擅长:python、mysql、java
<p>要最小限度地更改代码:</p>
<pre><code>leastFreq = [(counta, 'a'),(counte, 'e'),(counti, 'i'),(counto, 'o'),(countu, 'u')]
leastvowel = min(leastFreq)[1]
</code></pre>
<p>但是你应该改用<code>collections.Counter</code></p>
<pre><code>from collections import Counter
text = input("Enter text: ")
c = Counter(character for character in text if character in 'aeiou') #get counts of vowels
least_frequent = c.most_common()[-1]
</code></pre>
<p><code>least_frequent</code>将是一个类似于<code>('a', 1)</code>的元组</p>
<p>编辑:如果您想要所有最常用的项目,可以使用<code>itertools.groupby</code></p>
<pre><code>lestfreq=list(next(itertools.groupby(c.most_common()[::-1], key=lambda x:x[1]))[1])
</code></pre>
<p>这看起来很复杂,但它所说的只是按排序顺序获取一个列表,并获取所有具有相同第二个值的元组,然后将它们放入一个列表中。你知道吗</p>