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
有三种选择:
1. Zip地图
solution1 = map(list, zip(*l))
2. 列表理解
solution2 = [list(i) for i in zip(*l)]
3.For循环附加
solution3 = []
for i in zip(*l):
solution3.append((list(i)))
查看结果:
print(*solution1)
print(*solution2)
print(*solution3)
# [1, 4, 7], [2, 5, 8], [3, 6, 9]
其他回答
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]]
有三种选择:
1. Zip地图
solution1 = map(list, zip(*l))
2. 列表理解
solution2 = [list(i) for i in zip(*l)]
3.For循环附加
solution3 = []
for i in zip(*l):
solution3.append((list(i)))
查看结果:
print(*solution1)
print(*solution2)
print(*solution3)
# [1, 4, 7], [2, 5, 8], [3, 6, 9]
import numpy as np
r = list(map(list, np.transpose(l)))
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]]
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