flatten list of list through list comprehension

No-one has given the usual answer:

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

There are dupes of this question floating around StackOverflow.

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

One-liner:

>>> 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]

A readable version:

>>> 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']

Using nested list comprehension:(Going to be slow compared to 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']

An alternative solution using a generator:

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]