通过列表理解压扁名单列表通过列表理解压扁名单列表(flatten list of list thro

I am trying to flatten a list using list comprehension in python. My list is somewhat like

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

just for printing then individual item in this list of list I wrote this code

   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

Then I used the same logic to flatten my list through list comprehension I am getting the following error

    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

How can I flatten this type of list-of-list using using list comprehension ?

Answer 1:

>>> from collections import Iterable
>>> from itertools import chain

一内胆：

>>> list(chain.from_iterable(item if isinstance(item,Iterable) and
                    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):                                         #use `str` in py3.x 
...     if isinstance(x, Iterable) and not isinstance(x, basestring): 
...         return x
...     return [x]
... 
>>> 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))                                                                
[1, 2, 3, 4, 5, 6, 7, 8, 'foobar']

使用嵌套列表理解：（去相比要慢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:

没有人给通常的回答：

def flat(l):
  return [y for x in l for y in x]

有此问题漂浮的StackOverflow的受骗者。

Answer 3:

使用生成的替代解决方案：

import collections

def flatten(iterable):
    for item in iterable:
        if isinstance(item, collections.Iterable) and not isinstance(item, str):  # `basestring` < 3.x
            yield from item  # `for subitem in item: yield item` < 3.3
        else:
            yield item

>>> list(flatten([[1, 2, 3], [4, 5, 6], 7, 8]))
[1, 2, 3, 4, 5, 6, 7, 8]

Answer 4:

你试图通过一个号码，你不能这样做（因此错误）进行迭代。

如果你正在使用Python 2.7：

>>> from compiler.ast import flatten
>>> flatten(l)
[1, 2, 3, 4, 5, 6, 7, 8]

但是千万注意，模块现在已经过时，并在Python 3不复存在

Answer 5:

def nnl(nl):    # non nested list

    nn = []

    for x in nl:
        if type(x) == type(5):
            nn.append(x)

    if type(x) == type([]):
        n = nnl(x)

        for y in n:
            nn.append(y)
    return nn

print (nnl([[9, 4, 5], [3, 8,[5]], 6]))  # output: [9, 4, 5, 3, 8, 5, 6]

文章来源: flatten list of list through list comprehension