retrieve input of collections.Counter output

Question

Not sure if the title is correct but.

Lets say you have a list that would look like the output from a Counter object.

[(-3.0, 4), (-2.0, 1), (-1.0, 1), (0.0, 1), (1.0, 1), (2.0, 1), (3.0, 4)]

How could I go back and get the original list, as

[-3.0, -3.0, -3.0, -3.0, -2.0, -1.0, 0.0, 1.0, 2.0, 3.0, 3.0, 3.0, 3.0]

does the order matter? or can the list be sorted afterwards? — Jean-François Fabre, Jan 18 '18 at 16:16
@Jean-FrançoisFabre Nope. But it might do for someone else :) — Isbister, Jan 18 '18 at 16:17

score 3 · Accepted Answer · answered Jan 18 '18 at 16:06

3

list(Counter(dict(a)).elements())

Demo:

>>> from collections import Counter
>>> a = [(-3.0, 4), (-2.0, 1), (-1.0, 1), (0.0, 1), (1.0, 1), (2.0, 1), (3.0, 4)]
>>> list(Counter(dict(a)).elements())
[-3.0, -3.0, -3.0, -3.0, -2.0, -1.0, 0.0, 1.0, 2.0, 3.0, 3.0, 3.0, 3.0]

So if you actually do have a Counter, just ask it for its elements directly.

answered Jan 18 '18 at 16:06

Stefan Pochmann

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Python 3.6+ though. – Jeff Mercado Jan 18 '18 at 16:12
@JeffMercado Maybe, maybe not. – Stefan Pochmann Jan 18 '18 at 16:13
you're right! but what Jeff meant is that it doesn't preserve the order of the list. so it gives a different result in python < 3.6 where dicts are unordered. megets: `[0.0, 1.0, 2.0, 3.0, 3.0, 3.0, 3.0, -1.0, -3.0, -3.0, -3.0, -3.0, -2.0]` – Jean-François Fabre Jan 18 '18 at 16:14
well, the list can probably be sorted afterwards if order matters but order is the natural order of floats. – Jean-François Fabre Jan 18 '18 at 16:17
@Jean-FrançoisFabre Yeah I know, but I don't think it's clear whether order should be "preserved". I mean, he talks about the "original list", and who creates a Counter from a sorted list? Plus with 3.6+ it does preserve. – Stefan Pochmann Jan 18 '18 at 16:17
@Jean-FrançoisFabre Fair point, but that's just another reason to upgrade. :) Guido has promised that `dict` will from now on retain insertion order. OTOH, if the original list is sorted, we just need to wrap the `.elements()` call in sorted instead of `list`. – PM 2Ring Jan 18 '18 at 16:17
anyway I like that answer. – Jean-François Fabre Jan 18 '18 at 16:18
Me too. I was about to post it (but using `[*Counter(dict(a)).elements()]`) but I noticed that Stefan had ninja'd me. :) – PM 2Ring Jan 18 '18 at 16:20
@PM2Ring Ah yes, `[*...]`. Somehow I rarely think of that. Maybe that's my subconscious rejecting it for some reason :-) – Stefan Pochmann Jan 18 '18 at 16:23

score 3 · Answer 2 · answered Jan 18 '18 at 16:06

3

You can use the following nested comprehension:

lst = [(-3.0, 4), ..., (3.0, 4)]
[x for x, count in lst for _ in range(count)]
# [-3.0, -3.0, -3.0, -3.0, -2.0, -1.0, 0.0, 1.0, 2.0, 3.0, 3.0, 3.0, 3.0]

answered Jan 18 '18 at 16:06

user2390182

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score 2 · Answer 3 · answered Jan 18 '18 at 16:04

2

You can try this:

s = [(-3.0, 4), (-2.0, 1), (-1.0, 1), (0.0, 1), (1.0, 1), (2.0, 1), (3.0, 4)]
final_s = [i for b in [[a]*b for a, b in s] for i in b]

Output:

[-3.0, -3.0, -3.0, -3.0, -2.0, -1.0, 0.0, 1.0, 2.0, 3.0, 3.0, 3.0, 3.0]

answered Jan 18 '18 at 16:04

Ajax1234

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retrieve input of collections.Counter output

3 Answers3