Column to row vector
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How to extract and convert a column to a row vector so say I want to extract 5;10;9;25 how do I convert that to a row when I extract it?
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Ameer Hamza
2020년 11월 8일
You can use transpose()
x = [5;10;9;25];
y = x.'; % or y = transpose(x)
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Babak
2025년 8월 20일 16:34
Isn't there a function that could do this? Basically keep a row as a row and convert a column vector into a row. Ensuring if x was a row it wouldn't take the `transpose`
Babak
2025년 8월 20일 16:38
편집: DGM
2025년 8월 31일 10:28
x = [5;10;9;25]; % a column vector
reshape(x, 1, length(x)) % a row vector
y = [5 10 9 25]; % a row vector
reshape(y, 1, length(y)) % still a row vector
% create an anonymous function to do the job
f = @(x)reshape(x,1, length(x));
% when applied to the previous vectors, the results are as before
f(x)
f(y) % and another
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DGM
2025년 8월 31일 10:59
Edited to run the code for demonstration purposes; added comments.
Note that there are few cases where length() can be used incautiously without inviting problems. Unless you're making sure that the input is strictly a vector, you're waiting on a confusing error about the wrong number of elements.
If you want to explicitly specify the output size, use numel() instead of length(). The usefulness of the output may be zilch for a reshaped matrix, but at least it avoids the confusing error and returns a vector. Specifying a slack dimension as in @Stephen23's example does the same, and is more succinct.
x = [1 2 3 4].'; % a column vector or row vector
y = reshape(x,1,[]) % to a row vector
x = [1 3 5; 2 4 6]; % oops we didn't constrain the input dimensionality!
y = reshape(x,1,numel(x)) % but at least we got a vector without error
y = reshape(x,1,[]) % same thing, more succinct, easier to read/remember
y = reshape(x,1,length(x)) % you're trying to fit 6 elements in a 1x3 vector
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