i have a cell array cA and vector V with values
cA = {[11,21,3;14,5,63], [11,22], [33;95;66;7]};
V = [14 22 33 66];
i wanted to find which all values in V are there in cA and its position
output = { [ 0 0 0 1 0 0], [1 2], [3 0 4 0] }

 채택된 답변

Andrei Bobrov
Andrei Bobrov 2020년 3월 17일
편집: Andrei Bobrov 2020년 3월 17일

1 개 추천

out = cell(size(cA));
for i = 1:numel(out)
x = cA{i}';
[~,out{i}] = ismember(x(:)',V);
end
or with cellfun
[~,out] = cellfun(@(x)ismember(reshape(x',1,[]),V),cA,'un',0);

추가 답변 (1개)

Ahmed Anas
Ahmed Anas 2020년 3월 17일

0 개 추천

Dear, Always use matrix instead of cell arrays, working with matrix is really easy as compared to cell arrays. In the code below, i have split cA matrix into three small matrices and then checked the entities which are present in vector V. Afterwards compile the entire output matrix.
%%Splitting cA matrix into three consistent matrices
cA1 = [11,21,3;14,5,63]
cA2 = [11,22]
cA3 = [33;95;66;7]
V = [14 22 33 66];
%%Checking for entities of cA1
[Lia,Locb1] = ismember(cA1,V)
reshape(Locb1,1,[])
transpose(Locb1)
Locb1.'
Locb1(:)
Locb1=reshape(Locb1.',1,[])
%%Checking for entities of cA2
[Lia,Locb2] = ismember(cA2,V)
Locb2=reshape(Locb2,1,[])
%%Checking for entities of cA3
[Lia,Locb3] = ismember(cA3,V)
Locb3=reshape(Locb3,1,[])
Output=cell2mat({Locb1,Locb2,Locb3})

댓글 수: 2

Elysi Cochin
Elysi Cochin 2020년 3월 17일
편집: Elysi Cochin 2020년 3월 17일
the dimension of cell array changes. so how to keep the code unaltered for any dimension
sometimes
cA = {[11,22,3;44], [14,22]};
V = [14 22 33 66];
output = { [ 0 2 0 0 ], [1 2] }
I am unable to understand [11 22 3;44] in cA = {[11,22,3;44], [14,22]};
Its dimensions should be consistent

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카테고리

도움말 센터File Exchange에서 Matrices and Arrays에 대해 자세히 알아보기

질문:

2020년 3월 17일

편집:

2020년 3월 17일

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