파이썬에서 행렬 전치

파이썬에서 행렬 전치

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파이썬에 대한 행렬 전치 함수를 만들려고하는데 작동하지 않는 것 같습니다. 내가 가지고 있다고

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

나는 내 기능을 생각해 내고 싶다.

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

즉,이 2D 배열을 열과 ​​행으로 인쇄하려면 행을 열로, 열을 행으로 바꾸고 싶습니다.

나는 이것을 지금까지 만들었지 만 작동하지 않습니다

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

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파이썬 2 :

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

파이썬 3 :

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

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>>> 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']]

목록 생성기는 튜플 대신 목록 항목으로 새로운 2D 배열을 만듭니다.

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행이 같지 않으면 다음을 사용할 수도 있습니다 map.

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

편집 : Python 3에서는 map변경된 기능을 itertools.zip_longest대신 사용할 수 있습니다.
출처 : Python 3.0의 새로운 기능

>>> 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')]

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numpy를 사용하면 훨씬 쉽습니다.

>>> 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')

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원래 코드의 문제점은 transpose[t]행당 한 번이 아니라 모든 요소에서 초기화되었다는 것입니다 .

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.

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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)

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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 ] ) ) ]

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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=' '))

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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)

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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)

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#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'

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python 3 easy way with Comprehensions

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]))]

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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)

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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

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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

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`

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

`This function will return the transpose

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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(' ')

참고URL : https://stackoverflow.com/questions/4937491/matrix-transpose-in-python