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

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')]

>>> 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。

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map，itertools.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')]

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')

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。

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)

“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 ] ) ) ]

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=' '))

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)

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)

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]))]

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)

#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'

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)

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

`此函数将返回转置

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(' ')