Transpose a matrix in Python [duplicate]

Solution 1:

You can use zip with * to get transpose of a matrix:

>>> A = [[ 1, 2, 3],[ 4, 5, 6]]
>>> zip(*A)
[(1, 4), (2, 5), (3, 6)]
>>> lis  = [[1,2,3], 
... [4,5,6],
... [7,8,9]]
>>> zip(*lis)
[(1, 4, 7), (2, 5, 8), (3, 6, 9)]

If you want the returned list to be a list of lists:

>>> [list(x) for x in zip(*lis)]
[[1, 4, 7], [2, 5, 8], [3, 6, 9]]
#or
>>> map(list, zip(*lis))
[[1, 4, 7], [2, 5, 8], [3, 6, 9]]

Solution 2:

Is there a prize for being lazy and using the transpose function of NumPy arrays? ;)

import numpy as np

a = np.array([(1,2,3), (4,5,6)])

b = a.transpose()

Solution 3:

If we wanted to return the same matrix we would write:

return [[ m[row][col] for col in range(0,width) ] for row in range(0,height) ]

What this does is it iterates over a matrix m by going through each row and returning each element in each column. So the order would be like:

[[1,2,3],
[4,5,6],
[7,8,9]]

Now for question 3, we instead want to go column by column, returning each element in each row. So the order would be like:

[[1,4,7],
[2,5,8],
[3,6,9]]

Therefore just switch the order in which we iterate:

return [[ m[row][col] for row in range(0,height) ] for col in range(0,width) ]