<h3><code>melt</code>+<code>groupby</code>:</h3>
<pre><code>v = df.melt(id_vars=['family','V1','V2'], value_name='gr')
w = v.loc[v.gr != 'NO']
x = w.groupby(['family', 'gr']).sum()
(x.V2 / x.V1).reset_index(name='calc(sumV2/sumV1)')
</code></pre>
<p/>
<pre><code> family gr calc(sumV2/sumV1)
0 A all 6.0
1 A m1 6.0
2 A m12 6.0
3 A m14 2.0
4 B all 4.0
5 B m12 4.5
6 B m15 4.5
7 B m2 4.5
8 B m3 2.0
9 B m9 2.0
</code></pre>
<hr/>
<p>一种类似于<a href="https://stackoverflow.com/a/53323971/3483203">this answer</a>的方法,但其优点是完全矢量化,避免了<code>apply</code></p>
<hr/>
<p>性能:</p>
<pre><code>a = np.random.randint(1, 1000, (1_000_000, 7))
df = pd.DataFrame(a, columns=['family', 'V1', 'V2', 'gr_0', 'gr_1', 'gr_2', 'gr_3'])
df[['gr_0', 'gr_1', 'gr_2', 'gr_3']] = df[['gr_0', 'gr_1', 'gr_2', 'gr_3']].astype(str)
%%timeit
v = df.melt(id_vars=['family','V1','V2'], value_name='gr')
w = v.loc[v.gr != 'NO']
x = w.groupby(['family', 'gr']).sum()
(x.V2 / x.V1).reset_index(name='calc(sumV2/sumV1)')
2.71 s ± 32.9 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%%timeit
df_new = (df.melt(id_vars=['family','V1','V2']).groupby(['family','value'])
.apply(lambda x: x.V2.sum()/x.V1.sum())
.reset_index(name='calc(sumV2/sumV1)'))
df_new = df_new[df_new.value != 'NO'].reset_index(drop=True)
5min 24s ± 3.35 s per loop (mean ± std. dev. of 7 runs, 1 loop each)
</code></pre>