擅长:python、mysql、java
<p>与其定义一个修改全局变量的函数,不如创建一个直接返回解的函数。这样,它就不依赖于某个脆弱的全球国家。你知道吗</p>
<pre><code>def count_vowels(word):
vowels = {'a', 'e', 'i', 'o', 'u'} # I use a set here rather than a list because
# sets have very fast membership checks.
count = 0
for letter in word:
if letter in vowels:
count += 1
return count
</code></pre>
<p>然后你可以简单地做:</p>
<pre><code>num_vowels = count_vowels(name1)
</code></pre>
<hr/>
<p>可能值得注意的是<code>count_vowels</code>可以简化为一行:</p>
<pre><code>def count_vowels(word):
return sum(lett in {'a', 'e', 'i', 'o', 'u'} for lett in word)
</code></pre>