Finding row indexes in array

2017 4 4
2 个回答
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更新时间：2017 4 17
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0 个投票

I have an array
a=[0 0 1 0 ; 1 2 3 0; 1 0 3 4; 0 2 0 0 ]
& cell array like
b{1,1}=[4;3]
b{2,1}=[3;1;4]
I want to find elements of cell array in a. For example: for b{2,1} 1st, 3rd and 4th row of a should be displayed and 2nd should be zero.
Thanks in advance

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 采纳的回答

Andrei Bobrov
Andrei Bobrov 2017-4-4

0 个投票

a=[0 0 1 0 ; 1 2 3 0; 1 0 3 4; 0 2 0 0 ];
b{1,1}=[4;3];
b{2,1}=[3;1;4];
v = 1:size(a,1);
t = cell2mat(cellfun(@(x)ismember(v,x),b,'un',0))';
out = a.*permute(t,[1,3,2]);

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lucksBi
lucksBi 2017-4-4
Thank you so much. This is exactly what i want. Can I get final result in cell array other than 3D array?
Andrei Bobrov
Andrei Bobrov 2017-4-4
Yes.
a=[0 0 1 0 ; 1 2 3 0; 1 0 3 4; 0 2 0 0 ];
b{1,1}=[4;3];
b{2,1}=[3;1;4];
out = cellfun(@(x)a.*ismember((1:size(a,1))',x),b,'un',0);
lucksBi
lucksBi 2017-4-5
It is working fine on small matrix e.g i tried on 6x6 matrix size it is giving accurate results. But when i tried it on matrix of size 943x943, it gives following error. ??? Error using ==> times Matrix dimensions must agree.
Error in ==> @(x)a.*ismember((1:size(a,1))',x) Error in ==> file2 at 22 out=cellfun(@(x)a.*ismember((1:size(a,1))',x),b,'un',0);
can you please help on this.
Andrei Bobrov
Andrei Bobrov 2017-4-5
Please attach your data that gives an error.
lucksBi
lucksBi 2017-4-5
here are .mat files of a & b matrices
lucksBi
lucksBi 2017-4-17
hey can you please help on this matter? i am unable to fix this issue with with given mat files.

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更多回答（1 个）

Thorsten
Thorsten 2017-4-4

0 个投票

idx = cellfun(@(x) any(ismember(a, x)), b, 'UniformOutput', false);
rows = cellfun(@(x) a(:, x), idx, 'UniformOutput', false);
Or in one line:
rows = cellfun(@(x) a(:, any(ismember(a, x))), b, 'UniformOutput', false);

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lucksBi
lucksBi 2017-4-4
Thanks Alot. But it gives 4x2 array for b{1,1}, (changes no of columns in each cell) i want a 4x4 array with elements of rows that are member of b and other rows as zero.
Thorsten
Thorsten 2017-4-4
Like this?
idx = cellfun(@(x) any(ismember(a, x)), b, 'UniformOutput', false);
a0 = zeros(size(a));
for i = 1:numel(idx)
rows{i} = a0;
rows{i}(:,idx{i}) = a(:,idx{i});
end
lucksBi
lucksBi 2017-4-4
result should be like: result{1,1}= [0,0,0,0;0,0,0,0;1,0,3,4;0,2,0,0] result{1,2}= [0,0,1,0;0,0,0,0;1,0,3,4;0,2,0,0]

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