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 am 15 Jan. 2015
Bearbeitet: Youssef Khmou am 15 Jan. 2015

0 Stimmen

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)];

3 Kommentare

Alex Strongholm
Alex Strongholm am 15 Jan. 2015
That's it, thank you very much
Stephen23
Stephen23 am 15 Jan. 2015
Bearbeitet: Stephen23 am 18 Jan. 2015
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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Weitere Antworten (1)

Stephen23
Stephen23 am 15 Jan. 2015
Bearbeitet: Stephen23 am 16 Jan. 2015

1 Stimme

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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