擅长:python、mysql、java
<p>你可以这样做。你知道吗</p>
<pre><code>from itertools import groupby
A = [100,200,300,100,400,300]
B = [ 20, 65, 118, 10, 13,50]
C = ['dr1', 'dr2', 'dr3', 'dr4', 'dr5', 'dr6']
grps = sorted(zip(A, B, C), key=lambda x: (x[0], x[1]))
A, B, C = [], [] ,[]
for i, grp in groupby(grps, lambda x: x[0]):
subA, subB, subC = [], [] ,[]
for j in grp:
subA.append(j[0])
subB.append(j[1])
subC.append(j[2])
A.append(subA)
B.append(subB)
C.append(subC)
</code></pre>
<p>输出:</p>
<pre><code>>>> A
[[100, 100], [200], [300, 300], [400]]
>>> B
[[10, 20], [65], [50, 118], [13]]
>>> C
[['dr4', 'dr1'], ['dr2'], ['dr6', 'dr3'], ['dr5']]
</code></pre>