()-indexing must appear last in an index expression

Hello, I wrote this code

i=(find(A)~=0); % find indexes that verify this condition
% B  is a cell
B(i) (end:A(i),end:A(i))=0

Matlab displays this error : ()-indexing must appear last in an index expression .

Is there any other solution to do the same operation without working with loops?

Thanks in advance

Edit

Sorry, i wasn't clear.

Here is a better version

    i=(find(A)); % find indexes that verify this condition ( i is an array)
    % B  is a cell containing 160 matrixs
    B{i} (end:-1:A(i),end:-1:A(i))=0

The purpose is to change the indexes from the index find(A) to the end to 0 ( To be simple, I wrote A as a variable, but in reality ,it's if a cropped image hasn't the same size of the windows_size)

When I did these modifications, Matlab displayed : " expected one output from a curly brace or dot indexing expression but there were 148 results"

[Jan]

You forgot to mention the purpose of the command. Because it is no valid Matlab syntax, it is not clear what it should achieve. E.g. "end:A(i)" is most likely empty. If i is an array, "end:A(i)" is not even meaningful.

find(A) replies the indices of the non-zero values of A. Indices are greater than 0 in every case. Then find(A)~=0 is not useful or meaningful.

Please explain, what you want to achieve. Posting failing code does not clarify this.

[emar]

Sorry for not being explicit ! I edited the post

[Jan]

Now i is a [1 x 148] vector. Then B{i} are 148 elements of the cell. A(i) is a vector also. Then I still cannot guess the purpose of end:-1:A(i). You want to change the indices or the contents of B? What exactly should be set to 0?

[emar]

For example , if A(i) is non zero ( and A(i)= windows size - image size(i)), the borders from W(i) ( windows size) till the end of the image=0. Thus the code has to be written like

B{i} (W(i):end,W(i):end)=0

For the command end:-1:W(i), I found the code like that. It was an error.

So My question is what to do so B{i} (W(i):end,W(i):end)=0 is executed for each element of the array i ( without doing a loop on the elements of i )

[Jan]

The first part of this explanation is not clear already:

if A(i) is non zero

i is a vector. What does "A(i) is non zero" mean then? Which A(i)? all() or any or min or max?

Imagine A=[2, 3, 4, 0, 5]. Then i=find(A) replies i=[1,2,3,5]. A(i) is [2,3,4,5] in consequence and "A(i):end" is not useful. B{i} is a list of {B{1}, B{2}, B{3}, B{5}} and you want to set some elements to zero. But what is the meaning of setting (2:end) to 0, if (1:end) was set to 0 already?

Together with "To be simple, I wrote A as a variable, but in reality ,it's if a cropped image hasn't the same size of the windows_size" I'm completely lost. Perhaps Guillaume's answer hit your point. He is correct: There is no reason to implement this without a loop.

Answer 1

[Guillaume]

So My question is what to do ( without doing a loop on the elements of i )

You don't have a choice. There's no magic matlab syntax that can do what you want without a loop. Even arrayfun (which is arguably a loop) cannot help you in this case, since you've got an assignment.

indices = find(...)
for iidx = 1:numel(indices)
   index = indices(iidx);
   B{index}(W(index):end, W(index):end) = 0;
end

Until you've established that the loop is slow by actually timing it with the profile, there's no reason to avoid the loop. If it is a bottleneck, then you may have to resort to mex, or change your algorithm.