# How can I make following part of code without for loop?

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Shamsi Musayev on 13 May 2020
Commented: Rik on 13 May 2020
How can I make following part of code without for loop?:
for i=1:n_values
pre_matrix(i,J{i})=0
end
J{i} is a cell array that each cell includes some numbers that I want to change those elements in i-th row of pre_matrix. Each cell has different numbers and size of cells are different.
pre_matrix is very huge matrix, therefore it takes so long time to create matrix everytime. Is it possible to do same work without for loop?

Kelly Kearney on 13 May 2020
I'm having a hard time finding a solution that's faster than the original loop...
% Build test data
n_values = 10000;
nc = 100;
pre_matrix = rand(n_values,nc);
sz = size(pre_matrix);
J = cell(1,n_values);
for ii = 1:n_values
n0 = randi(nc,1);
J{ii} = randi(nc,1,n0);
end
% The original loop
tic
for i=1:n_values
pre_matrix(i,J{i})=0;
end
toc
% Indexing directly, but with loop to get row indices (Rik's suggestion)
tic
I = cell(size(J));
for ii = 1:n_values
I{ii} = ii.*ones(size(J{ii}));
end
idx = sub2ind(sz, cat(2, I{:}), cat(2, J{:}));
pre_matrix(idx) = 0;
toc
% Indexing directly, no explicit loops
tic
I = cellfun(@(a,b) ones(size(a)).*b, J, num2cell(1:n_values), 'uni', 0);
idx = sub2ind(sz, cat(2, I{:}), cat(2, J{:}));
pre_matrix(idx) = 0;
toc
Results:
Elapsed time is 0.017597 seconds. <- loop
Elapsed time is 0.047797 seconds. <- indexing with loop
Elapsed time is 0.119106 seconds. <- indexing with cellfun
My guess here is perhaps the culprit is that missing semicolon in the loop? If you're printing out the huge matrix every iteration, that's definitely going to take some time! But the loop itself is pretty fast.
##### 2 CommentsShowHide 1 older comment
Rik on 13 May 2020
Good catch. I guess this is why you want mlint to give you warnings about missing semicolons.

Rik on 13 May 2020
Your issue is probably not with the loop itself, but with the modifications to the large matrix. You should probably generate a list of indices and do the modification in one go.
sz=size(pre_matrix);
ind_list=cell(1,n_values);
for i=1:n_values
ind_list{i}=sub2ind(sz,i*ones(size(J{i})),J{i});
end
ind_list=cell2mat(ind_list);%untested code, check this line
pre_matrix(ind_list)=0;
Shamsi Musayev on 13 May 2020
Yeah, it was the problem. For each step of the for loop matrix was modified and new matrix was created, hence, it took so long time. Your code worked very well. Thanks :)