Python：嵌套列表中的第一个元素

Question

我想要一个只包含嵌套列表中第一个元素的列表。

这个嵌套列表L看起来像这样：

L =[ [[0,1,2],[3,4,5],[6,7,8]], [[0,1,2],[3,4,5],[6,7,8]], [[0,1,2],[3,4,5],[6,7,8]] ]

for l in L:
 for t in l:
  R.append(t[0])
print 'R=', R

输出结果是 R= [0, 3, 6, 0, 3, 6, 0, 3, 6]，但我想要得到一个分开的结果，像这样：

R= [[0, 3, 6], [0, 3, 6], [0, 3, 6]]

我也尝试过用列表推导式，比如 [[R.append(t[0]) for t in l] for l in L]，但这给出的结果是 [[None, None, None], [None, None, None], [None, None, None]]

这有什么问题呢？

Answer 1

你想要的输出是一个嵌套列表。你可以像这样“手动”嵌套它们：

L =[ [[0,1,2],[3,4,5],[6,7,8]], [[0,1,2],[3,4,5],[6,7,8]], [[0,1,2],[3,4,5],[6,7,8]] ]
R = []
for l in L:
    R2 = []
    for t in l:
        r2.append(t[0])
    R.append(R2)
print 'R=', R

Answer 2

L =[ [[0,1,2],[3,4,5],[6,7,8]], [[0,1,2],[3,4,5],[6,7,8]], [[0,1,2],[3,4,5],[6,7,8]] ]
R=[]
for l in L:
 temp=[]
 for t in l:
  temp.append(t[0])
 R.append(temp)
print 'R=', R

输出结果：

R= [[0, 3, 6], [0, 3, 6], [0, 3, 6]]

Answer 3

你可以使用numpy数组和转置函数。

import numpy as np
L = [ [[0,1,2],[3,4,5],[6,7,8]], [[0,1,2],[3,4,5],[6,7,8]], [[0,1,2],[3,4,5],[6,7,8]] ]
Lnumpy = np.array(L)
Ltransposed = Lnumpy.transpose(0, 2, 1) # Order of axis

现在的输出是

[[[0 3 6]
  [1 4 7]
  [2 5 8]]

 [[0 3 6]
  [1 4 7]
  [2 5 8]]

 [[0 3 6]
  [1 4 7]
  [2 5 8]]]

现在你不需要每个成员的第一个元素，只需要第一个成员。

print(Ltransposed[0][0]) 现在会给你 [0, 3, 6]。

然后

for i in ltr:
    print(ltr[0][0])

输出

[0 3 6]
[0 3 6]
[0 3 6]

顺便提一下，还有一种方法可以使用zip...(这里是针对Python 3的...)

print(list(zip(*Ltransposed))[0])

结果是一样的。如果你需要列表，可以再转换回来... list()...

Answer 4

你可以这样做：

>>> L = [ [[0,1,2],[3,4,5],[6,7,8]], [[0,1,2],[3,4,5],[6,7,8]], [[0,1,2],[3,4,5],[6,7,8]] ]
>>> R = [ [x[0] for x in sl ] for sl in L ]
>>> print R
[[0, 3, 6], [0, 3, 6], [0, 3, 6]]

Answer 5

你的解决方案返回了 [[None, None, None], [None, None, None], [None, None, None]]，这是因为 append 这个方法返回的值是 None。把它换成 t[0] 应该就能解决问题。

你需要的内容是：

R = [[t[0] for t in l] for l in L]