Python中的矩阵转置

问题:Python中的矩阵转置

我正在尝试为python创建矩阵转置函数,但似乎无法使其工作。说我有

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

我想提出我的职能

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

因此,换句话说,如果我要将此2D数组打印为列和行,我希望将行变成列,将列变成行。

我到目前为止已经做到了,但是没有用

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

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

回答 0

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')]

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')]

回答 1

>>> 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']]

列表生成器使用列表项而不是元组创建一个新的2d数组。

>>> 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.


回答 2

如果行数不相等,也可以使用map

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

编辑:在Python 3中mapitertools.zip_longest可以改用已更改的功能:
来源:Python 3.0中的新增功能

>>> 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')]

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')]

回答 3

使用numpy容易得多:

>>> 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')

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')

回答 4

原始代码的问题在于,您transpose[t]在每个元素上都进行了初始化,而不是每行只初始化一次:

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

尽管有更多的Python方式可以完成相同的工作,包括@JF的zip应用程序,但这种方法仍然有效。

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.


回答 5

要完成JF Sebastian的答案,如果您有长度不一的列表,请查看ActiveState上的出色文章。简而言之:

内置函数zip可以执行类似的工作,但是会将结果截断为最短列表的长度,因此之后原始数据中的某些元素可能会丢失。

要处理具有不同长度的列表,请使用:

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)

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)

回答 6

“最佳”答案已经提交,但是我想我要补充一下,您可以使用嵌套列表推导,如Python教程中所示

这是获取转置数组的方法:

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

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 ] ) ) ]

回答 7

这将保留矩形形状,以便随后的转置将获得正确的结果:

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

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=' '))

回答 8

您可以像下面这样使用列表理解来尝试

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)

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)


回答 9

如果要转置像A = np.array([[1,2 ,, [3,4]])这样的矩阵,则可以简单地使用AT,但是对于像a = [1,2],aT的向量不返回移调!并且您需要使用a.reshape(-1,1),如下所示

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)

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)

回答 10

您可以简单地使用python理解来做到这一点。

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]))]

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]))]

回答 11

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)
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)

回答 12

#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'
#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'

回答 13

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)
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)

回答 14

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
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

回答 15

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
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

回答 16

`

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

`此函数将返回转置

`

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

`This function will return the transpose


回答 17

Python程式转置矩阵:

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(' ')

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(' ')