计算字符串中每个字母的频率

问题描述

这是来自 pyschools 的问题.

我确实做对了，但我猜会有更简单的方法.这是最简单的方法吗?

def countLetters(word):信函={}单词中的字母:字母字典[字母] = 0单词中的字母:字母字典[字母] += 1返回信函

这应该是这样的:

>>>countLetters('谷歌'){'e':1，'g':2，'l':1，'o':2}

解决方案

在 2.7+ 中:

导入集合字母 = collections.Counter('google')

更早的版本(2.5+，现在已经很古老了):

导入集合字母 = collections.defaultdict(int)单词中的字母:字母[字母] += 1

This is a question from pyschools.

I did get it right, but I'm guessing that there would be a simpler method. Is this the simplest way to do this?

def countLetters(word):
    letterdict={}
    for letter in word:
        letterdict[letter] = 0
    for letter in word:
        letterdict[letter] += 1
    return letterdict

This should look something like this:

>>> countLetters('google')
{'e': 1, 'g': 2, 'l': 1, 'o': 2}

解决方案

In 2.7+:

import collections
letters = collections.Counter('google')

Earlier (2.5+, that's ancient by now):

import collections
letters = collections.defaultdict(int)
for letter in word:
    letters[letter] += 1

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