How do I make cell filttering

Dear all, I facing problem with cell array. For example; I have cell
a= { [1 2 3 4] [7 4] [ 3 5] [ 3 6 7 4]}
How I know each array repeat it number 4 in it and take first element in array?
Results1= {[1 2 3 4] [7 4] [ 3 6 7 4]};
Results2= [ 1 7 3];
Thanks…

3 个评论

Jan
Jan 2017-1-27
编辑:Jan 2017-1-27
Please do not bump a thread without providing new information. Getting no answer is a secure indicator of a question, which is not clear.
What does "repeat it number 4 in it and take first element in array" mean? I do not see the relation between this and the shown wanted result. Do you want to find all arrays, which contain the element 4?
Please don't add comments as Answers.
I don't fully understand your question. Why is [3 5] removed for Results1 and why is Results2 not [1 7 3 3]?
Thanks for replying.... [3 5] removed coz does not include number 4. I want only the arrays including number 4 as shown in Results1. and then take the first element from each array including number 4 as shown in Results2. Thanks...

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 采纳的回答

At a guess, you want all the arrays that contain the number 4:
a = {[1 2 3 4] [7 4] [ 3 5] [ 3 6 7 4]}
Results1 = a(cellfun(@(v) ismember(4, v), a))
And the first value of each element of Results1:
Results2 = cellfun(@(v) v(1), Results1)
If cellfun is too complex for you, the same with a loop:
a = {[1 2 3 4] [7 4] [ 3 5] [ 3 6 7 4]}
contains4 = false(size(a));
firstelement = zeros(size(a));
for idx = 1:numel(a)
contains4(idx) = ismember(4, a{idx});
firstelement(idx) = a{idx}(1);
end
Results1 = a(contains4)
Results2 = firstelement(contains4)

1 个评论

Funny. I've posted nearly the same code in the opposite order. Well, I will vote for your code, because I'm convinced it is efficient :-)

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更多回答(2 个)

Jan
Jan 2017-1-27
编辑:Jan 2017-1-27
A bold guess: Get all arrays, which contain the value 4 and copy their first element:
a = {[1 2 3 4], [7 4], [3 5], [3 6 7 4]};
n = numel(a);
found = false(1, n);
Result2 = zeros(1, n); % Pre-allocate
iResult = 0;
for k = 1:n
if any(a{k} == 4)
found(k) = true;
Result2(k) = a{k}(1);
end
end
Result1 = a(found);
Result2 = Result2(1:nnz(found));
If this does what you need, here a compact version:
Result1 = a(cellfun(@(x) any(x==4), a));
Result2 = cellfun(@(x) x(1), Result1);
skysky2000
skysky2000 2017-1-27

0 个投票

Many thanks Guillaume and Jan . its work. Please, Another question; I have one vector (c) and one cell array (H), how can i separate cells from H depend on c vector. this is just simple vector( my real vector about 200 elements. c=[4 8 7] H= {[ 2 3 4] [5 6 3 4][[5 7] [5 6 8] [ 2 1 9 7] [9 7]}
results should be : Re1= cells including first elements in c (4). Re2= cells including second element in c (8). Re3= cells including third element in c (7).
Re1={[ 2 3 4] [5 6 3 4]}; Re2={[5 6 8]}; Re3={[ 2 1 9 7] [9 7]} . . . .
Many thanks....

1 个评论

Jan
Jan 2017-1-27
编辑:Jan 2017-1-27
You can create a loop around the given code.

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