One liner for extend loop python

Question

I have the following code lines:

records = []

for future in futures:
  records.extends(future.result())

each future returns a list.

How can I write the above code but in one liner?

records = [future.result() for future in futures]

would result in a list inside list.

I have millions of records, i would rather not flat it after creating lists inside list

Have you tried `records.extends(future.result() for future in futures)`? — Akshat Mahajan, Jun 08 '17 at 17:49
I just got this answer below - but this way I need to declare records = [] in another line. this will make it 2 lines — Dejell, Jun 08 '17 at 17:49
@Dejell Don't assume there is always a reasonable one-liner for every loop. I find your original loop slightly clearer than the accepted oneliner (although I wouldn't object to it in a code review). — chepner, Jun 08 '17 at 18:17
@Dejell I updated my answer, check it and see if it works for you, I made a little research and I will apreciate your feedback — Damian Lattenero, Jun 08 '17 at 19:24

score 9 · Accepted Answer · answered Jun 08 '17 at 17:50

9

records = [r for future in futures for r in future.result()]

answered Jun 08 '17 at 17:50

Błotosmętek

inspectorG4dget · Answer 2 · 2017-06-08T18:04:32.870

2

There are many ways to do this:

Use itertools.chain:
records = list(itertools.chain.from_iterable(future.result() for future in futures))
Use the itertools consume recipe:
records = collections.deque((records.extend(future.result()) for future in futures), maxlen=0)
Use a throw-away list:
[records.extend(future.result()) for future in futures]. records will now have all the required content, and you will have temporarily made a list of Nones

You could also do functools.reduce(operator.add, (future.result() for future in futures)), but that wouldn't scale very well

edited Jun 08 '17 at 18:04

answered Jun 08 '17 at 17:49

inspectorG4dget

regarding the third suggestion - I will need to declare records in another line – Dejell Jun 08 '17 at 17:50
2

Never a great fan of using a list comprehension for side effects. Just use a loop - `[print(x) for x in iterable]` - ugh! Vote for number 1. Doesn't number 2 just discard everything? – AChampion Jun 08 '17 at 17:55
@AChampion: you're right. I was just trying to cover all the bases. Thanks for the bugreport on #2. Fixed now – inspectorG4dget Jun 08 '17 at 18:05
@AChampion You were right about the comment in my answer, I edited it to not to make the same mistake again – Damian Lattenero Jun 08 '17 at 19:17
@inspectorG4dget Thanks a lot for sharing, that helped me a lot to make a better answer! plus one for ya – Damian Lattenero Jun 08 '17 at 19:38

Damian Lattenero · Answer 3 · 2017-06-08T19:35:22.917

1

I think this is what you are looking for

import functools
records = functools.reduce(lambda res, future: res + future.result()), futures, [])

edited Jun 08 '17 at 19:35

answered Jun 08 '17 at 17:45

Damian Lattenero

but then I need to declare records in a different line. that's 2 lines: records = [] first – Dejell Jun 08 '17 at 17:48
This is an abuse of a list comprehension, using it for the side effect of its expression rather than actually using the list of `None`s it creates. – chepner Jun 08 '17 at 18:15
@chepner how could I improve it? – Damian Lattenero Jun 08 '17 at 18:17
@chepner, do you know a way to return the list inside the lambda? – Damian Lattenero Jun 08 '17 at 18:43
If you were going to use `reduce`, you would use it like shown in inspectorG4dget's answer. – chepner Jun 08 '17 at 18:54
@chepner I think I got it, I have to ask a question my self in this site haaha, but yes, it works just fine :) – Damian Lattenero Jun 08 '17 at 19:16

3 Answers3