Transpose list of lists

Question

Let's take:

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

The result I'm looking for is

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

and not

r = [(1, 4, 7), (2, 5, 8), (3, 6, 9)]

Answer 1

Python 3:

# short circuits at shortest nested list if table is jagged:
list(map(list, zip(*l)))

# discards no data if jagged and fills short nested lists with None
list(map(list, itertools.zip_longest(*l, fillvalue=None)))

Python 2:

map(list, zip(*l))

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

Explanation:

There are two things we need to know to understand what's going on:

1. The signature of zip: zip(*iterables) This means zip expects an arbitrary number of arguments each of which must be iterable. E.g. zip([1, 2], [3, 4], [5, 6]).
2. Unpacked argument lists: Given a sequence of arguments args, f(*args) will call f such that each element in args is a separate positional argument of f.
3. itertools.zip_longest does not discard any data if the number of elements of the nested lists are not the same (homogenous), and instead fills in the shorter nested lists then zips them up.

Coming back to the input from the question l = [[1, 2, 3], [4, 5, 6], [7, 8, 9]], zip(*l) would be equivalent to zip([1, 2, 3], [4, 5, 6], [7, 8, 9]). The rest is just making sure the result is a list of lists instead of a list of tuples.

Answer 2

Equivalently to Jena's solution:

>>> l=[[1,2,3],[4,5,6],[7,8,9]]
>>> [list(i) for i in zip(*l)]
... [[1, 4, 7], [2, 5, 8], [3, 6, 9]]

Answer 3

One way to do it is with NumPy transpose. For a list, a:

>>> import numpy as np
>>> np.array(l).T.tolist()
[[1, 4, 7], [2, 5, 8], [3, 6, 9]]

Or another one without zip (python < 3):

>>> map(list, map(None, *l))
[[1, 4, 7], [2, 5, 8], [3, 6, 9]]

Or for python >= 3:

>>> list(map(lambda *x: list(x), *l))
[[1, 4, 7], [2, 5, 8], [3, 6, 9]]

Answer 4

just for fun, valid rectangles and assuming that m[0] exists

>>> m = [[1,2,3],[4,5,6],[7,8,9]]
>>> [[row[i] for row in m] for i in range(len(m[0]))]
[[1, 4, 7], [2, 5, 8], [3, 6, 9]]

Answer 5

Methods 1 and 2 work in Python 2 or 3, and they work on ragged, rectangular 2D lists. That means the inner lists do not need to have the same lengths as each other (ragged) or as the outer lists (rectangular). The other methods, well, it's complicated.

the setup

import itertools
import six

list_list = [[1,2,3], [4,5,6, 6.1, 6.2, 6.3], [7,8,9]]

---

method 1 — map(), zip_longest()

>>> list(map(list, six.moves.zip_longest(*list_list, fillvalue='-')))
[[1, 4, 7], [2, 5, 8], [3, 6, 9], ['-', 6.1, '-'], ['-', 6.2, '-'], ['-', 6.3, '-']]

six.moves.zip_longest() becomes

• itertools.izip_longest() in Python 2
• itertools.zip_longest() in Python 3

The default fillvalue is None. Thanks to @jena's answer, where map() is changing the inner tuples to lists. Here it is turning iterators into lists. Thanks to @Oregano's and @badp's comments.

In Python 3, pass the result through list() to get the same 2D list as method 2.

---

method 2 — list comprehension, zip_longest()

>>> [list(row) for row in six.moves.zip_longest(*list_list, fillvalue='-')]
[[1, 4, 7], [2, 5, 8], [3, 6, 9], ['-', 6.1, '-'], ['-', 6.2, '-'], ['-', 6.3, '-']]

The @inspectorG4dget alternative.

---

method 3 — map() of map() — broken in Python 3.6

>>> map(list, map(None, *list_list))
[[1, 4, 7], [2, 5, 8], [3, 6, 9], [None, 6.1, None], [None, 6.2, None], [None, 6.3, None]]

This extraordinarily compact @SiggyF second alternative works with ragged 2D lists, unlike his first code which uses numpy to transpose and pass through ragged lists. But None has to be the fill value. (No, the None passed to the inner map() is not the fill value. It means there is no function to process each column. The columns are just passed through to the outer map() which converts them from tuples to lists.)

Somewhere in Python 3, map() stopped putting up with all this abuse: the first parameter cannot be None, and ragged iterators are just truncated to the shortest. The other methods still work because this only applies to the inner map().

---

method 4 — map() of map() revisited

>>> list(map(list, map(lambda *args: args, *list_list)))
[[1, 4, 7], [2, 5, 8], [3, 6, 9]]   // Python 2.7
[[1, 4, 7], [2, 5, 8], [3, 6, 9], [None, 6.1, None], [None, 6.2, None], [None, 6.3, None]] // 3.6+

Alas the ragged rows do NOT become ragged columns in Python 3, they are just truncated. Boo hoo progress.

Answer 6

Three options to choose from:

1. Map with Zip

solution1 = map(list, zip(*l))

2. List Comprehension

solution2 = [list(i) for i in zip(*l)]

3. For Loop Appending

solution3 = []
for i in zip(*l):
    solution3.append((list(i)))

And to view the results:

print(*solution1)
print(*solution2)
print(*solution3)

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

Answer 7

import numpy as np
r = list(map(list, np.transpose(l)))

Answer 8

Maybe not the most elegant solution, but here's a solution using nested while loops:

def transpose(lst):
    newlist = []
    i = 0
    while i < len(lst):
        j = 0
        colvec = []
        while j < len(lst):
            colvec.append(lst[j][i])
            j = j + 1
        newlist.append(colvec)
        i = i + 1
    return newlist

Answer 9

more_itertools.unzip() is easy to read, and it also works with generators.

import more_itertools
l = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
r = more_itertools.unzip(l) # a tuple of generators.
r = list(map(list, r))      # a list of lists

or equivalently

import more_itertools
l = more_itertools.chunked(range(1,10), 3)
r = more_itertools.unzip(l) # a tuple of generators.
r = list(map(list, r))      # a list of lists

Answer 10

matrix = [[1,2,3],
          [1,2,3],
          [1,2,3],
          [1,2,3],
          [1,2,3],
          [1,2,3],
          [1,2,3]]
    
rows = len(matrix)
cols = len(matrix[0])

transposed = []
while len(transposed) < cols:
    transposed.append([])
    while len(transposed[-1]) < rows:
        transposed[-1].append(0)

for i in range(rows):
    for j in range(cols):
        transposed[j][i] = matrix[i][j]

for i in transposed:
    print(i)

Answer 11

One more way for square matrix. No numpy, nor itertools, use (effective) in-place elements exchange.

def transpose(m):
    for i in range(1, len(m)):
        for j in range(i):
            m[i][j], m[j][i] = m[j][i], m[i][j]

Answer 12

Just for fun: If you then want to make them all into dicts.

In [1]: l = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
   ...: fruits = ["Apple", "Pear", "Peach",]
   ...: [dict(zip(fruits, j)) for j in [list(i) for i in zip(*l)]]
Out[1]:
[{'Apple': 1, 'Pear': 4, 'Peach': 7},
 {'Apple': 2, 'Pear': 5, 'Peach': 8},
 {'Apple': 3, 'Pear': 6, 'Peach': 9}]

Answer 13

Here is a solution for transposing a list of lists that is not necessarily square:

maxCol = len(l[0])
for row in l:
    rowLength = len(row)
    if rowLength > maxCol:
        maxCol = rowLength
lTrans = []
for colIndex in range(maxCol):
    lTrans.append([])
    for row in l:
        if colIndex < len(row):
            lTrans[colIndex].append(row[colIndex])

Answer 14

    #Import functions from library
    from numpy import size, array
    #Transpose a 2D list
    def transpose_list_2d(list_in_mat):
        list_out_mat = []
        array_in_mat = array(list_in_mat)
        array_out_mat = array_in_mat.T
        nb_lines = size(array_out_mat, 0)
        for i_line_out in range(0, nb_lines):
            array_out_line = array_out_mat[i_line_out]
            list_out_line = list(array_out_line)
            list_out_mat.append(list_out_line)
        return list_out_mat