“% each element of ROI, tar and cor is 128×128
for i=1:47*2
if i<=48
ROI{i} = logical(cor{i}) ;
else
ROI{i} = logical(tar{i-47}) ;
end
end"

this seems right.

Look at the doc for cellfun. The example for covariance is very similar to what you are trying to do. The function you want is called mtimes (for matrix multiply).

–Loren

Cell1={[1 2 3; 4 5 6; 7 8 9]...
[9 8 7; 6 5 4; 3 2 1]}; 

Cell2={[1 3 5; 2 4 6; 1 5 9]...
[9 5 1; 6 4 2; 5 3 1]}; 

I want to multiply the contents of Cell1 and Cell2 to produce a third cell array that contains the product of the respective matrices in Cell1 and Cell2. So I should end up with:

Cell3={[8 26 44; 20 62 104; 32 98 164]…
[164 98 32; 104 62 20; 44 26 8]};

I can’t seem to figure out how to use cellfun to accomplish this.

Thanks,
Rima

I know how to do it with cellfun if your original data are in a cell array already.

a = num2cell(1:10);
b = cellfun(@(x)[sin(x), cos(x)], num2cell(a), 'UniformOutput',false)

If they are the way you say in your post, you can instead use arrayfun.

a = num2cell(1:10);
b = arrayfun(@(x)[sin(x), cos(x)], a, 'UniformOutput',false)

I am not sure either of these is faster however, especially when returning 3×3 matrices. Another possibility is to vectorize it and return 3x3xlength(vector) in a 3-D array. You might be able to vectorize that without using cellfun or arrayfun (perhaps bsxfun could be helpful?) but you didn’t say what the relationship is between the input A and the output B.

–Loren

a = 1:10;

I’d like to create cell array b, where each cell contains a vector [sin(a) cos(a)] for each value of a. So you’d have something like

b={[sin(1) cos(1)] [sin(2) cos(2)] [sin(3) cos(3)]...[sin(10) cos(10)]};

Is there a way to do this with cellfun without having to use a for loop? The size of a can be a vector upto 200,000 values, and b will actually be a cell array of 3×3 matrices for each of those values.

thanks,
Rima

uniformoutput true means that cellfun puts the output into a regular array (a logical vector in your second example). False means that either the size or shape of each output of cellfun may differ, so instead of trying to put the outputs into a single regular array, the output from each individual cell goes into its own cell in a new cell array.

–Loren

ans =

[1] [0] [0]

>> cellfun(@(x)~ischar(x), x,’UniformOutput’,true)

ans =

1 0 0

I would transpose the first test array, run unique on each column and fill the end values or each column with []. I would make sure I preallocated space for the result, and then I would transpose the array back.

–Loren

test = {[1,2,3],[5,6],[2,3],[7,6],[2,3],[4,4];
[1,2,4],[1,1],[2,3],[4,5],[3,4],[];
[1,3,4],[2,3],[4,4],[4,4],[5,3],[2,3];
}

Now i want that each entry in a row is unique and that all empty cells located at the end of a row. In that example it should afterwards look like :

test = {[1,2,3],[5,6],[2,3],[7,6],[4,4],[];
[1,2,4],[1,1],[2,3],[4,5],[3,4],[];
[1,3,4],[2,3],[4,4],[5,3],[],[];
}

I have found a lot of usefull commands but it doesn´t work.

I hope you can help me.

Lirave

I don’t understand why the data you have is in cells in the first place and what the range of valuesis. Why aren’t they in a numeric array (3-D, i.e., 128x128xP)? I think they’d be much easier to deal with there. Is P 47^2?

Second, why do you want to write this without a for loop? You can generally write most cell operations using cellfun instead, but it’s not necessarily faster, and sometimes slower at the moment.

Why not have 2 for loops, one from 1-48, and one from 49-94? That would cover the cases in your if statement, I believe. Or two correctly indexed expressions using cellfun?

I haven’t tried it, but something like this might work. I think I have the indexing a bit wrong – I didn’t look at that part carefully.

anon = @(x) logical(x);
ROI = cellfun(anon,cor(1:48),'UniformOutput',false);
ROI(49:94) = cellfun(anon, tar(2:47),'UniformOutput',false);

–Loren