数一数字符串中的字母、数字和标点符号

Answer 1

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)

Answer 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)

Answer 3

对于更精简/更快的版本，还有

count=lambda l1，l2:sum（[1表示l1中的x，如果x表示l2中的x]）

例如：

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

使用一套应该会让你的速度有所提高。

另外，取决于字符串的大小，我认为您应该通过过滤器获得内存优势。