Question

以下代码在每次当前字母为辅音而下一个字母为元音时，将字符串作为输入并将其拆分为字符串列表。这很慢，我该如何改善？谢谢

def count_syllables(word):
    word= list(word)
    vowels= {"a":1,"e":2,"i":3,"o":4,"u":5,"y":6,"j":7}
    s = word[0]
    syllables = [ ]
    count = 1
    l_max = len(word)
    while count != l_max :
        if word[count] not in vowels and word[count-1] in vowels:
            syllables.append(s)
            s = word[count]
            count += 1
        else:
            s += word[count]
            count += 1
    syllables.append(s)

    value1 = len(syllables)
    value2 = syllables[-1]
    return (value1, value2)

Answer 1

您可以尝试使用正则表达式

import re

def count_syllables(word):
    syllables = re.findall("[bcdfghjklmnpqrstvwxy]+[aeiouy]*", word)
    return len(syllables), syllables

Answer 2

尝试一下

def count_syl(word:str):
    vowels = set("eyuioa")
    syllables = []
    count = 0
    i = 0
    while i < len(word) - 1:
        if word[i] in vowels and word[i + 1] not in vowels:
            syb = word[count:i+1]
            count = i + 1
            syllables.append(syb)
        i += 1
    syllables.append(word[count:])
    return (len(syllables), syllables)

带条件将字符串拆分为多个字符串

2 个答案: