Given A=[1 3 5 7 9] and B=[2 4 6 8], how can I create C=[1 2 3 4 5 6 7 8 9]?

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Youssef  Khmou
Youssef Khmou 2015년 1월 15일
편집: Youssef Khmou 2015년 1월 15일

0 개 추천

This question is general due to the variation of array dimensions, however for a particular case you described, vectors A and B can be mixed by single loop, so the following scheme is valid only when dim(A)=dim(B)+1 as in the example :
A=[1 3 5 7 9];
B=[2 4 6 8];
n=min(length(A),length(B));
C=[];
for t=1:n
C=[C A(t) B(t)];
end
C=[C A(end)];

댓글 수: 3

Alex Strongholm
Alex Strongholm 2015년 1월 15일
That's it, thank you very much
Stephen23
Stephen23 2015년 1월 15일
편집: Stephen23 2015년 1월 18일
This answer is very poor use of MATLAB.
The use of a for loop and concatenating scalar values onto the end of with every iteration is poor coding practice in MATLAB. If the arrays are large, then this will be slow as MATLAB keeps expanding the array and copying it to new memory. One solution is to preallocate the array.
For a much neater and simpler solution see my answer below.

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추가 답변 (1개)

Stephen23
Stephen23 2015년 1월 15일
편집: Stephen23 2015년 1월 16일

1 개 추천

This can be done simply using indexing, without any loops:
>> A = [1,3,5,7,9];
>> B = [2,4,6,8];
>> C(1:2:2*numel(A)) = A;
>> C(2:2:end) = B
C =
1 2 3 4 5 6 7 8 9
This solution also assumes that numel(A)==numel(B)+1.
Most importantly, for larger arrays this code will be much faster than the accepted solution, so it is the most universal solution.

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도움말 센터File Exchange에서 Loops and Conditional Statements에 대해 자세히 알아보기

질문:

2015년 1월 15일

편집:

2015년 1월 18일

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