Python 如何找出列表中以元音开头的单词？
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/12833512/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
How to find out the words begin with vowels in a list?
提问by user1718826
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)
My code doesn't right, and it will print out all the words in the original list.
我的代码不对，它会打印出原始列表中的所有单词。
采纳答案by John La Rooy
words = ['apple', 'orange', 'pear', 'milk', 'otter', 'snake','iguana','tiger','eagle']
for word in words:
if word[0] in 'aeiou':
print(word)
You can also use a list comprehension like this
您还可以使用这样的列表理解
words_starting_with_vowel = [word for word in words if word[0] in 'aeiou']
回答by st0le
if vowel [0]=='a,e':
words.append(vowel)
You are appending it to the original list here. It should be your vowellist.
您将其附加到此处的原始列表中。它应该是你的vowel清单。
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)
Using List comprehension
使用列表理解
vowel = [word for word in words if word[0] in "aeiou"]
Using filter
使用 filter
vowel = filter(lambda x : x[0] in "aeiou",words)
回答by K Z
Here is a one-liner answer with list comprehension:
这是一个带有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']
回答by georg
Good python reads almost like natural language:
好的 Python 读起来几乎像自然语言：
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)]
The problem with w[0]solution is that it doesn't work with empty words (doesn't matter in this particular example, but important in real-life tasks like parsing user input).
w[0]解决方案的问题在于它不适用于空词（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）。
