我正在嘗試獲取一個(gè)字符串并從超過 4 個(gè)字符的單詞中刪除元音。有沒有更有效的方法來編寫這段代碼？(1) 從字符串中創(chuàng)建一個(gè)數(shù)組。(2) 遍歷一個(gè)數(shù)組并從超過 4 個(gè)字符的字符串中刪除元音。(3) 將數(shù)組中的字符串連接到新字符串。謝謝！def abbreviate_sentence(sent): split_string = sent.split() for words in split_string: abbrev = [words.replace("a", "").replace("e", "").replace("i", "").replace("o", "").replace("u", "") if len(words) > 4 else words for words in split_string] sentence = " ".join(abbrev) return sentenceprint(abbreviate_sentence("follow the yellow brick road")) # => "fllw the yllw brck road"
2 回答
臨摹微笑
TA貢獻(xiàn)1982條經(jīng)驗(yàn) 獲得超2個(gè)贊
您可以避免使用外部for，因?yàn)槟呀?jīng)在內(nèi)部進(jìn)行了迭代。另一方面，您可以將多個(gè)replaces 替換為另一個(gè)列表推導(dǎo)式，該推導(dǎo)式將嵌套在現(xiàn)有推導(dǎo)式中。
# for words in split_string: <- This line is not required
vowels = 'aeiou'
abbrev = [''.join([x for x in words if x.lower() not in vowels]) if len(words) > 4 else words for words in split_string]
sentence = " ".join(abbrev)
return sentence
或者將字符串部分的形成抽象到一個(gè)新函數(shù)中，這可能會(huì)增加它的可讀性：
def form_word(words):
vowels = 'aeiou'
return ''.join([x for x in words if x.lower() not in vowels])
def abbreviate_sentence(sent):
split_string = sent.split()
abbrev = [form_word(words) if len(words) > 4 else words for words in split_string]
sentence = " ".join(abbrev)
return sentence
夢里花落0921
TA貢獻(xiàn)1772條經(jīng)驗(yàn) 獲得超6個(gè)贊
奧斯汀的解決方案或以下解決方案都應(yīng)該有效。我認(rèn)為兩者在計(jì)算上都不會(huì)比您現(xiàn)在擁有的效率高得多，因此我將重點(diǎn)放在可讀性和合理性上。
def abbreviate_sentence(sent):
abbrev = []
for word in sent.split():
if len(word) > 4:
abbrev.append(words.replace("a", "").replace("e", "").replace("i", "").replace("o", "").replace("u", ""))
else:
abbrev.append(word)
return " ".join(abbrev)
print(abbreviate_sentence("follow the yellow brick road"))
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