic and toc for finding how long the first for-loop alone is taking to execute. Add tic before the for loop and toc after it. If the time taken for 50 images is about 104 seconds, then it would be significantly more for 1200 images.

beginner:referencing a cell containing a zero matlab

So far I have got this code:
clear all; % clears all variables from your workspace
close all; % closes all figure windows
clc; % clears command window
%%=============================================
%%define number of trials and subjects
%%=============================================
njump=81;
nsbj=6;
%%
%%==============================================
%%defining size of cell that will be created
%%==============================================
data=cell(njump,1);
%%
%%==============================================
%%defining gravity and frame rate per second (fps)
%%==============================================
fps=0.001; %frames per second
g=-9.81; %acceleration
%%
%%==============================================
%%read in excel data in CSV format
%%===============================================
for i=1:njump;
x=sprintf('Trial%02d.csv',i),sprintf('Trial%02d',i),'A1:E7000';;% jump data
data{i}=xlsread(x,'A1:E7000');
%%===============================================
%%defining total no. of frames and time of flight
%tnf=total number of frames equal to zero
%n = nnz(X) returns the number of nonzero elements in matrix X.
%%===============================================
% myMax(i) = nanmax(data{i}(:,5));
% vals = find(data{i}(:,5) > (myMax(i) - 10));
% pointsInAir = numel(vals,i);
tnf(i,1) = size(data{i,1},1) - nnz(data{i,1}(:,5)); %number of zeros
tof(i)=tnf(i)*fps; %Time of flight is equal to this
jh(i,1)=(-g*(tof(i).^2)/8); %defining jump height
%%=================================================
%%to find average power first use "find" function to find the first zero in
%%Fz, have the cell referenced
%%then use nanmean for average power(av_pwr)
%%use nanmin for peak power (peak_pwr)
%%=================================================
n = 1; % the following call to find retrieves only the n first element(s) found.
ref= find(data{i,1}(:,5)==0, n, 'first');
% ref=find(data(:,5)==0);
peak_pwr(i,1) = nanmin (data {i,1}(1:ref,5)); %preak power in coloumn E1 in all data with reference to cell found
av_pwr(i,1)=nanmean(data {i,1}(1:ref,5));%average power in coloumn E1 in all data with reference to cell found
%%==================================================
%%Plot the results onto a time vs jump height, time vs average power and
%%time vs peak power
However the part that is hard is trying to find the first zero in column E which is the 5th column to use as a reference cell. I want to use this reference cell so that I can do my average and peak power calcs. that use the numbers before this zero.
In this case ref is empty so you cannot access the first element.
If you think that ref should not be empty you need to go back further to see where things go wrong. Otherwise, you can use something like:
if any(ref)
%Do something
else
%Return the default value/do alternative action
end
It could help to have an example of what's in data. I have created one which might be similar to yours :
data{1,1}=magic(6)-10
Now in this matrix, column 5 actually does have a zero element so ref= find(data{1,1}(:,5)==0); ref(1) actually works and retrieves the first index of a zero element. However, if it didn't, you would be trying to access the first element of an empty matrix.
Try instead using the second (and perhaps third) arguments of find to achieve this :
n = 1;
% the following call to find retrieves only the n first element(s) found.
ref= find(data{1,1}(:,5)==0, n, 'first');
The rest of your code seems like it should work, although from the looks of it i have a feeling your loop (i take it you are using a loop for i) could maybe be vectorized.
Hope this helps :)
Tepp

Output loop result in Matlab

Hi have this code and I don't know how to put the output result with every pixel.I think the output code are not well defined.
EDIT:
I'm going to try to explain the code:
% I have an image
imagen1=imread('recor.tif');
imagen2= double(imagen1);
band1= imagen2(:,:,1);
% I preallocate the result (the image size is 64*89*6)
yvan2= zeros(61,89,1);
% For every pixel of the image, I want to get one result (each one is different).
for i = 1 : nfiles
for j = 1 : nrows
for i = 1:numel(band1)
% I'm doing this because I've to multiply the results of this interpolation with that result a2ldb1y= ldcm_1(:,1). This vector has a length of 2151x1 and I need to muliply the result of the interpolation for (101:267) position on the vector, this is the reason because I'm doing the interpolation since 101 to 267 (also because I don't have that values).
interplan= interp1(van1,man2,t2,'spline');
ma(96) = banda1a(i); % I said 96, because I want to do an interpollation
end
van1= [101 96 266]';
mos1= ma(134);
van2= [0 pos1 0];
t= 101:267;
t2= t';
xi= 101:1:267;
xi2=xi';
interplan= interp1(van1,van2,t2,'spline');
% After this, I 'prepare' the vector.
out=zeros(2151,1)
out(101:267) = interplan;
% And then, I do all this operation (are important for the result)
a2ldb1y= ldcm_1(:,1);
a2ldsum_pesos1= sum(a2ldb1y);
a2l7dout1_a= a2ldb1y.*out;
a2l7dout1_b= a2l7dout1_a./a2ldsum_pesos1;
a2l7dout1_c= sum(a2l7dout1_b);
% And the result a2l7dout1_c I want it for every pixel (the results are different because every pixel has a different value...)
**yvan2(:,:,1)= [a2l7dout1_c];**
end
end
Thanks in advance,
I'm shooting in the dark here, but I think you're looking for:
yvan2(i, j, 1)= a2l7dout1_c;
instead of:
yvan2(:,:,1)= [a2l7dout1_c];
and thus your output should be stored in the variable yvan2 after the loops are done.
P.S
Some issues in your code:
Why do you have two loops using the same iteration variable i? Your calculations are probably incorrect since i is being modified by two for loops.
Why do you even need the second loop? Each iteration overruns the value of ma(134) set by the previous iteration. You can just replace the entire loop with:
ma(134) = banda1a(numel(band1))
You shouldn't be using the names i and j for loop variables. They are already reserved for the imaginary unit (that is, sqrt(-1)), so MATLAB needs extra processing time for name resolution. You'd rather use other loop variable names instead, even ii and jj.