擅长:python、mysql、java
<p>使用<code>itertools.groupby</code></p>
<p><strong>例如:</strong></p>
<pre><code>from itertools import groupby
issuelist = [["k1","v1"],["k2","v2"],["k3","v1"],["k4","v2"],["k5","v2"],["k6","v3"],["k7","v1"]]
result = {k: [i[0] for i in v ]for k, v in groupby(sorted(issuelist, key=lambda x: x[1]), lambda x: x[1])}
print(result)
print(list(result.values()))
</code></pre>
<p><strong>输出:</strong></p>
<pre><code>{'v1': ['k1', 'k3', 'k7'], 'v2': ['k2', 'k4', 'k5'], 'v3': ['k6']}
[['k1', 'k3', 'k7'], ['k2', 'k4', 'k5'], ['k6']]
</code></pre>