擅长:python、mysql、java
<p>解决这个问题的合理方法是使用字典来存储每个名字的职业列表。例如,如果您有以下设置:</p>
<pre><code>data = [("Rolt12", "musician"), ("Rolt1", "dancer"), ("Rolt1", "actor"), ("Rolt14", "singer")]
</code></pre>
<p>您可以使用以下代码为每个姓名创建职业列表:</p>
<pre><code>occupations = {}
for name, occupation in data:
if name not in occupations:
occupations[name] = []
occupations[name].append(occupation)
</code></pre>
<p>或者,更习惯地说:</p>
<pre><code>import collections
occupations = collections.defaultdict(list)
for name, occupation in data:
occupations[name].append(occupation)
</code></pre>
<p>然后,您可以迭代字典以打印所需的数据:</p>
<pre><code>for name, all_occupations in occupations.items():
occupations_string = "; ".join(all_occupations)
print(f"{name} is a {occupations_string}")
</code></pre>