How do I get a count of each of the words in my list?

Question

I have a list:

 X = ['raz', 'dwa', 'raz', 'trzy', 'dwa', 'raz', 'trzy', 'cztery']

and want Output:

{'cztery': 1, 'dwa': 2, 'raz': 3, 'trzy': 2}

@Brian Can you write an answer demonstrating that? – tacaswell Oct 10 '13 at 18:42 — tacaswell, Oct 10 '13 at 18:42
@tcaswell: I just wrote an answer using `groupby`. – Matthias Oct 10 '13 at 18:51 — Matthias, Oct 10 '13 at 18:51

score 4 · Answer 1 · 2013-10-10T18:54:58.287

4

You can use a dictionary comprehension:

>>> lst = ['raz', 'dwa', 'raz', 'trzy', 'dwa', 'raz', 'trzy', 'cztery']
>>> {x:lst.count(x) for x in set(lst)}
{'raz': 3, 'cztery': 1, 'dwa': 2, 'trzy': 2}
>>>

or, you can use collections.Counter:

>>> from collections import Counter
>>> Counter(lst)
Counter({'raz': 3, 'dwa': 2, 'trzy': 2, 'cztery': 1})
>>>

The second solution is probably what you want because it does the same thing as the comprehension, is more efficient, and also uses the Counter class, which comes with a lot of great tools (e.g. most_common and subtract).

edited Oct 10 '13 at 18:54

answered Oct 10 '13 at 18:41

@Downvoter - Please leave a reason you downvoted. I care a lot about the quality of my work. Hence, if you have found something wrong with my post, please let me know so I can address it. – Oct 10 '13 at 20:32

score 1 · Answer 2 · answered Oct 10 '13 at 18:45

1

Another method using stdlib:

In [6]: from collections import Counter
In [7]: l = ['raz', 'dwa', 'raz', 'trzy', 'dwa', 'raz', 'trzy', 'cztery']
In [8]: dict(Counter(l).items())
Out[8]: {'cztery': 1, 'dwa': 2, 'raz': 3, 'trzy': 2}

answered Oct 10 '13 at 18:45

Demian Brecht

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Err, downvoted for what reason? Gotta love the SE community at times :P – Demian Brecht Oct 10 '13 at 23:04

score 0 · Answer 3 · answered Oct 10 '13 at 18:50

0

Just demonstrating the version with groupby that Brian suggested.

import itertools
data = ['raz', 'dwa', 'raz', 'trzy', 'dwa', 'raz', 'trzy', 'cztery']
print({k:len(list(v)) for k, v in itertools.groupby(sorted(data))})

answered Oct 10 '13 at 18:50

Matthias

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This is `O(NLogN)` in complexity. – Ashwini Chaudhary Oct 10 '13 at 18:58
It was a demonstration. I didn't say it was a good solution. :-) – Matthias Oct 10 '13 at 19:12

How do I get a count of each of the words in my list?

3 Answers3