Howto transpose multidimensional array in place

Question

how to transpose a 2D matrix in place?

@aaa: Transposing a non-square matrix in place does not make a lot of sense. — Jens, Apr 21 '10 at 07:29
Though depending on the representation (for example using a single array of N*M dimensions) it could be done. — Matthieu M., Apr 21 '10 at 13:52
This was pretty much "beaten to death" in [this post](http://stackoverflow.com/questions/3062974/in-place-permutation-of-a-array-follows-this-rule) — NealB, Nov 08 '11 at 20:39
@Jens It makes a lot of sense if you store all matrix data in one single array. — Waruyama, Jan 15 '20 at 10:15

score 7 · Answer 1 · answered Apr 21 '10 at 21:37

7

Wikipedia had an article In-place matrix transposition. The article covers non-square matrices.

http://en.wikipedia.org/wiki/In-place_matrix_transposition

answered Apr 21 '10 at 21:37

dlb

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score 6 · Answer 2 · answered Apr 21 '10 at 06:23

6

for (int i=0; i<n; i++) {
  for (int j=0; j<i; j++) {
    temp = a[i][j];
    a[i][j] = a[j][i];
    a[j][i] = temp;
  }
}

answered Apr 21 '10 at 06:23

Simon Nickerson

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WARNING! This is only correct for square arrays. See @dlb's Wikipedia link below for a non-square matrix implementation. – Mayank Apr 07 '14 at 06:04

score 4 · Answer 3 · answered Apr 21 '10 at 06:23

4

You have not specified a language, but generally, what you do is:

let a be your array.
for each i,j with i<j switch a[i,j] with a[j,i]

answered Apr 21 '10 at 06:23

Jens

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This too is only correct for a square matrix. Handling non-square matrices is surprisingly difficult. – Waruyama Jan 15 '20 at 10:14

codaddict · Answer 4 · 2014-07-09T09:11:44.763

2

To get the transpose of a square matrix we need to consider elements above the main diagonal or below it and swap each with its reflection along the main diagonal:

for i->0 to N-1
 for j->i+1 to N-1
  swap matrix[i][j] with matrix[j][i]

edited Jul 09 '14 at 09:11

answered Apr 21 '10 at 06:25

codaddict

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score 1 · Answer 5 · edited May 07 '15 at 06:58

1

for(i=0;i<N;i++)
  for(j=0;j<N;j++)
    if(i!=j && j>i)
      {
        temp=a[i][j];
        a[i][j]=a[j][i];
        a[j][i]=temp;
      }

(N is the size of your array)

edited May 07 '15 at 06:58

Maroun

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answered May 07 '15 at 06:12

dimmat

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Biggum · Answer 6 · 2013-08-26T12:56:20.210

0

in c#

string[,] Value;
//fill Value

//create transposed array
ValueAux = new string[Value.GetLength(1),Value.GetLength(0)];
for (i = 0; i < Value.GetLength(0); i++)
{
  for (j = 0; j < Value.GetLength(1); j++)
  {
    Valueaux[j, i] = Value[i, j];
  }
}

The result is in ValueAux

edited Aug 26 '13 at 12:56

answered Aug 23 '13 at 11:39

Biggum

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score 0 · Answer 7 · answered Apr 21 '10 at 06:38

0

Why bother ? Just swap indices in any access statement.

answered Apr 21 '10 at 06:38

High Performance Mark

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There is a serious performance penalty for accessing 2D arrays in the "wrong" order - it's often better to pay the price of a transpose in order to get the benefits of contiguous memory access (unit stride). – Paul R Apr 21 '10 at 07:29

score -1 · Answer 8 · answered Apr 27 '13 at 03:22

-1

This seems to work well:

function transpose(a)
{
  return Object.keys(a[0]).map(function (c) { return a.map(function (r) { return r[c]; }); });
}

answered Apr 27 '13 at 03:22

JWally

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Howto transpose multidimensional array in place

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