Question

words = ['apple', 'orange', 'pear', 'milk', 'otter', 'snake', 'iguana',
         'tiger', 'eagle']
vowel=[]
for vowel in words:
    if vowel [0]=='a,e':
        words.append(vowel)
    print (words)

我的代码不对，它会打印出原始列表中的所有单词。

Answer 1

words = ['apple', 'orange', 'pear', 'milk', 'otter', 'snake','iguana','tiger','eagle']
for word in words:
    if word[0] in 'aeiou':
        print(word)

你也可以使用像这样的列表理解

words_starting_with_vowel = [word for word in words if word[0] in 'aeiou']

Answer 2

以下是列表理解的单行答案：

>>> print [w for w in words if w[0] in 'aeiou']
['apple', 'orange', 'otter', 'iguana', 'eagle']

Answer 3

好的python几乎就像自然语言：

vowel = 'a', 'e', 'i', 'o', 'u'
words = 'apple', 'orange', 'pear', 'milk', 'otter', 'snake', 'iguana', 'tiger', 'eagle'
print [w for w in words if w.startswith(vowel)]

w[0]解决方案的问题在于它不适用于空单词（在此特定示例中无关紧要，但在解析用户输入等现实任务中很重要）。

Answer 4

if vowel [0]=='a,e':
        words.append(vowel)

您将此处附加到原始列表中。它应该是您的vowel列表。

words = ['apple', 'orange', 'pear', 'milk', 'otter', 'snake','iguana','tiger','eagle']
vowel=[]
for word in words:
    if word[0] in "aeiou":
        vowel.append(word)
print (vowel)

使用列表理解

vowel = [word for word in words if word[0] in "aeiou"]

使用filter

vowel = filter(lambda x : x[0] in "aeiou",words)

Answer 5


res = [] 
list_vowel = "aeiou"
for sub in words: 
   flag = False
   
   # checking for begin char 
   for ele in list_vowel: 
       if sub.startswith(ele): 
           flag = True
           break
   if flag: 
       res.append(sub) 

# printing result 
list_vowel = str(res)
print(list_vowel)```

如何在列表中找出以元音开头的单词？

5 个答案: