返回的Python字典

Question

I'm trying to create dictionary with the count of each letter appearing in a list of words. The method count_letters_v2(word_list) , print the letters but does not return the dictionary. Why? And how do I fix this? Thank you.

def count_letters_v2(word_list):
    count = {}
    for word in word_list:
        print word
        for letter in word:
            print letter
            if letter not in count:
                count[letter] = 1
            else:
                count[letter] = count[letter] + 1
    return count


def main():
    count_letters_v2(['a','short','list','of','words'])


if __name__ == '__main__':
    main()

Answer 1

It does return the dictionary. That's what return count does. Do you want to print out the dictionary? Then change main() to

def main():
    print count_letters_v2(['a','short','list','of','words'])

---

For the record, there's a Counter object (a subclass of dict so can do all the same things) in the standard library that will do this all for you.

from collections import Counter
def count_letters_v3(word_list):
    return Counter(''.join(word_list))

print count_letters_v3(['a','short','list','of','words'])

Output:

Counter({'a': 1,
         'd': 1,
         'f': 1,
         'h': 1,
         'i': 1,
         'l': 1,
         'o': 3,
         'r': 2,
         's': 3,
         't': 2,
         'w': 1})

Answer 2

As said, you code works but you didn't do anytinhg with the return value of the function.

That said, I can think of some improvements though. First, the get method of dict will make the case of a new letter cleaner, setting the default value to 0 :

...
count = {}
for word in word_list:
    for letter in word:
        count[letter] = count.get(letter, 0) + 1
...

Otherwise, you can use a Counter object from collections:

from collections import Counter

def count_letters_v2(word_list):
    count = Counter()
    for word in word_list:
       count.update(word)
    return count

If you have a very long list of words, you shouldn't use str.join since it builds new string. The chain_from_iterable method of itertools module will chain the letters for free:

from collections import Counter
from itertools import chain

def count_letters_v2(word_list):
    return Counter(chain.from_iterable(word_list))