我正在寻找一种最有效的方法来计算列表中的字母数。我需要像
这样的东西word=[h e l l o]
for i in alphabet:
for j in word:
if j==i:
## do something
字母应该是西班牙语字母,即英文字母,包括特殊字符'ñ'。
我已经考虑过以[[a,0],[b,1],...]的形式创建一对对象列表,但我想有一种更有效/更干净的方式。
答案 0 :(得分:2)
它实际上并不是一个欺骗,因为你想要过滤只计算某个集合中的字符,你可以使用Counter dict进行计数,并使用一组允许的字符进行过滤:
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)
如果你真的只想要总和:
total = sum(s in allowed for s in word)
或使用功能性方法:
total = sum(1 for _ in filter(allowed.__contains__, word))
对于任何方法,使用过滤器会更快一些:
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
如果您想要不区分大小写的匹配项,请使用 ascii_letters 并添加'ñÑ'
:
from string import ascii_letters
allowed = set(ascii_letters+ 'ñÑ')
答案 1 :(得分:0)
这很简单:
import collections
print collections.Counter("señor")
打印:
Counter({'s': 1, 'r': 1, 'e': 1, '\xa4': 1, 'o': 1})