通过列表理解拼合列表

Question

我正在尝试使用python中的列表理解来扁平化列表。我的清单有点像

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

只是为了打印，然后在此列表中的单个项目，我编写了此代码

   def flat(listoflist):

     for item in listoflist:

             if type(item) != list:

                     print item

             else:

                     for num in item:

                             print num  

>>> flat(list1)

1

2

3

4

5

6

7

8

然后我使用相同的逻辑通过列表理解来整理列表，但出现以下错误

    list2 = [item if type(item) != list else num for num in item for item in list1]

    Traceback (most recent call last):

    File "<stdin>", line 1, in <module>

    TypeError: 'int' object is not iterable

如何使用列表理解来展平这种类型的列表？

Answer 1

>>> from collections import Iterable>>> from itertools import chain单线：>>> list(chain.from_iterable(item if isinstance(item,Iterable) and&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; not isinstance(item, basestring) else [item] for item in lis))[1, 2, 3, 4, 5, 6, 7, 8]可读版本：>>> def func(x):&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;#use `str` in py3.x&nbsp;...&nbsp; &nbsp; &nbsp;if isinstance(x, Iterable) and not isinstance(x, basestring):&nbsp;...&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;return x...&nbsp; &nbsp; &nbsp;return [x]...&nbsp;>>> list(chain.from_iterable(func(x) for x in lis))[1, 2, 3, 4, 5, 6, 7, 8]#works for strings as well>>> lis = [[1, 2, 3], [4, 5, 6], 7, 8, "foobar"]>>> list(chain.from_iterable(func(x) for x in lis))&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;[1, 2, 3, 4, 5, 6, 7, 8, 'foobar']使用嵌套列表理解：（与相比，速度会很慢itertools.chain）：>>> [ele for item in (func(x) for x in lis) for ele in item][1, 2, 3, 4, 5, 6, 7, 8, 'foobar']

Answer 2

没有人给出通常的答案：def&nbsp;flat(l): &nbsp;&nbsp;return&nbsp;[y&nbsp;for&nbsp;x&nbsp;in&nbsp;l&nbsp;for&nbsp;y&nbsp;in&nbsp;x]在StackOverflow周围有这个问题的虚假信息。

Answer 3

您正在尝试遍历一个数字，而这个数字是您无法做到的（因此会出现错误）。如果您使用的是python 2.7：>>> from compiler.ast import flatten>>> flatten(l)[1, 2, 3, 4, 5, 6, 7, 8]但请注意，该模块现已弃用，并且在Python 3中不再存在