[dgouldin]

It's not pretty, but you can do it:

  >>> sum([['a', 'b', 'c'], ['d', 'e', 'f']], [])
  ['a', 'b', 'c', 'd', 'e', 'f']

This makes use of the optional start argument and list add operator. However, Python's docs suggest using itertools.chain instead:

http://docs.python.org/library/functions.html#sum

  >>> import itertools
  >>> [l for l in itertools.chain(*[['a', 'b', 'c'], ['d', 'e', 'f']])]
  ['a', 'b', 'c', 'd', 'e', 'f']

(Of course you lose the benefit of a generator by using a list comprehension, but this is just an example.)

[jacobolus]

Your second example is easier to just write as:

  >>> from itertools import chain
  >>> list_of_lists = [['a', 'b', 'c'], ['d', 'e', 'f']]
  >>> list(chain(*list_of_lists))
  ['a', 'b', 'c', 'd', 'e', 'f']

That is, list(«foo») is clearer and more idiomatic than [x for x in «foo»].

[scottbessler]

I may be missing something, but why isn't skipping your list comprehension maintaining the benefit of using a generator?

   >>> from itertools import chain
   >>> chain(*[['a', 'b', 'c'], ['d', 'e', 'f']])

[dgouldin]

It is, but I don't get to show you the result of the chain that way. :)