计算没有元音的字符串长度(Count length of string without vowels)

计算没有元音的字符串长度(Count length of string without vowels)

我有一个字符串： -

a = 'abcderui'

我想计算字符串的长度减去元音。 实现这一目标的最佳目标是什么？

目前我在做： -

a = list(a) for i in a: if i in vowels: continue count +=1

I've a string:-

a = 'abcderui'

I want to count the length of the string a minus the vowels. What would be the fasstest wat to implement that?

Currently I'm doing:-

a = list(a) for i in a: if i in vowels: continue count +=1

最满意答案

如果字符串不是很大，那么str.translate是最好的选择：

>>> a = 'abcderui' >>> len(a.translate(None, 'aeiou')) 4

时序：

>>> a = 'abcderui'*100 >>> %timeit len(a.translate(None, 'aeiou')) 1000000 loops, best of 3: 1.86 µs per loop >>> %timeit sum(1 for c in a if c not in 'aeiou') 10000 loops, best of 3: 53.2 µs per loop >>> %timeit len(nonvowels.findall(a)) 10000 loops, best of 3: 65.3 µs per loop >>> %timeit len(vowels.sub('', a)) 10000 loops, best of 3: 72 µs per loop

If strings are not huge then str.translate is the best option:

>>> a = 'abcderui' >>> len(a.translate(None, 'aeiou')) 4

Timings:

>>> a = 'abcderui'*100 >>> %timeit len(a.translate(None, 'aeiou')) 1000000 loops, best of 3: 1.86 µs per loop >>> %timeit sum(1 for c in a if c not in 'aeiou') 10000 loops, best of 3: 53.2 µs per loop >>> %timeit len(nonvowels.findall(a)) 10000 loops, best of 3: 65.3 µs per loop >>> %timeit len(vowels.sub('', a)) 10000 loops, best of 3: 72 µs per loop