Counting letters numbers and punctuation in a string

Question

How can I write a program that counts letters, numbers and punctuation(separately) in a string?

What have you tried so far? Please show that you have put some effort into trying to answer this question yourself. — Greg Hewgill, Aug 06 '11 at 22:12

score 8 · Answer 1 · answered Jan 09 '13 at 06:42

8

For a slightly more condensed / faster version, there is also

count = lambda l1,l2: sum([1 for x in l1 if x in l2])

so for example:

count = lambda l1,l2: sum([1 for x in l1 if x in l2])

In [11]: s = 'abcd!!!'

In [12]: count(s,set(string.punctuation))                                                                                                      
Out[12]: 3

using a set should get you a speed boost somewhat.

also depending on the size of the string I think you should get a memory benefit over the filter as well.

answered Jan 09 '13 at 06:42

Eiyrioü von Kauyf

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I like this solution better. The top answer is a little more muddled by lambdas. – bozdoz May 27 '13 at 22:26
You don't need a list comprehension here, a plain generator expression is not only shorter but also more efficient. – Frerich Raabe Mar 12 '15 at 14:28
but not faster. generators are slower because they ... generate. hence why i said faster – Eiyrioü von Kauyf Mar 14 '15 at 06:38
even more concisely you could generate a regex for punctuation/numbers/letters - it would be faster than anything else here (depending on how they implemented the regex) – Eiyrioü von Kauyf Mar 14 '15 at 06:42

BrainStorm · Answer 2 · 2011-08-07T22:00:07.343

4

import string
a = "I'm not gonna post my homework as question on OS again, I'm not gonna..."

count = lambda l1, l2: len(list(filter(lambda c: c in l2, l1)))

a_chars =  count(a, string.ascii_letters)
a_punct = count(a, string.punctuation)

edited Aug 07 '11 at 22:00

answered Aug 06 '11 at 22:36

BrainStorm

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`sum` should be `len` I think. And also, there is parenthesis mismatch. – utdemir Aug 06 '11 at 23:44
you're right, i wrote in about 10 seconds 'cause i was in a hurry, fixed =P – BrainStorm Aug 07 '11 at 06:28
Also a note, in Python 3+, filter returns a filter object and it doesn't support len. Maybe `list(filter(...))` is a better for compatibility. – utdemir Aug 07 '11 at 10:48
@utdemir nice to know, I'm still too lazy to install and use Py3 but i know i have to ^^' – BrainStorm Aug 07 '11 at 22:01

score 2 · Answer 3 · answered Aug 06 '11 at 22:36

2

count_chars = ".arPZ"
string = "Phillip S. is doing a really good job."
counts = tuple(string.count(c) for c in count_chars)

print counts

(2, 2, 1, 1, 0)

answered Aug 06 '11 at 22:36

Niklas R

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score 1 · Answer 4 · answered Aug 06 '11 at 23:57

>>> import string
>>> import operator
>>> import functools
>>> a = "This, is an example string. 42 is the best number!"
>>> letters = string.ascii_letters
>>> digits = string.digits
>>> punctuation = string.punctuation
>>> letter_count = len(filter(functools.partial(operator.contains, letters), a))
>>> letter_count
36
>>> digit_count = len(filter(functools.partial(operator.contains, digits), a))
>>> digit_count
2
>>> punctuation_count = len(filter(functools.partial(operator.contains, punctuation), a))
>>> punctuation_count
3

http://docs.python.org/library/string.html

http://docs.python.org/library/operator.html#operator.contains

http://docs.python.org/library/functools.html#functools.partial

http://docs.python.org/library/functions.html#len

http://docs.python.org/library/functions.html#filter

score 0 · Answer 5 · answered Aug 06 '11 at 22:16

0

To loop over a string you can use a for-loop:

for c in "this is a test string with punctuation ,.;!":
    print c

outputs:

t
h
i
s
...

Now, all you have to do is count the occurrences...

answered Aug 06 '11 at 22:16

Fredrik Pihl

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a = "I'm not gonna post my homework as question on OS again." – BrainStorm Aug 06 '11 at 22:29

Counting letters numbers and punctuation in a string

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