Python：嵌套列表中的第一个元素

Question

我想要一个仅包含嵌套列表中第一个元素的列表。嵌套列表L如下：

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]，但我想得到一个分离的结果，如下所示：

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

我还尝试了通过列表推导式 [[R.append(t[0]) for t in l] for l in L]，但是得到的结果是 [[None, None, None], [None, None, None], [None, None, None]]，这是什么问题？

- lara

Answer 1

你的解决方案返回了[[None, None, None], [None, None, None], [None, None, None]]，因为方法append返回值None。将其替换为t[0]即可达到目的。

你需要寻找的内容是：

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

- Hetzroni

Answer 2

你可以像这样做：

你可以像这样做：

>>> 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]]

- isedev

Answer 3

你可以使用numpy数组的转置函数。

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

输出现在是

[[[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成员，只需要第一个成员。

print(Ltransposed[0][0]) 现在给出的是 [0, 3, 6]。

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

输出

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

仅供参考，还有使用zip的可能性...(这里是Python 3...)

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

给你同样的东西。如果你需要列表，将其转换回来...list()...

- Daniel Malachov

Answer 4

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：

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

- g4ur4v

Answer 5

你想要输出一个嵌套列表。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

- Mathieu Schopfer