index position of the 1
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Hi,
I have the following cell array
TEST = [1] [] [] [0] []
I want to get the index position of the 1 not the 0. thanks :) !
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採用された回答
Walter Roberson
2016 年 11 月 7 日
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) )
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その他の回答 (2 件)
KSSV
2016 年 11 月 7 日
TEST = {[1] [] [] [0] []} ;
% index position of 1
idx = find([TEST{:}] == 1)
1 件のコメント
Walter Roberson
2016 年 11 月 7 日
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.
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