让我们来:
l = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
我想要的结果是
r = [[1, 4, 7], [2, 5, 8], [3, 6, 9]]
而不是
r = [(1, 4, 7), (2, 5, 8), (3, 6, 9)]
让我们来:
l = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
我想要的结果是
r = [[1, 4, 7], [2, 5, 8], [3, 6, 9]]
而不是
r = [(1, 4, 7), (2, 5, 8), (3, 6, 9)]
当前回答
也许不是最优雅的解决方案,但这里有一个使用嵌套while循环的解决方案:
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
其他回答
等价于耶拿的解决方案:
>>> 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]]
下面是一个不一定是平方的列表的转置的解决方案:
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])
一种方法是用NumPy转置。如需列出清单,请填写:
>>> import numpy as np
>>> np.array(l).T.tolist()
[[1, 4, 7], [2, 5, 8], [3, 6, 9]]
或者另一个没有zip (python < 3)的:
>>> map(list, map(None, *l))
[[1, 4, 7], [2, 5, 8], [3, 6, 9]]
或者对于python >= 3:
>>> list(map(lambda *x: list(x), *l))
[[1, 4, 7], [2, 5, 8], [3, 6, 9]]
#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
也许不是最优雅的解决方案,但这里有一个使用嵌套while循环的解决方案:
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