CODEWITHSUNDEEP
pythonlistcharacter
D1X
D1X

Reputation: 5444

Using alphabet as counter in a loop

I am looking for the most efficient way to count the number of letters in a list. I need something like

word=[h e l l o]

for i in alphabet:
   for j in word:
      if j==i:
         ## do something

Where alphabet should be the spanish alphabet, that is the english alphabet including the special character 'ñ'.

I have thought about creating a list of pairs in the form of [[a, 0], [b,1], ...] but I suppose there is a more efficient/clean way.

Upvotes: 1

Views: 823

Answers (2)

Padraic Cunningham
Padraic Cunningham

Reputation: 180391

It is not actually a dupe as you want to filter to only count characters from a certain set, you can use a Counter dict to do the counting and a set of allowed characters to filter by:

word = ["h", "e", "l", "l", "o"]

from collections import Counter
from string import ascii_lowercase

# create a set of the characters you want to count.
allowed = set(ascii_lowercase + 'ñ')

# use a Counter dict to get the counts, only counting chars that are in the allowed set.
counts = Counter(s for s in word if s in allowed)

If you actually just want the total sum:

total = sum(s in allowed for s in word)

Or using a functional approach:

total = sum(1 for _ in filter(allowed.__contains__, word))

Using filter is going to be a bit faster for any approach:

In [31]: from collections import Counter
    ...: from string import ascii_lowercase, digits
    ...: from random import choice
    ...: 

In [32]: chars = [choice(digits+ascii_lowercase+'ñ') for _ in range(100000)]

In [33]: timeit Counter(s for s in chars if s in allowed)

100 loops, best of 3: 36.8 ms per loop


In [34]: timeit Counter(filter(allowed.__contains__, chars))
10 loops, best of 3: 31.7 ms per loop

In [35]: timeit sum(s in allowed for s in chars)
10 loops, best of 3: 35.4 ms per loop

In [36]: timeit sum(1 for _ in filter(allowed.__contains__, chars))

100 loops, best of 3: 32 ms per loop

If you want a case insensitive match, use ascii_letters and add 'ñÑ':

from string import ascii_letters

allowed = set(ascii_letters+ 'ñÑ')

Upvotes: 2

unwind
unwind

Reputation: 399793

This is pretty easy:

import collections
print collections.Counter("señor")

This prints:

Counter({'s': 1, 'r': 1, 'e': 1, '\xa4': 1, 'o': 1})

Upvotes: 0

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