Python: no. of occurences of each character in a string

Question

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how to get the number of occurrences of each character using python

What is the best way to obtain the count of each character in a string and store it(I'm using a dictionary for this - can this choice make a big difference?)? A couple of ways that I thought of:

1.

for character in string:
    if character in characterCountsDict:
        characterCountsDict[character] += 1
    else:
        characterCountsDict[character] = 1

2.

character = 0
while character < 127:
    characterCountsDict[str(unichr(character))] = string.count(str(unichr(character))
    character += 1

I think the second method is better... But is either of them good? Is there a much better way to do this?

score 10 · Answer 1 · answered Oct 01 '12 at 13:26

10

>>> from collections import Counter
>>> Counter("asdasdff")
Counter({'a': 2, 's': 2, 'd': 2, 'f': 2})

Note that you can use Counter object like a dict.

answered Oct 01 '12 at 13:26

defuz

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oh wait....for a 20MB string it's actually taking more time(30 sec) than the method 2(12 sec) ? – Jayanth Koushik Oct 01 '12 at 13:32
You generate an input string something like `"asdasd" * 200`, right? – defuz Oct 01 '12 at 13:44
I read a file into a string: striing = file.read() (the file is about 20MB) – Jayanth Koushik Oct 01 '12 at 13:47
Hmmm, interesting. Important question: what kind of data you are going to processed? Are you sure you only need the characters in the 0-127? – defuz Oct 01 '12 at 13:56
yes only characters in 0-127. The file I'm using for testing is the GPL duplicated a bunch of times. – Jayanth Koushik Oct 01 '12 at 13:58
In this case, your method is preferred. Just seems odd transformation `str()` – defuz Oct 01 '12 at 14:16
See https://gist.github.com/3812061 :) – defuz Oct 01 '12 at 14:20
1

Also, if you need only printable characters, just use `{c: string.count(c) for c in printable}` where `printable` is imported from module `string` – defuz Oct 01 '12 at 14:28

georg · Accepted Answer · 2012-10-01T14:56:26.463

2

If you're interested in the most efficient way, it appears to be like this:

from collections import defaultdict

def count_chars(s):
    res = defaultdict(int)
    for char in s:
        res[char] += 1
    return res

Timings:

from collections import Counter, defaultdict

def test_counter(s):
    return Counter(s)

def test_get(s):
    res = {}
    for char in s:
        res[char] = res.get(char, 0) + 1
    return res

def test_in(s):
    res = {}
    for char in s:
        if char in res:
            res[char] += 1
        else:
            res[char] = 1
    return res

def test_defaultdict(s):
    res = defaultdict(int)
    for char in s:
        res[char] += 1
    return res


s = open('/usr/share/dict/words').read()
#eof

import timeit

test = lambda f: timeit.timeit(f + '(s)', setup, number=10)
setup = open(__file__).read().split("#eof")[0]
results = ['%.4f %s' % (test(f), f) for f in dir() if f.startswith('test_')]
print  '\n'.join(sorted(results))

Results:

0.8053 test_defaultdict
1.3628 test_in
1.6773 test_get
2.3877 test_counter

edited Oct 01 '12 at 14:56

answered Oct 01 '12 at 14:51

georg

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thank you :) definitely overkilled the answer :) – Jayanth Koushik Oct 01 '12 at 15:49
1

@JayanthKoushik: ;) that wondered me since long time... that's why. – georg Oct 01 '12 at 16:30
Wow, I can't believe counter is performing so bad , since Counter is a dict – wim May 31 '13 at 07:45
@wim: Counter uses `self.get` when updating itself, so its performance should be similar to `test_get` – georg May 31 '13 at 08:46

Python: no. of occurences of each character in a string

2 Answers2