Unique lists from a list

Question

Given a list I need to return a list of lists of unique items. I'm looking to see if there is a more Pythonic way than what I came up with:

def unique_lists(l):
    m = {}
    for x in l:
        m[x] = (m[x] if m.get(x) != None else []) + [x]
    return [x for x in m.values()]    

print(unique_lists([1,2,2,3,4,5,5,5,6,7,8,8,9]))

Output:

[[1], [2, 2], [3], [4], [5, 5, 5], [6], [7], [8, 8], [9]]

Thank you for Counter, groupby and defaultdict solutions! I learned something new today. — Yuriy Zubarev, Mar 30 '12 at 04:21

John La Rooy · Accepted Answer · 2012-03-30T04:16:45.073

9

>>> L=[1,2,2,3,4,5,5,5,6,7,8,8,9]
>>> from collections import Counter
>>> [[k]*v for k,v in Counter(L).items()]
[[1], [2, 2], [3], [4], [5, 5, 5], [6], [7], [8, 8], [9]]

edited Mar 30 '12 at 04:16

answered Mar 30 '12 at 04:04

John La Rooy

295,403
53
369
502

1

I originally posted a solution that was one line shorter by simply using the list's built in count method, but gnibbler makes the excellent point that list.count() is O(n), making my algorithm O(n^2). +1 – Nolen Royalty Mar 30 '12 at 04:08
3

There's no reason that the creation of the `Counter` can't be folded into the list comprehension here: `[[k]*v for k, v in Counter(L).items()]`. – Karl Knechtel Mar 30 '12 at 04:13
1

My rule of thumb is to automatically elide temporaries that are only written once and read once, unless they represent significant complexity ("flat is better than nested"). – Karl Knechtel Mar 30 '12 at 05:49

score 2 · Answer 2 · answered Mar 30 '12 at 04:15

2

Using default dict.

>>> from collections import defaultdict
>>> b = defaultdict(list)
>>> a = [1,2,2,3,4,5,5,5,6,7,8,8,9]
>>> for x in a:
...     b[x].append(x)
...
>>> b.values()
[[1], [2, 2], [3], [4], [5, 5, 5], [6], [7], [8, 8], [9]]

answered Mar 30 '12 at 04:15

Amjith

22,626
14
43
38

score 0 · Answer 3 · answered Nov 08 '16 at 07:23

I find the build in function set() useful very often:

lst=[1,2,2,3,4,5,5,5,6,7,8,8,9]

def all_eq_elms(lst, elm):
    while True:
        try:
            yield lst.pop(lst.index(elm))
        except:
            break

[[e for e in all_eq_elms(lst,elm)] for elm in set(lst)]

Out[43]: [[1], [2, 2], [3], [4], [5, 5, 5], [6], [7], [8, 8], [9]]

Unique lists from a list

3 Answers3

Linked