x in 'aeiou' and y in 'aeiou'，有没有更加优雅的写法？

class Solution:
    def vowelStrings(self, words: List[str], queries: List[List[int]]) -> List[int]:
        acc_arr = list(accumulate(words, lambda acc, word: acc + (word[0] in 'aeiou' and word[-1] in 'aeiou'), initial=0))
        
        res = []
        for l, r in queries:
            res.append(acc_arr[r + 1] - acc_arr[l])
        return res

我觉得 word[0] in 'aeiou' and word[-1] in 'aeiou'不够优雅，有没有好的写法？

Supplement 1 · Apr 3, 2023

优化后的代码如下：

class Solution:
    def vowelStrings(self, words: List[str], queries: List[List[int]]) -> List[int]:
        acc_arr = list(accumulate(words, lambda acc, word: acc + all(c in 'aeiou' for c in {word[0], word[-1]}), initial=0))
        return [acc_arr[r + 1] - acc_arr[l] for l, r in queries]

5 replies • 2023-05-29 16:49:07 +08:00

NessajCN

Apr 3, 2023

你重点关注错了
word[0] in 'aeiou' and word[-1] in 'aeiou' 优不优雅无所谓的，对性能没啥影响
但是下面的 for append 不怎么优雅，建议用 list comprehension 或 map

JasonLaw

Apr 3, 2023 via iPhone

@NessajCN #1 对，下面的 for loop 可以替换为 lost comprehension

featureoverload

Apr 3, 2023

`all([c in 'aeiou' for c in (x, y)])`

JasonLaw

Apr 3, 2023

@featureoverload #3 谢谢，不过可以优化成`all(c in 'aeiou' for c in (x, y))`

XueXianqi

May 29, 2023

@featureoverload

`all([c in 'aeiou' for c in (x, y)])`看起来优雅，但是性能考虑得不够周全
Python 中有个叫“短路运算”的，`a and b`，如果 a 不成立，就不去运算 b 了，显然这里的 all 用得就不够优雅了
用 all 还需要注意的是：不能完全代替 and ，比如说：
`if a.hasattr("name") and a.name.startswith("A"):`

这里就不能用 all ，如果 a 没有 name 这个 attribute ，就会报错。