My code works but every time it runs through line 13 it writes on the command window :"Warning: Integer operands are required for colon operator
when used as index ".
The relevant part of my code looks like that:
filename = uigetfile;
obj = mmreader(filename);
nFrames=obj.NumberOfFrames;
for k = 1 : nFrames
this_frame = read(obj, k);
thisfig = figure();
thisax = axes('Parent', thisfig);
image(this_frame, 'Parent', thisax);
if k==1
handle=imrect;
pos=handle.getPosition;
end
partOf=this_frame(pos(2):pos(2)+pos(4),pos(1):pos(1)+pos(3));%this is line 13
vector(k)=mean2(partOf);
title(thisax, sprintf('Frame #%d', k));
end
Why this warning appears and can i ignore it?
It's probably because one or more of the following: pos(2), pos(2)+pos(4), pos(1) and pos(1)+pos(3) are not integers, which indices are supposed to be. You may want to use the round function to round them up to integer values.
Maayan,
The problem seems to occur from the values of your pos vector (and the value of the calculations you make to pos vector values).
This is the solution quoted from MathWorks(MATLAB) website:
http://www.mathworks.com/support/solutions/en/data/1-FA9A2S/?solution=1-FA9A2S
Modify the index computations using the FIX, FLOOR, CEIL, or ROUND functions to ensure that the indices are integers. You can test if a variable contains an integer by comparing the variable to the output of the ROUND function operating on that variable when MATLAB is in debug mode on the line containing the variable.
Related
So, I have a series of display functions, ranging from x1 to x7. These all contain both strings and variables like:
x1 = ['The result of the scalar multiplication of V and U: ',num2str(scalar_uv)];
x2 = similar to above but with for example a value on the cross multiplication of the two scalars.
Instead of printing out each one through:
disp(x1);
disp(x2);
disp(x3);
I thought it would be possible to print them all out through a for loop or perhaps a nested for loop but I just can't figure out how to do it. I preferably don't want straight up solutions (I won't say no to them) but rather some hints or tips of possible.
A simple example solution would be to make a cell array and loop through it, or use celldisp() to display it. But if you want to print nicely, i.e. formatted specifically, to the command window you can use the fprintf function and format in line breaks. For example:
for displayValue = {x1, x2, x3, x4}
fprintf('%s\n', displayValue{1});
end
If you want more formatting options, such as precision, or fieldwidth, the formatspec code (%s in the example) has many configurations. You can see them on the fprintf helpdoc. The \n just tells the fprintf function to create a newline when it prints.
Instead of creating seven different variables (x1...x7), just create a cell array to hold all your strings:
x{1} = ['The result of the scalar multiplication of V and U: ',num2str(scalar_uv)];
x{2} = ['Some other statement with a value at the end: ',num2str(somevar)];
Now you can write a loop:
for iX = 1:length(x)
disp(x{iX})
end
Or use cellfun to display them without a for loop:
cellfun(#disp,x)
If you really want to keep them named x1...x7, then you can use an eval statement to get your variable names:
for iX = 1:7
disp(eval(['x' num2str(iX)]));
end
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
If I want to construct a double for loop, but for index k I don't want it to be continuous index but like [1,2,4,7]. I tried to do the following, and it did not work.
for i=1:100
for k=1:2:4:7;
b(i)=i*k;
end
end
Anyone could help me deal with that?
When you want to construct an array, you don't want to necessarily use the colon (:) operator unless you want to create a range of values (like you do when you want i to be all values between 1 and 100), instead you want to use square brackets ([]) with comma separators to explicitly create an array of discrete values.
k = [1,2,4,7];
Now that you do this you can specify your loop like you had it with this small substitution for the values of k
kvalues = [1,2,4,7];
for i = 1:100
for k = 1:numel(kvalues)
b(i) = i * kvalues(k);
end
end
Notice that I have defined kvalues once outside of the loop so that it is not created every iteration through the outer loop (thanks to #dfri for pointing this oversight out)
The way that you have your loop written, you're actually over-writing the value of b(i) every time through the inner loop. I'm not sure if that's what you intended to do. If it is, then you can reduce your loop to the following:
b = k(end) * (1:100);
Otherwise, if you meant to have it be b(i,k) = i * k, you could rewrite this with bsxfun.
b = bsxfun(#mtimes, [1,2,4,7], (1:100).');
The syntax k=1:2:4:7 will not work as you intend. Usually we make use of the "two colon" syntax to describe a non-default (1) step size from the given start to end values, k = start:stepSize:end. Using an additional colon here even yields a Matlab warning ("third colon probably not intended?").
One possible workaround is to let your "non-continuous" indices reside in a vector and extract members of this vector in the inner for loop as follows
nonContIndex = [1; 2; 4; 7];
numIndices = numel(nonContIndex);
b = zeros(100,numIndices);
for i=1:100
for k=1:numIndices
b(i,k)=i*nonContIndex(k);
end
end
As noted by comments and the other answer: if b is simply a vector, your original loop will overwrite the b(i):th entry for each run of the inner loop. The above assumes that you in fact want a 2D matrix as a result.
I am working on a code to extract my AR(1)-GARCH(1) parameter, which I estimated using an AR(1)-GJR(1,1) model to individual matrices so that I can use them as variables in my calculations. As I have 16 time series variables, I combine the code with a loop in the following way:
for i=1:nIndices
AA_ARCH(:,i) = cell2mat(fit{i}.Variance.ARCH)';
end;
My problem is that for some variables is are no for AA_ARCH(:,i) the dimension is lower than nIndices. Naturally, when I try to export the estimates in the loop which specified the dimension of (:,i) and nIndices matlab reports a dimension mismatch. I would like to tell Matlab to replace the NaN with 0 instead of leaving the spot empty so that it is able to produce a (1,nIndices) matrix from AA_ARCH.
I thought of something like the this:
fit{i}.Variance.Leverage(isnan(fit{i}.Variance.Leverage))=0
but I wasn't able to combine this part with the previous code.
I would be very happy about any hints!
Best, Carolin
UPDATE:
Here is a fully a runnable version of my code which produces my problem. Notice that the code produces a dimension mismatch error because there is no ARCH and GARCH estimate in the fit.gjr(1,1) for time series 1. For these missing values I would like to have 0 as a placeholder in the extracted matrix.
returns = randn(2,750)';
T = size(returns,1);
nIndices = 2;
model = arima('AR', NaN, 'Variance', gjr(1,1));
residuals = NaN(T, nIndices);
variances = NaN(T, nIndices);
fit = cell(nIndices,1);
options = optimset('fmincon');
options = optimset(options, 'Display' , 'off', 'Diagnostics', 'off', ...
'Algorithm', 'sqp', 'TolCon' , 1e-7);
for i = 1:nIndices
fit{i} = estimate(model, returns(:,i), 'print', false, 'options', options);
[residuals(:,i), variances(:,i)] = infer(fit{i}, returns(:,i));
end
for i=1:nIndices
AA_beta(:,i) = cell2mat(fit{i}.AR)';
AA_GARCH(:,i) = cell2mat(fit{i}.Variance.GARCH)';
AA_ARCH(:,i) = cell2mat(fit{i}.Variance.ARCH)';
AA_Leverage(:,i) = cell2mat(fit{i}.Variance.Leverage)';
end;
I have some general things to say about the code, but first a solution to your problem:
You can put a simple if/else structure in your loop to handle the case of an empty array:
for ind1=1:nIndices
AA_beta(:,ind1) = cell2mat(fit{ind1}.AR)'; %//'
%// GARCH
if isempty(cell2mat(fit{ind1}.Variance.GARCH)') %//'
AA_GARCH(1,ind1) = 0;
else
AA_GARCH(:,ind1) = cell2mat(fit{ind1}.Variance.GARCH)'; %//'
end
%// ARCH (same exact code, should probably be exported to a function)
if isempty(cell2mat(fit{ind1}.Variance.ARCH)') %//'
AA_ARCH(1,ind1) = 0;
else
AA_ARCH(:,ind1) = cell2mat(fit{ind1}.Variance.ARCH)'; %//'
end
AA_Leverage(:,ind1) = cell2mat(fit{ind1}.Variance.Leverage)'; %//'
end;
Side note: I initially tried something like this: soz = #(A)isempty(A)*0+~isempty(A)*A; as an inline replacement for the if/else, but it turns out that MATLAB doesn't handle [] + 0 the way I wanted (it results in [] instead of 0; unlike other languages like JS).
As for the other things I have to say:
I am a firm supporter of the notion that one shouldn't use i,j as loop indices, as this may cause compatibility problems in some cases where complex numbers are involved (e.g. if you loop index is i then 1*i now refers to the loop index instead of to the square root of -1).
Part of your problem was that the arrays you were writing into weren't preallocated - which also means the correct datatype was unknown to MATLAB at the time of their creation. Besides the obvious performance hit this entails, it could also result in errors like the one you encountered here. If, for example, you used cells for AA_beta etc. then they could contain empty values, which you could later replace with whichever placeholder your heart desired using a combination of cellfun and isempty. Bottom line: lint (aka the colorful square on the top right of the editor window) is your friend - don't ignore it :)
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.