index position of the 1

Hi,

I have the following cell array

TEST = [1] [] [] [0] []

I want to get the index position of the 1 not the 0. thanks :) !

Answer 1

[Walter Roberson]

find( cellfun(@(C) numel(C) == 1 && C == 1, TEST) )

In the special case where each entry is only empty or a scalar, then

find( cellfun(@(C) C == 1, TEST) )

If you want to find a 1 anywhere in the cell:

find( cellfun(@(C) any(C == 1), TEST) )

Answer 2

[KSSV]

TEST = {[1] [] [] [0] []} ;
% index position of 1
idx = find([TEST{:}] == 1)

[Walter Roberson]

If you switch to

TEST = {[0] [] [] [1] []} ;

then [TEST{:}] would be [0 1] and you would return an index of 2, but the actual index should be 4.

Answer 3

[khaled hassan]

thank you a lot :) !!!!!!!!!!!!! it worked perfectly