Flattening a shallow list in Python

Question

Is there a simple way to flatten a list of iterables with a list comprehension, or failing that, what would you all consider to be the best way to flatten a shallow list like this, balancing performance and readability?

I tried to flatten such a list with a nested list comprehension, like this:

[image for image in menuitem for menuitem in list_of_menuitems]

But I get in trouble of the NameError variety there, because the name 'menuitem' is not defined. After googling and looking around on Stack Overflow, I got the desired results with a reduce statement:

reduce(list.__add__, map(lambda x: list(x), list_of_menuitems))

But this method is fairly unreadable because I need that list(x) call there because x is a Django QuerySet object.

Conclusion:

Thanks to everyone who contributed to this question. Here is a summary of what I learned. I'm also making this a community wiki in case others want to add to or correct these observations.

My original reduce statement is redundant and is better written this way:

>>> reduce(list.__add__, (list(mi) for mi in list_of_menuitems))

This is the correct syntax for a nested list comprehension (Brilliant summary dF!):

>>> [image for mi in list_of_menuitems for image in mi]

But neither of these methods are as efficient as using itertools.chain:

>>> from itertools import chain
>>> list(chain(*list_of_menuitems))

And as @cdleary notes, it's probably better style to avoid * operator magic by using chain.from_iterable like so:

>>> chain = itertools.chain.from_iterable([[1,2],[3],[5,89],[],[6]])
>>> print(list(chain))
>>> [1, 2, 3, 5, 89, 6]

Answer 1

If you're just looking to iterate over a flattened version of the data structure and don't need an indexable sequence, consider itertools.chain and company.

>>> list_of_menuitems = [['image00', 'image01'], ['image10'], []]
>>> import itertools
>>> chain = itertools.chain(*list_of_menuitems)
>>> print(list(chain))
['image00', 'image01', 'image10']

It will work on anything that's iterable, which should include Django's iterable QuerySets, which it appears that you're using in the question.

Edit: This is probably as good as a reduce anyway, because reduce will have the same overhead copying the items into the list that's being extended. chain will only incur this (same) overhead if you run list(chain) at the end.

Meta-Edit: Actually, it's less overhead than the question's proposed solution, because you throw away the temporary lists you create when you extend the original with the temporary.

Edit: As J.F. Sebastian says itertools.chain.from_iterable avoids the unpacking and you should use that to avoid * magic, but the timeit app shows negligible performance difference.