combine two double cells into 1
이전 댓글 표시
i have two Double[] cells with size of 1*n; every element is has some numbers in it . like :
C1{1} = [1 2 3 4]
C2{1} = [5 2 3 7]
.
.
.
C1{n} = [1 2]
C2{n} = [4 5]
how can i have a cell like this :
C{1} = [1 2 3 4 5 7]
.
.
.
C{n} = [1 2 4 5]
i`m totally new to matlab and sorry if its already been answered , but i search a lot and couldn`t find the answer .
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추가 답변 (3개)
If you are in doubt, start with a loop:
C1 = {[1 2 3 4], [1 2]};
C2 = {[5 2 3 7], [4 5]};
C = cell(size(C1));
for iC = 1:numel(C)
C{iC} = unique([C1{iC}, C2{iC}]);
end
댓글 수: 3
+1 Note that the example has only the unique values (so is not just the concatenation):
C{iC} = unique([C1{iC}, C2{iC}]);
Jan
2017년 8월 6일
Thanks, Stephen. I hit the Submit button too early.
Hamid Salari
2017년 8월 6일
Star Strider
2017년 8월 6일
편집: Star Strider
2017년 8월 6일
This works, and should work with your entire ‘C1’ and ‘C2’:
C1{1} = [1 2 3 4];
C1{2} = [1 2];
C2{1} = [5 2 3 7];
C2{2} = [4 5];
C3 = cellfun(@union, C1, C2, 'Uni',0);
C3{1} % Display Result
C3{2} % Display Result
ans =
1 2 3 4 5 7
ans =
1 2 4 5
EDIT — Added output (the ‘ans’ variables).
dpb
2017년 8월 6일
That's a little towards the advanced-beginner side... :)
>> C=cellfun(@(c1,c2) unique([c1 c2]),C1,C2,'uniform',0);
>> C{:}
C =
1 2 3 4 5 7
C =
1 2 4 5
>>
The @ symbol is defining an "anonymous function", with two arguments; cellfun passes the content of its two cell array arguments to the defined function cell-by-cell and unique does what it sounds like which is to return the unique values in its argument; the [] simply concatenate the two vectors.
댓글 수: 3
Hamid Salari
2017년 8월 6일
Jan
2017년 8월 7일
Anonymous functions are very handy, but tend to be slow e.g. in cellfun.
dpb
2017년 8월 7일
Note Star S's use of UNION over UNIQUE, though...saves a step of the concatenation.
I've never done specific timing; where the syntax is easy I'll write the CELLFUN solution as above and only if it turns out to be a bottleneck worry about what generally would be small differences in run time. Then again, I'm not actively consulting any longer so don't in general ever have large datasets where it would ever be an issue so "caveat emptor" reigneth, methinks. Keep Jan's note in mind if you're still waiting next week for the prompt to come back... :)
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