<p>使用<code>itertools.product</code>:</p>
<pre><code>from itertools import product
l1 = ['A', 'B']
l2 = ['C', 'D', 'E']
for c1, c2 in product(l1, l2):
df[f'{c1}{c2}'] = df[c1].mul(df[c2])
id A B C D E AC \
2017-12 93426 0.687377 -4.000753 -3.191796 0.235393 0.007100 -2.193967
2017-12 93428 0.240590 -4.000753 -3.191796 0.235393 0.007100 -0.767914
2017-12 93429 0.052937 -4.000753 -3.191796 0.235393 0.007100 -0.168964
2017-12 93434 0.910938 -4.000753 -3.191796 0.235393 0.007100 -2.907528
2017-12 93436 0.137670 -4.000753 -3.191796 0.235393 0.007100 -0.439415
2018-01 93426 3.362003 -2.997135 -2.029331 1.016955 0.011298 -6.822617
2018-01 93428 1.330341 -2.997135 -2.029331 1.016955 0.011298 -2.699702
2018-01 93429 1.579284 -2.997135 -2.029331 1.016955 0.011298 -3.204890
AD AE BC BD BE
2017-12 0.161804 0.004880 12.769587 -0.941749 -0.028405
2017-12 0.056633 0.001708 12.769587 -0.941749 -0.028405
2017-12 0.012461 0.000376 12.769587 -0.941749 -0.028405
2017-12 0.214428 0.006468 12.769587 -0.941749 -0.028405
2017-12 0.032407 0.000977 12.769587 -0.941749 -0.028405
2018-01 3.419006 0.037984 6.082179 -3.047951 -0.033862
2018-01 1.352897 0.015030 6.082179 -3.047951 -0.033862
2018-01 1.606061 0.017843 6.082179 -3.047951 -0.033862
</code></pre>
<hr/>
<p><strong>详细信息</strong>:</p>
<p><code>itertools.product</code>给出了两个列表的组合,因此我们迭代这些组合并创建列:</p>
<pre><code>list(product(l1, l2))
[('A', 'C'), ('A', 'D'), ('A', 'E'), ('B', 'C'), ('B', 'D'), ('B', 'E')]
</code></pre>
<p><strong>来自OC的有用编辑</strong></p>
<p>我使用的是Python3.4,必须使用.format函数</p>
<pre><code>df['{c1}{c2}'.format(c1=c1, c2=c2)]
</code></pre>