matlab: find and replace elements of matrices within a cell array

2
votes

I have a cell array that contains multiple matrices of different size. I'd like to find & replace all elements of the matrices by a condition, e.g. replace all 1's with 0's. I found a temporary solution from find and replace values in cell array, but it seems way more complicated that it should be:

Example:

A = {[1 2;3 4] [1 2 3;4 5 6;7 8 9]}
replacement = 1:9;
replacement(replacement==1)=0;
A = cellfun(@(x) replacement(x) ,A,'UniformOutput',false)
A{:}

ans =

 0     2
 3     4

ans =

 0     2     3
 4     5     6
 7     8     9

So it works but I feel like this should be doable without first specifying a list of replacement values and then "exchanging" all elements. (I have to do this a lot and with more complex conditions). Any suggestions?

2
matlabcell-arraymatrix-indexing

2 Answers

2
votes

One approach would be with elementwise-multiplication with the mask of such 1s -

cellfun(@(x) (x~=1).*x, A, 'uni',0)

Sample run -

>> celldisp(A)  % Input cell array
A{1} =
     3     2
     1     4
A{2} =
     7     1     3
     4     5     1
     7     8     9
>> C = cellfun(@(x) (x~=1).*x, A, 'uni',0);
>> celldisp(C)
C{1} =
     3     2
     0     4
C{2} =
     7     0     3
     4     5     0
     7     8     9

Generic case : To make it generic that could replace any number by any other number, we need a slight modification, like so -

function out = replace_cell_array(A, oldnum, newnum)

out = cellfun(@(x) x+(x==oldnum).*(newnum-oldnum), A, 'uni',0);

Sample runs -

>> A = {[1 2;3 5] [1 2 3;5 5 3;7 8 9]}; % Input cell array
>> celldisp(A)
A{1} =
     1     2
     3     5
A{2} =
     1     2     3
     5     5     3
     7     8     9
>> celldisp(replace_cell_array(A,1,0)) % replace 1s with 0s
ans{1} =
     0     2
     3     5
ans{2} =
     0     2     3
     5     5     3
     7     8     9
>> celldisp(replace_cell_array(A,3,4)) % replace 3s with 4s
ans{1} =
     1     2
     4     5
ans{2} =
     1     2     4
     5     5     4
     7     8     9
1
votes

Using cellfun with an anonymous function has the limitation that an anonymous function can only contain a single statement. So it cannot assign a value to an entry of the matrix (unless you resort to ugly, not recommended tricks).

To avoid that you can use a for loop. This is not necessarily slower than cellfun. In fact, it may be a little faster, and is arguably more readable:

A = {[1 2;3 4] [1 2 3;4 5 6;7 8 9]}
repl_source = 1;
repl_target = 0;
for k = 1:numel(A)
    A{k}(A{k}==repl_source) = repl_target;
end