Transpose a matrix in Python

To transpose a matrix, you turn its rows into columns: the first row becomes the first column, and so on. A matrix with 2 rows and 3 columns becomes one with 3 rows and 2 columns.

The program

matrix = [
    [1, 2, 3],
    [4, 5, 6],
]

rows = len(matrix)
cols = len(matrix[0])

transposed = []
for c in range(cols):
    new_row = []
    for r in range(rows):
        new_row.append(matrix[r][c])
    transposed.append(new_row)

for row in transposed:
    print(row)

[1, 4]
[2, 5]
[3, 6]

How it works

• matrix[r][c] is the item in row r, column c.
• The outer loop goes through the columns. For each column, the inner loop collects the item from every row, and that list becomes one new row.
• So the new item at [c][r] is the old item at [r][c].
• This code assumes every row has the same length, as rows in a matrix should. It also needs at least one row: for an empty list, matrix[0] raises an IndexError.

Another way: zip(*matrix)

*matrix passes each row to zip() as a separate argument. zip() then takes the first item of every row, then the second item of every row, and so on. Those groups are exactly the columns:

matrix = [
    [1, 2, 3],
    [4, 5, 6],
]

transposed = [list(column) for column in zip(*matrix)]
print(transposed)
print([list(column) for column in zip(*[])])

[[1, 4], [2, 5], [3, 6]]
[]

zip() gives tuples, so list(column) turns each one back into a list. An empty matrix gives an empty result here, with no error. If rows have different lengths, zip() stops at the shortest row without any error, so extra items are dropped.