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
matrix = [[1,2,3],
[1,2,3],
[1,2,3],
[1,2,3],
[1,2,3],
[1,2,3],
[1,2,3]]
rows = len(matrix)
cols = len(matrix[0])
transposed = []
while len(transposed) < cols:
transposed.append([])
while len(transposed[-1]) < rows:
transposed[-1].append(0)
for i in range(rows):
for j in range(cols):
transposed[j][i] = matrix[i][j]
for i in transposed:
print(i)
import numpy as np
r = list(map(list, np.transpose(l)))
More_itertools.unzip()很容易阅读,它也可以用于生成器。
import more_itertools
l = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
r = more_itertools.unzip(l) # a tuple of generators.
r = list(map(list, r)) # a list of lists
或者同样的
import more_itertools
l = more_itertools.chunked(range(1,10), 3)
r = more_itertools.unzip(l) # a tuple of generators.
r = list(map(list, r)) # a list of lists
只是为了好玩,有效的矩形假设m[0]存在
>>> m = [[1,2,3],[4,5,6],[7,8,9]]
>>> [[row[i] for row in m] for i in range(len(m[0]))]
[[1, 4, 7], [2, 5, 8], [3, 6, 9]]
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]]