转换并加入多个嵌套列表列表答案

【问题标题】：Convert and join multiple nested list of lists转换并加入多个嵌套列表列表
【发布时间】：2017-02-23 16:37:40
【问题描述】：

我有以下嵌套列表：

a = [[[[-79.43402638260521, -1.69184588855758], [-79.4339722432865, -1.691845844583909], [-79.43397178076256, -1.691851284533779],
    [-79.43395283944169, -1.692053292637794], [-79.43395281911414, -1.692054736321033], [-79.43395535750368, -1.692093535418117],
    [-79.43390444734398, -1.69223087834723], [-79.43390428016939, -1.692231372437897], [-79.43374523144152, -1.692750043925838],
    [-79.4340256570161, -1.692750271834557], [-79.43402638260521, -1.69184588855758]]], [[[-79.43381375958064, -1.691845715849684],
    [-79.43312765678151, -1.691845158387183], [-79.4331269307764, -1.692749541273626], [-79.43354270912953, -1.692749879305633],
    [-79.43364983051107, -1.692588468489809], [-79.4336510738479, -1.692585646334773], [-79.43371548446397, -1.692327269168548],
    [-79.43380554258165, -1.692094789340216], [-79.43380615195998, -1.692091785860122], [-79.43381375958064, -1.691845715849684]]]]

我会怎样去：

a = [[-79.43402638260521, -1.69184588855758], [-79.4339722432865, -1.691845844583909], [-79.43397178076256, -1.691851284533779], 
[-79.43395283944169, -1.692053292637794], [-79.43395281911414, -1.692054736321033], [-79.43395535750368, -1.692093535418117], 
[-79.43390444734398, -1.69223087834723], [-79.43390428016939, -1.692231372437897], [-79.43374523144152, -1.692750043925838], 
[-79.4340256570161, -1.692750271834557], [-79.43402638260521, -1.69184588855758], [-79.43381375958064, -1.691845715849684], 
[-79.43312765678151, -1.691845158387183], [-79.4331269307764, -1.692749541273626], [-79.43354270912953, -1.692749879305633], 
[-79.43364983051107, -1.692588468489809], [-79.4336510738479, -1.692585646334773], [-79.43371548446397, -1.692327269168548], 
[-79.43380554258165, -1.692094789340216], [-79.43380615195998, -1.692091785860122], [-79.43381375958064, -1.691845715849684]]

所以基本上删除了嵌套列表和每个列表开头和结尾的双括号。我已经尝试了以下和 ittertools 没有成功：

flatten = lambda list: [item for sublist in list for item in sublist]

注意：len(a) == 2

非常感谢！

【问题讨论】：

在你的特殊情况下，a[0][0]?
@CarlesMitjans，@Arman，len(a) == 2
@falsetru 你是对的，没注意到
@falsetru，确切地说，这就是为什么 a[0][0] 在这种情况下不起作用

标签： python list nested

【解决方案1】：

这样就可以了。

a = a[0][0] + a[1][0]

这可以扩展为，

a = sum([a[i][0] for i in range(len(a))], [])

【讨论】：

这在我看来是最明智的解决方案。如果有两个以上的元素很容易，它也可以适应工作。
没错！如果格式一致，这是最简单的方法。 @sphericalcowboy
你们能评论下投票的原因吗？这不是一个“错误”的答案。
是的。在列表的情况下，它有效，但它不是正确的方法。 :) 我会编辑它。
感谢您的回答。这确实似乎是最合乎逻辑的解决方案，因为它不依赖于其他库。

【解决方案2】：

使用itertools.chain.from_iterable：

list(itertools.chain.from_iterable(xs[0] for xs in a))

上面类似于：a[0][0] + a[1][0] + ...（没有导致列表创建的串联）

>>> a = [[[[-79.43402638260521, -1.69184588855758],
...        [-79.4339722432865, -1.691845844583909],
...        [-79.43397178076256, -1.691851284533779],
...        [-79.43395283944169, -1.692053292637794],
...        [-79.43395281911414, -1.692054736321033],
...        [-79.43395535750368, -1.692093535418117],
...        [-79.43390444734398, -1.69223087834723],
...        [-79.43390428016939, -1.692231372437897],
...        [-79.43374523144152, -1.692750043925838],
...        [-79.4340256570161, -1.692750271834557],
...        [-79.43402638260521, -1.69184588855758]]],
...      [[[-79.43381375958064, -1.691845715849684],
...        [-79.43312765678151, -1.691845158387183],
...        [-79.4331269307764, -1.692749541273626],
...        [-79.43354270912953, -1.692749879305633],
...        [-79.43364983051107, -1.692588468489809],
...        [-79.4336510738479, -1.692585646334773],
...        [-79.43371548446397, -1.692327269168548],
...        [-79.43380554258165, -1.692094789340216],
...        [-79.43380615195998, -1.692091785860122],
...        [-79.43381375958064, -1.691845715849684]]]]
>>> 
>>> import itertools
>>> import pprint
>>> b = list(itertools.chain.from_iterable(xs[0] for xs in a))
>>> pprint.pprint(b)

结果：

[[-79.43402638260521, -1.69184588855758],
 [-79.4339722432865, -1.691845844583909],
 [-79.43397178076256, -1.691851284533779],
 [-79.43395283944169, -1.692053292637794],
 [-79.43395281911414, -1.692054736321033],
 [-79.43395535750368, -1.692093535418117],
 [-79.43390444734398, -1.69223087834723],
 [-79.43390428016939, -1.692231372437897],
 [-79.43374523144152, -1.692750043925838],
 [-79.4340256570161, -1.692750271834557],
 [-79.43402638260521, -1.69184588855758],
 [-79.43381375958064, -1.691845715849684],
 [-79.43312765678151, -1.691845158387183],
 [-79.4331269307764, -1.692749541273626],
 [-79.43354270912953, -1.692749879305633],
 [-79.43364983051107, -1.692588468489809],
 [-79.4336510738479, -1.692585646334773],
 [-79.43371548446397, -1.692327269168548],
 [-79.43380554258165, -1.692094789340216],
 [-79.43380615195998, -1.692091785860122],
 [-79.43381375958064, -1.691845715849684]]

【讨论】：

【解决方案3】：

假设你的列表不是任意嵌套的，但你只是想更深一层，你可以选择一些非常简单的东西，比如

a = [item for sublist in a for subsublist in sublist for item in subsublist]

演示

>>> a = [[[[-79.43402638260521, -1.69184588855758], [-79.4339722432865, -1.691845844583909], [-79.43397178076256, -1.691851284533779],
[-79.43395283944169, -1.692053292637794], [-79.43395281911414, -1.692054736321033], [-79.43395535750368, -1.692093535418117],
[-79.43390444734398, -1.69223087834723], [-79.43390428016939, -1.692231372437897], [-79.43374523144152, -1.692750043925838],
[-79.4340256570161, -1.692750271834557], [-79.43402638260521, -1.69184588855758]]], [[[-79.43381375958064, -1.691845715849684],
[-79.43312765678151, -1.691845158387183], [-79.4331269307764, -1.692749541273626], [-79.43354270912953, -1.692749879305633],
[-79.43364983051107, -1.692588468489809], [-79.4336510738479, -1.692585646334773], [-79.43371548446397, -1.692327269168548],
[-79.43380554258165, -1.692094789340216], [-79.43380615195998, -1.692091785860122], [-79.43381375958064, -1.691845715849684]]]]
>>> desired = [[-79.43402638260521, -1.69184588855758], [-79.4339722432865, -1.691845844583909], [-79.43397178076256, -1.691851284533779], 
[-79.43395283944169, -1.692053292637794], [-79.43395281911414, -1.692054736321033], [-79.43395535750368, -1.692093535418117], 
[-79.43390444734398, -1.69223087834723], [-79.43390428016939, -1.692231372437897], [-79.43374523144152, -1.692750043925838], 
[-79.4340256570161, -1.692750271834557], [-79.43402638260521, -1.69184588855758], [-79.43381375958064, -1.691845715849684], 
[-79.43312765678151, -1.691845158387183], [-79.4331269307764, -1.692749541273626], [-79.43354270912953, -1.692749879305633], 
[-79.43364983051107, -1.692588468489809], [-79.4336510738479, -1.692585646334773], [-79.43371548446397, -1.692327269168548], 
[-79.43380554258165, -1.692094789340216], [-79.43380615195998, -1.692091785860122], [-79.43381375958064, -1.691845715849684]]

>>> a = [item for sublist in a for subsublist in sublist for item in subsublist]
>>> a == desired
True

但是，正如 falsetru 所指出的，您的示例中有一个空的嵌套级别，在这种情况下，可以很好地清理解决方案 as they have shown。

【讨论】：

【解决方案4】：

你可以试试：

[z for x in a for y in x for z in y]

为了证明这一点：

len([y for x in a for y in x]) == 21

您可以使用pprint 正确查看：

pprint.pprint(a)

输出：

[[[[-79.43402638260521, -1.69184588855758],
   [-79.4339722432865, -1.691845844583909],
   [-79.43397178076256, -1.691851284533779],
   [-79.43395283944169, -1.692053292637794],
   [-79.43395281911414, -1.692054736321033],
   [-79.43395535750368, -1.692093535418117],
   [-79.43390444734398, -1.69223087834723],
   [-79.43390428016939, -1.692231372437897],
   [-79.43374523144152, -1.692750043925838],
   [-79.4340256570161, -1.692750271834557],
   [-79.43402638260521, -1.69184588855758]]],
 [[[-79.43381375958064, -1.691845715849684],
   [-79.43312765678151, -1.691845158387183],
   [-79.4331269307764, -1.692749541273626],
   [-79.43354270912953, -1.692749879305633],
   [-79.43364983051107, -1.692588468489809],
   [-79.4336510738479, -1.692585646334773],
   [-79.43371548446397, -1.692327269168548],
   [-79.43380554258165, -1.692094789340216],
   [-79.43380615195998, -1.692091785860122],
   [-79.43381375958064, -1.691845715849684]]]]

pprint.pprint([z for x in a for y in x for z in y])

输出：

[[-79.43402638260521, -1.69184588855758],
 [-79.4339722432865, -1.691845844583909],
 [-79.43397178076256, -1.691851284533779],
 [-79.43395283944169, -1.692053292637794],
 [-79.43395281911414, -1.692054736321033],
 [-79.43395535750368, -1.692093535418117],
 [-79.43390444734398, -1.69223087834723],
 [-79.43390428016939, -1.692231372437897],
 [-79.43374523144152, -1.692750043925838],
 [-79.4340256570161, -1.692750271834557],
 [-79.43402638260521, -1.69184588855758],
 [-79.43381375958064, -1.691845715849684],
 [-79.43312765678151, -1.691845158387183],
 [-79.4331269307764, -1.692749541273626],
 [-79.43354270912953, -1.692749879305633],
 [-79.43364983051107, -1.692588468489809],
 [-79.4336510738479, -1.692585646334773],
 [-79.43371548446397, -1.692327269168548],
 [-79.43380554258165, -1.692094789340216],
 [-79.43380615195998, -1.692091785860122],
 [-79.43381375958064, -1.691845715849684]]

【讨论】：

【解决方案5】：

您需要遍历嵌套列表并加入它们。

一个简单的方法：

def flatten1(list_of_lists):
  "Flattens one level of lists."
  result = []
  for sub_list in list_of_lists:
    result.extend(sub_list)
  return result

一个聪明的方法，利用你可以添加列表的事实：

flatten1 = lambda(list_of_lists): sum(list_of_lists, [])

现在，你可以a[0] = flatten1(a[0])。

【讨论】：