Matrix Transpose in Python

Question

I am trying to create a matrix transpose function for python but I can't seem to make it work. Say I have

theArray = [['a','b','c'],['d','e','f'],['g','h','i']]

and I want my function to come up with

newArray = [['a','d','g'],['b','e','h'],['c', 'f', 'i']]

So in other words, if I were to print this 2D array as columns and rows I would like the rows to turn into columns and columns into rows.

I made this so far but it doesn't work

def matrixTranspose(anArray):
    transposed = [None]*len(anArray[0])
    for t in range(len(anArray)):
        for tt in range(len(anArray[t])):
            transposed[t] = [None]*len(anArray)
            transposed[t][tt] = anArray[tt][t]
    print transposed

jfs · Accepted Answer · 2018-04-18T07:39:14.023

346

Python 2:

>>> theArray = [['a','b','c'],['d','e','f'],['g','h','i']]
>>> zip(*theArray)
[('a', 'd', 'g'), ('b', 'e', 'h'), ('c', 'f', 'i')]

Python 3:

>>> [*zip(*theArray)]
[('a', 'd', 'g'), ('b', 'e', 'h'), ('c', 'f', 'i')]

edited Apr 18 '18 at 07:39

answered Feb 08 '11 at 19:41

jfs

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if you're going to iterate through the results, `izip` from `itertools` can save memory for large arrays. – Antony Hatchkins Mar 28 '13 at 08:38
1

How would you have it return a list for the sub lists? Like `[['a', 'b', 'g'], ['d', 'e', 'h'], ['c', 'f', 'i']]` instead of `[('a', 'd', 'g'), ('b', 'e', 'h'), ('c', 'f', 'i')]`? – acollection_ Feb 28 '16 at 03:15
20

@acollection_: `map(list, zip(*theArray))`. – jfs Feb 28 '16 at 12:45
Why the asterisk? – FreelanceConsultant Oct 20 '16 at 16:38
@user3728501 [What does asterisk * mean in Python?](http://stackoverflow.com/q/400739/4279) – jfs Oct 20 '16 at 17:14
@J.F.Sebastian I read that but unfortunately I didn't understand any of it – FreelanceConsultant Oct 20 '16 at 17:23
@user3728501 have you just read the question, its duplicate and their more than a dozen answers in 8 minutes? Or have you seen them before? [The argument unpacking is discussed even in the official Python tutorial](https://docs.python.org/3/tutorial/controlflow.html#unpacking-argument-lists). If it is still unclear; pick an answer that you understand the most and ask a separate question about the unclear parts. – jfs Oct 20 '16 at 18:01
1

@AntonyHatchkins This is not needed with Python 3.0 and above. There, `zip` already returns an iterator: https://docs.python.org/3.0/whatsnew/3.0.html#views-and-iterators-instead-of-lists – xuiqzy May 09 '20 at 13:06
1

@xuiqzy It's not that I'm not aware of it, but that's true. – Antony Hatchkins May 09 '20 at 17:30

score 68 · Answer 2 · answered Jul 08 '12 at 23:05

68

>>> theArray = [['a','b','c'],['d','e','f'],['g','h','i']]
>>> [list(i) for i in zip(*theArray)]
[['a', 'd', 'g'], ['b', 'e', 'h'], ['c', 'f', 'i']]

the list generator creates a new 2d array with list items instead of tuples.

answered Jul 08 '12 at 23:05

sqwerl

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This is the way to go if you want to assign the result to a variable (as opposed to, e.g., iterating over it directly) — assuming you want lists instead of tuples, as mentioned. – ASL Oct 17 '16 at 19:34
Another option (as implied by the comments in the accepted answer) would be: `list(map(list, zip(*theArray)))` – ASL Oct 18 '16 at 12:42

bigjim · Answer 3 · 2013-01-03T16:04:50.357

If your rows are not equal you can also use map:

>>> uneven = [['a','b','c'],['d','e'],['g','h','i']]
>>> map(None,*uneven)
[('a', 'd', 'g'), ('b', 'e', 'h'), ('c', None, 'i')]

Edit: In Python 3 the functionality of map changed, itertools.zip_longest can be used instead:
Source: What’s New In Python 3.0

>>> import itertools
>>> uneven = [['a','b','c'],['d','e'],['g','h','i']]
>>> list(itertools.zip_longest(*uneven))
[('a', 'd', 'g'), ('b', 'e', 'h'), ('c', None, 'i')]

score 20 · Answer 4 · answered Dec 08 '14 at 05:34

Much easier with numpy:

>>> arr = np.array([[1,2,3],[4,5,6],[7,8,9]])
>>> arr
array([[1, 2, 3],
       [4, 5, 6],
       [7, 8, 9]])
>>> arr.T
array([[1, 4, 7],
       [2, 5, 8],
       [3, 6, 9]])
>>> theArray = np.array([['a','b','c'],['d','e','f'],['g','h','i']])
>>> theArray 
array([['a', 'b', 'c'],
       ['d', 'e', 'f'],
       ['g', 'h', 'i']], 
      dtype='|S1')
>>> theArray.T
array([['a', 'd', 'g'],
       ['b', 'e', 'h'],
       ['c', 'f', 'i']], 
      dtype='|S1')

score 9 · Answer 5 · answered Feb 08 '11 at 19:48

The problem with your original code was that you initialized transpose[t] at every element, rather than just once per row:

def matrixTranspose(anArray):
    transposed = [None]*len(anArray[0])
    for t in range(len(anArray)):
        transposed[t] = [None]*len(anArray)
        for tt in range(len(anArray[t])):
            transposed[t][tt] = anArray[tt][t]
    print transposed

This works, though there are more Pythonic ways to accomplish the same things, including @J.F.'s zip application.

Note that this implementation doesn't work with matrices that have different numbers of columns and rows — Vector, Nov 10 '19 at 03:18

score 5 · Answer 6 · answered Jun 12 '13 at 21:47

The "best" answer has already been submitted, but I thought I would add that you can use nested list comprehensions, as seen in the Python Tutorial.

Here is how you could get a transposed array:

def matrixTranspose( matrix ):
    if not matrix: return []
    return [ [ row[ i ] for row in matrix ] for i in range( len( matrix[ 0 ] ) ) ]

score 5 · Answer 7 · answered Mar 08 '12 at 18:05

5

To complete J.F. Sebastian's answer, if you have a list of lists with different lengths, check out this great post from ActiveState. In short:

The built-in function zip does a similar job, but truncates the result to the length of the shortest list, so some elements from the original data may be lost afterwards.

To handle list of lists with different lengths, use:

def transposed(lists):
   if not lists: return []
   return map(lambda *row: list(row), *lists)

def transposed2(lists, defval=0):
   if not lists: return []
   return map(lambda *row: [elem or defval for elem in row], *lists)

answered Mar 08 '12 at 18:05

Franck Dernoncourt

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That's good catch. However, matrices doesn't have lists with different lengths. – Olli Oct 04 '12 at 08:07
It depends on how they are stored. – Franck Dernoncourt Mar 09 '13 at 00:59

score 2 · Answer 8 · answered Feb 03 '16 at 18:51

2

This one will preserve rectangular shape, so that subsequent transposes will get the right result:

import itertools
def transpose(list_of_lists):
  return list(itertools.izip_longest(*list_of_lists,fillvalue=' '))

answered Feb 03 '16 at 18:51

Vanuan

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sharif_42 · Answer 9 · 2020-02-10T11:17:27.460

2

you can try this with list comprehension like the following

matrix = [['a','b','c'],['d','e','f'],['g','h','i']] n = len(matrix) transpose = [[row[i] for row in matrix] for i in range(n)] print (transpose)

edited Feb 10 '20 at 11:17

answered Feb 10 '20 at 11:10

sharif_42

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score 1 · Answer 10 · answered Jul 05 '19 at 06:57

If you want to transpose a matrix like A = np.array([[1,2],[3,4]]), then you can simply use A.T, but for a vector like a = [1,2], a.T does not return a transpose! and you need to use a.reshape(-1, 1), as below

import numpy as np
a = np.array([1,2])
print('a.T not transposing Python!\n','a = ',a,'\n','a.T = ', a.T)
print('Transpose of vector a is: \n',a.reshape(-1, 1))

A = np.array([[1,2],[3,4]])
print('Transpose of matrix A is: \n',A.T)

score 1 · Answer 11 · edited Jan 22 '20 at 09:54

1

You may do it simply using python comprehension.

arr = [
    ['a', 'b', 'c'], 
    ['d', 'e', 'f'], 
    ['g', 'h', 'i']
]
transpose = [[arr[y][x] for y in range(len(arr))] for x in range(len(arr[0]))]

edited Jan 22 '20 at 09:54

pylover

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answered Sep 27 '19 at 21:50

rasoul poordelan

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Although this may be a correct answer. Two lines of code isn't very useful without an explanation of what and how it solves the original question. Please provide details to your answer. – RyanNerd Sep 27 '19 at 22:29
1

when posting a new answer to an old question, expectations are high. Please don't post an inferior solution to the ones already posted – Jean-François Fabre Sep 28 '19 at 09:05

score -1 · Answer 12 · edited Oct 04 '12 at 08:13

-1

def matrixTranspose(anArray):
  transposed = [None]*len(anArray[0])

  for i in range(len(transposed)):
    transposed[i] = [None]*len(transposed)

  for t in range(len(anArray)):
    for tt in range(len(anArray[t])):            
        transposed[t][tt] = anArray[tt][t]
  return transposed

theArray = [['a','b','c'],['d','e','f'],['g','h','i']]

print matrixTranspose(theArray)

edited Oct 04 '12 at 08:13

Olli

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answered Feb 08 '11 at 19:51

Asterisk

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score -1 · Answer 13 · edited Jun 02 '21 at 07:31

import  numpy as np #Import Numpy 

m=int(input("Enter row")) #Input Number of row

n=int(input("Enter column")) #Input number of column

a=[] #Blank Matrix

for i in range(m): #Row Input

    b=[] #Blank List

    for j in range(n):#column Input

        j=int(input("Enter Number in Pocket ["+str(i)+"]["+str(j)+"]")) #sow Row Column Number 

        b.append(j) #addVlaue to list

    a.append(b)#Add List To Matrix

a=np.array(a)#convert 1matrix as Numpy

b=a.transpose()#transpose Using Numpy

print(a) #Print Matrix 

print(b)#print Transpose Matrix

score -2 · Answer 14 · answered Apr 27 '14 at 18:21

#generate matrix
matrix=[]
m=input('enter number of rows, m = ')
n=input('enter number of columns, n = ')
for i in range(m):
    matrix.append([])
    for j in range(n):
        elem=input('enter element: ')
        matrix[i].append(elem)

#print matrix
for i in range(m):
    for j in range(n):
        print matrix[i][j],
    print '\n'

#generate transpose
transpose=[]
for j in range(n):
    transpose.append([])
    for i in range (m):
        ent=matrix[i][j]
        transpose[j].append(ent)

#print transpose
for i in range (n):
    for j in range (m):
        print transpose[i][j],
    print '\n'

score -3 · Answer 15 · edited Mar 30 '15 at 06:30

a=[]
def showmatrix (a,m,n):
    for i in range (m):
        for j in range (n):
            k=int(input("enter the number")
            a.append(k)      
print (a[i][j]),

print('\t')


def showtranspose(a,m,n):
    for j in range(n):
        for i in range(m):
            print(a[i][j]),
        print('\t')

a=((89,45,50),(130,120,40),(69,79,57),(78,4,8))
print("given matrix of order 4x3 is :")
showmatrix(a,4,3)


print("Transpose matrix is:")
showtranspose(a,4,3)

score -3 · Answer 16 · answered Mar 13 '16 at 19:16

-3

def transpose(matrix):
   x=0
   trans=[]
   b=len(matrix[0])
   while b!=0:
       trans.append([])
       b-=1
   for list in matrix:
       for element in list:
          trans[x].append(element)
          x+=1
       x=0
   return trans

answered Mar 13 '16 at 19:16

mohammad hassan jafari

1
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score -3 · Answer 17 · answered Aug 14 '17 at 21:58

-3

def transpose(matrix):
    listOfLists = []
    for row in range(len(matrix[0])):
        colList = []
        for col in range(len(matrix)):
            colList.append(matrix[col][row])
    listOfLists.append(colList)

    return listOfLists

answered Aug 14 '17 at 21:58

Ravneet Singh

1

Its a simple implementation for a transpose, though there are libraries like mentioned in other answers are also available. – Ravneet Singh Aug 14 '17 at 22:00

score -3 · Answer 18 · answered Aug 19 '17 at 18:10

-3

`

def transpose(m):
    return(list(map(list,list(zip(*m)))))

`This function will return the transpose

answered Aug 19 '17 at 18:10

user2412711

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score -3 · Answer 19 · edited Feb 14 '18 at 15:05

-3

Python Program to transpose matrix:

row,col = map(int,input().split())
matrix = list()

for i in range(row):
    r = list(map(int,input().split()))
    matrix.append(r)

trans = [[0 for y in range(row)]for x in range(col)]

for i in range(len(matrix[0])):
    for j in range(len(matrix)):
        trans[i][j] = matrix[j][i]     

for i in range(len(trans)):
    for j in range(len(trans[0])):
        print(trans[i][j],end=' ')
    print(' ')

edited Feb 14 '18 at 15:05

JHBonarius

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answered Feb 14 '18 at 14:13

Manish Rana

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This is not useful! – tripulse May 08 '19 at 04:28

Matrix Transpose in Python

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