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yakov
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Switch columns in a matrix without a loop

Asked by yakov
on 13 Apr 2013
Latest activity Commented on by sami elahj on 10 Jan 2016
How can I switch specific columns in a matrix (for example every 4th column), in one command without using a loop? Is it even possible?

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2 Answers

Answer by Florian
on 13 Apr 2013

Hi Yakov,
please specify, since in a matrix with 10 columns you will have for 3 columns at index 3,6,9 but only 2 columns with index mod 5, namely 5, 10.
Hence could you please specify exactly which columns you want to switch. Or do you want to mix all columns randomly?
Regards, Florian

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Hi Florian, for example, if I choose to switch every third column, in a matrix of 10 columns I'd want to switch the 3,6,9 columns only, and if I choose to switch every fourth column I'd want to switch the 4 and 8 columns. I want to switch them with a different column vector of my choosing, but all of them with the same vector.

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Answer by Ahmed A. Selman on 13 Apr 2013

for exchanging two columns 3 - 9 :
A = randi(20,4,10); % A random 4 by 10 matrix of integers
A(1,:)=[1 2 3 4 5 6 7 8 9 10]; % the first row only is replaced for clarity
B=A; % the new matrix, same as A for now
B(:,3)=A(:,9); % puts 3 of B with 9 of A
B(:,9)=A(:,3); % puts 9 B with 3 A
B has columns 3 and 9 exchanged with A 9 and 3, and A is intact. A similar method to exchange columns 3 - 6 - 9 - 3. So if you want to use same A, use A=B.
To exchange rows use B(row_no,:)=A(new_row,:)..etc

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