在循環中使用字母作爲計數器

Question

我正在尋找計算列表中字母數量的最有效方法。我需要這樣的東西

word=[h e l l o] 

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

凡字母應該是西班牙字母，即包括特殊字符「N」英文字母。

我曾想過以[[a，0]，[b，1]，...]的形式創建對列表，但我想有更高效/乾淨的方式。

Answer 1

它實際上不是一個傻瓜，你要進行過濾，只算從一組特定的字符，你可以使用Counter字典做計數和一組允許字符過濾：

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) 

如果你其實只是想總和：

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

或使用功能的方法：

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

使用過濾將是一個更快一點對於任何方法：

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 

如果您想要不區分大小寫的匹配項，請使用ascii_letters並添加'ñÑ'：

from string import ascii_letters 

allowed = set(ascii_letters+ 'ñÑ') 

Answer 2

這是很容易的：

import collections 
print collections.Counter("señor") 

此打印：

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