Return multiple lists from comprehension in python

Question

Is there any way in Python to return multiple lists from a comprehension?

I want to do something to the effect of:

x,y = [i,(-1*j) for (i,j) in enumerate(range(10))]
# x = [0 .. 9]
# y = [0 .. -9]

that's a dumb example, but I'm just wondering if it's possible.

Probably but I can't think of a practical reason of why you would want to. — kylieCatt
– kylieCatt, Commented Mar 1, 2016 at 0:55
@IanAuld you cant see a practical reason for wanting to seperate your x's and y's ? (maybe for plotting in certain libs?) — Joran Beasley
– Joran Beasley, Commented Mar 1, 2016 at 0:57
It would probably be more readable and more maintainable to do it separately. There's no need to stick everything on one line. — kylieCatt
– kylieCatt, Commented Mar 1, 2016 at 0:59

Joran Beasley · Accepted Answer · 2016-03-01 00:57:36Z

2

x,y =zip(* [(i,(-1*j)) for (i,j) in enumerate(range(10))] )

you just unzip the list

xy = [(1,2),(3,4),(5,6)]
x,y = zip(*xy)
# x = (1,3,5)
# y = (2,4,6)

edited Mar 1, 2016 at 0:57

answered Mar 1, 2016 at 0:53

Joran Beasley

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5 Comments

kylieCatt Over a year ago

The return types of both of these are tuples not lists

wim Over a year ago

it will give tuples, not lists, but it's the right idea

kylieCatt Over a year ago

Sort of. A major difference being mutable != immutable.

Joran Beasley Over a year ago

maybe /.... usually people don't need mutable for this type of problem ... although its easy enough to make it mutable ...

ShadowRanger Over a year ago

Just wrap with map(list, ), as in, x, y = map(list, zip(...)) and it will produce lists, not a big deal.

dawg · Accepted Answer · 2016-03-01 01:14:35Z

1

You can skip the list comprehension:

>>> x,y=range(0,10), range(0,-10,-1)
>>> x
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> y
[0, -1, -2, -3, -4, -5, -6, -7, -8, -9]

Or you could do:

>>> x,y=map(list, zip(*[(e,-e) for e in range(10)]))

edited Mar 1, 2016 at 1:14

answered Mar 1, 2016 at 1:03

dawg

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1 Comment

ShadowRanger Over a year ago

Note: range only returns list on Py2; on Py3, it's a special range object that would need to be wrapped in the list constructor to make an actual, mutable list. But yes, if the problem is as simple as given, best to avoid the complexity of generating paired values and unpairing them when it's easier to generate unpaired values directly. And if mutability is not needed, using the raw Py3 range would often be better than a mutable list (Py2 xrange isn't quite as good as a drop-in replacement, even if you don't need mutability).

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