unique count in (list of smaller lists)

Question

Is there a pythonic way to count the elements in a list of lists preferably using collections?

lol = [[1,2,3],[4,2],[5,1,6]]

Out:

1: 2
2: 2
3: 1
4: 1
5: 1
6: 1

*preferably using collections*. Presumably you already looked and found [`collections.Counter()`](https://docs.python.org/2/library/collections.html#collections.Counter) then? What problems did you encounter making that work? — Martijn Pieters, Jun 05 '14 at 12:15
`sum(map(Counter, lol), Counter())`, short but slower compared to itertools version. — Ashwini Chaudhary, Jun 05 '14 at 12:19
@sundarnatarajサンダーナタラジ Sure, but also point out that this is less efficient in comparison to itertools version. — Ashwini Chaudhary, Jun 05 '14 at 12:23
@MartijnPieters I dont think this is a duplicate as the other question is asking for count of a single element at a time. Not all of them! — Abhishek Thakur, Jun 05 '14 at 12:29
@AbhishekThakur: you'll need to address 200 OK about that one; I voted to close as 'unclear what you are asking' because your question lacked any detail as to where you were stuck. — Martijn Pieters, Jun 05 '14 at 12:32
@AbhishekThakur Did you see Imran's [answer](http://stackoverflow.com/a/5828150/846892) there? None of the answers here add nothing new to it. — Ashwini Chaudhary, Jun 05 '14 at 12:32
@200OK, I think there's a case to be made that this is not a duplicate _of that question_. That question has an answer to this question, but that doesn't necessarily make this a duplicate of that question. However I am certain that this has been asked before. — senderle, Jun 05 '14 at 12:34
It's also almost a duplicate of [this question](http://stackoverflow.com/q/15151762/577088), and is an exact (imo) duplicate of [this question](http://stackoverflow.com/q/23098468/577088). — senderle, Jun 05 '14 at 12:39

sundar nataraj · Accepted Answer · 2014-06-05T13:32:51.683

3

from collections import Counter
import itertools
a= [[1,2,3],[4,2],[5,1,6]]

print Counter(itertools.chain(*a))

#output Counter({1: 2, 2: 2, 3: 1, 4: 1, 5: 1, 6: 1})

b=Counter(itertools.chain(*a))
for key,val in b.iteritems():
    print key,':',val

output:

Other way of doing this but less efficient compared to itertools( thanks to 200OK)

a= [[1,2,3],[4,2],[5,1,6]]
sum(map(Counter, a), Counter())
#output {1: 2, 2: 2, 3: 1, 4: 1, 5: 1, 6: 1}

edited Jun 05 '14 at 13:32

answered Jun 05 '14 at 12:15

sundar nataraj

8,524
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You don't need `list(...)` – sloth Jun 05 '14 at 12:17
`isinstance(Counter(), dict)` – Ashwini Chaudhary Jun 05 '14 at 12:27
@200OK what is above comment. i dint get it. – sundar nataraj Jun 05 '14 at 12:28
@sundarnatarajサンダーナタラジ `Counter` is a subclass of `dict`, so instead of `b=dict(Counter(itertools.chain(*a))) for ... in b` you can simply `for key,val in Counter(itertools.chain(*a))` without creating another `dict`. – sloth Jun 05 '14 at 13:01

score 0 · Answer 2 · answered Jun 05 '14 at 12:16

0

from collections import Counter
import itertools
lol = [[1,2,3],[4,2],[5,1,6]]
Counter(itertools.chain.from_iterable(lol))

Output

Counter({1: 2, 2: 2, 3: 1, 4: 1, 5: 1, 6: 1})

answered Jun 05 '14 at 12:16

Cory Kramer

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unique count in (list of smaller lists)

2 Answers2