error in returning widths over iterated findpeaks

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Teddy Fisher
Teddy Fisher am 15 Jan. 2020
Beantwortet: Raunak Gupta am 23 Jan. 2020
Hi,
I am trying to use findpeaks over a matrix, x, column by column. I am using a for loop as below:
for k=1:size(x,2)
[P{k},locs{k},widths,proms{k}]=findpeaks(x(:,k))
end
Everything is working fine except for returning widths. P, locs and proms all return as cell arrays with all the data for the peaks in each column. However, widths only returns the widths of the peaks for the last column. When I try to edit it by making widths return as a cell array:
for k=1:size(x,2)
[P{k},locs{k},widths{k},proms{k}]=findpeaks(x(:,k))
end
I get the error message "Cell contents assignment to a non-cell array object"
Is there a way to get this function to return all the widths of all the peaks for each column?
  2 Kommentare
Image Analyst
Image Analyst am 15 Jan. 2020
Bearbeitet: Image Analyst am 15 Jan. 2020
Can you make it easy for us to help you by attaching your x in a .mat file with the paper clip icon?
save('answers.mat', 'x')
What I tried just worked fine:
x = 100 * rand(20, 5);
for k=1:size(x,2)
fprintf('\n============================\nGetting data for column %d...\n', k);
[P{k},locs{k},widths{k},proms{k}]=findpeaks(x(:,k))
end
Teddy Fisher
Teddy Fisher am 15 Jan. 2020
Yes here it is attached.
I'm not sure what that line you added in does, could you explain that a bit?

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Raunak Gupta
Raunak Gupta am 23 Jan. 2020
Hi,
I tried with the Release R2017b that is mentioned in question and with the latest Release that is R2019b but didn’t find any issues while running the code. Still I guess you may try pre declare the variables that are returned from findpeaks as below.
P = cell(1,size(deltaFoFCopy,2));
locs = cell(1,size(deltaFoFCopy,2));
widths = cell(1,size(deltaFoFCopy,2));
proms = cell(1,size(deltaFoFCopy,2));
Also, from the data file I can see that some columns are entirely zero so for those only empty vectors will return. Therefore, for some indexes in cell array the value will be empty vectors.

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