Python：嵌套列表中的第一个元素

我想要一个只包含嵌套列表的第一个元素的列表。嵌套表L，它的样子：Python：嵌套列表中的第一个元素

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

输出是R= [0, 3, 6, 0, 3, 6, 0, 3, 6]但我想得到这样一个分离的结果是：

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

我也通过列表理解试过像[[R.append(t[0]) for t in l] for l in L]但这给出[[None, None, None], [None, None, None], [None, None, None]]

出了什么问题？

来源

2014-09-12 lara

您的解决方案返回[[None, None, None], [None, None, None], [None, None, None]]因为append返回值None的方法。用t[0]取代它应该可以做到。

什么你要找的是：

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

来源

2014-09-12 10:58:42 Hetzroni

你可以做这样的：

>>> 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]]

来源

2014-09-12 10:54:55 isedev

@ isedev谢谢！ – lara 2014-09-12 11:07:19

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]]

来源

2014-09-12 10:57:54 g4ur4v

你所要的输出是一个嵌套列表。您可以嵌套他们“手”是这样的：

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

来源

2014-09-12 10:58:17 MathieuS

Python：嵌套列表中的第一个元素

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