Question

我的代码到目前为止，但由于我迷失了它并没有做任何接近我想要它做的事情：

vowels = 'a','e','i','o','u','y'
#Consider 'y' as a vowel

input = input("Enter a sentence: ")

words = input.split()
if vowels == words[0]:
    print(words)

所以对于这样的输入：

"this is a really weird test"

我希望它只打印：

this, is, a, test

因为它们只包含1个元音。

Answer 1

试试这个：

vowels = set(('a','e','i','o','u','y'))

def count_vowels(word):
    return sum(letter in vowels for letter in word)

my_string = "this is a really weird test"

def get_words(my_string):
    for word in my_string.split():
        if count_vowels(word) == 1:
            print word

结果：

>>> get_words(my_string)
this
is
a
test

Answer 2

这是另一种选择：

import re

words = 'This sentence contains a bunch of cool words'

for word in words.split():
    if len(re.findall('[aeiouy]', word)) == 1:
        print word

输出：

This
a
bunch
of
words

Answer 3

您可以将所有元音翻译成单个元音并计算该元音：

import string
trans = string.maketrans('aeiouy','aaaaaa')
strs = 'this is a really weird test'
print [word for word in strs.split() if word.translate(trans).count('a') == 1]

Answer 4

>>> s = "this is a really weird test"
>>> [w for w in s.split() if len(w) - len(w.translate(None, "aeiouy")) == 1]
['this', 'is', 'a', 'test']

不确定是否需要没有元音的单词。如果是，请将== 1替换为< 2

Answer 5

如果您检查了下一个字符是空格，则可以使用一个for循环将子字符串保存到字符串数组中。他们为每个子串，检查是否只有一个a，e，i，o，u（元音），如果是，则添加到另一个数组

aFTER THAT，from另一个数组，用空格和逗号

连接所有字符串

Answer 6

试试这个：

vowels = ('a','e','i','o','u','y')
words = [i for i in input('Enter a sentence ').split() if i != '']
interesting = [word for word in words if sum(1 for char in word if char in vowel) == 1]

Answer 7

我在这里找到了很多不错的代码，我想展示我的丑陋代码：

v = 'aoeuiy'
o = 'oooooo'

sentence = 'i found so much nice code here'

words = sentence.split()

trans = str.maketrans(v,o)

for word in words:
    if not word.translate(trans).count('o') >1:
        print(word)

Answer 8

^{我发现你缺乏正则表达式令人不安。}

这是一个普通的正则表达式解决方案(ideone)：

import re

str = "this is a really weird test"

words = re.findall(r"\b[^aeiouy\W]*[aeiouy][^aeiouy\W]*\b", str)

print(words)

如何打印只有1个元音的单词？

8 个答案: