One-liner to merge lists?

Sun Feb 27 16:49:24 EST 2022

On 27/02/2022 17:28, Chris Angelico wrote:
> On 2022年2月28日 at 03:24, MRAB <python at mrabarnett.plus.com> wrote:
>>>> On 2022年02月27日 08:51, Barry Scott wrote:
>>>>>>>>>> On 22 Feb 2022, at 09:30, Chris Angelico <rosuav at gmail.com> wrote:
>>>>>>>> On 2022年2月22日 at 20:24, Frank Millman <frank at chagford.com <mailto:frank at chagford.com>> wrote:
>>>>>>>>>> Hi all
>>>>>>>>>> I think this should be a simple one-liner, but I cannot figure it out.
>>>>>>>>>> I have a dictionary with a number of keys, where each value is a single
>>>>> list -
>>>>>>>>>>>>> d = {1: ['aaa', 'bbb', 'ccc'], 2: ['fff', 'ggg']}
>>>>>>>>>> I want to combine all values into a single list -
>>>>>>>>>>>>> ans = ['aaa', 'bbb', 'ccc', 'fff', 'ggg']
>>>>>>>>>> I can do this -
>>>>>>>>>>>>> a = []
>>>>>>>> for v in d.values():
>>>>> ... a.extend(v)
>>>>> ...
>>>>>>>> a
>>>>> ['aaa', 'bbb', 'ccc', 'fff', 'ggg']
>>>>>>>>>> I can also do this -
>>>>>>>>>>>>> from itertools import chain
>>>>>>>> a = list(chain(*d.values()))
>>>>>>>> a
>>>>> ['aaa', 'bbb', 'ccc', 'fff', 'ggg']
>>>>>>>>>>>>>>>>>> Is there a simpler way?
>>>>>>>>>>>>> itertools.chain is a good option, as it scales well to arbitrary
>>>> numbers of lists (and you're guaranteed to iterate over them all just
>>>> once as you construct the list). But if you know that the lists aren't
>>>> too large or too numerous, here's another method that works:
>>>>>>>>>>> sum(d.values(), [])
>>>> ['aaa', 'bbb', 'ccc', 'fff', 'ggg']
>>>>>>>> It's simply adding all the lists together, though you have to tell it
>>>> that you don't want a numeric summation.
>>>>>> If you code is performance sensitive do not use sum() as it creates lots of tmp list that are deleted.
>>>>>> I have an outstanding ticket at work to replace all use of sum() on lists as when we profiled it
>>> stands out as a slow operation. We have a lots of list of list that we need to flatten.
>>>>> I think that 'sum' uses '__add__' but not '__iadd__'.
>>>> If it copied the given value, if present, and then used '__iadd__', if
>> present, wouldn't that speed it up?
>> It's hardly relevant. If you care about speed, use chain(), like
> everyone's been saying all along :)

Not everyone. I believe (*) that __iadd__ is faster, and again, you can
spell it
functools.reduce(operator.iconcat, list_of_lists, [])
(*) Actual measurements left as an exercise ;)