# Remove values from cell array based on condition

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lucksBi 2017 年 8 月 16 日
コメント済み: lucksBi 2017 年 8 月 16 日
Hey
I have two cell arrays like this:
A= {[2,3,4,5,6]; [1,3,4,5,6,7,8] }
B= {NaN, 1,1,-0.9,0.8,[],[]; NaN, NaN, 0.9,-1,NaN,0.8,0.2}
I want to find indexes of values less than 0 and of NaN values in B like here in first row of B we will get 1 (for NaN) and 4 for (values less than zero) and for these indexes i want to remove values placed in A like value at index 1 in A is 2 so it will be removed and also 5 will be removed. Same for row 2. New matrix A will look like this:
NewA= {[3,4,6]; [4,7,8]}

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### 採用された回答

Guillaume 2017 年 8 月 16 日
I can't help but feel that you're doing something very wrong if you want to do what you're asking. Your problem is very ill-defined: your A is a 2x1 cell array, your B is a 2x7 cell array, so you're asking to match column indices of cell array B with column indices of the contents of column 1 of A. That does not sound right.
Furthermore, what happens if a NaN or negative number occurs at an index larger than the number of columns in the corresponding A cell?
One way to do what you want:
tokeep = num2cell(cellfun(@(c) isempty(c) || (~isnan(c) && c>=0), B), 2);
newA = cellfun(@(a, keep) a(keep(1:numel(a))), A, tokeep, 'UniformOutput', false)
This will do what you want but as said, it does sound like something in your logic is broken.
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lucksBi 2017 年 8 月 16 日
No values in B are also based on A using some mathematics so logic is correct in this case. Thanks for helping.

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### その他の回答 (1 件)

José-Luis 2017 年 8 月 16 日
A= {[2,3,4,5,6]; [1,3,4,5,6,7,8] }
B= {NaN, 1,1,-0.9,0.8,[],[]; NaN, NaN, 0.9,-1,NaN,0.8,0.2}
result = cell(size(A));
dummy = cellfun(@(x) ~isempty(x) && ~isnan(x) && x >= 0 ,B,'UniformOutput', false);
for ii = 1:size(A,1)
result{ii} = A{ii}([dummy{ii,:}])
end
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lucksBi 2017 年 8 月 16 日
This also works. Thankyou so much for your answer.

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