<p>有一种方法-</p>
<pre><code>a = df.col2.values
df.col2 = np.take(['others','PC','Ln'], np.in1d(a,lis1) + 2*np.in1d(a,lis2))
</code></pre>
<p>分步运行示例-</p>
<pre><code># Input dataframe
In [206]: df
Out[206]:
col1 col2
0 1 A
1 2 A
2 3 S
3 4 O
4 5 S
5 6 P
# Extract out col2 values
In [207]: a = df.col2.values
# Form an indexing array based on where we have matches in lis1 or lis2 or neither
In [208]: idx = np.in1d(a,lis1) + 2*np.in1d(a,lis2)
In [209]: idx
Out[209]: array([1, 1, 2, 2, 2, 0])
# Index into a list of new strings with those indices
In [210]: newvals = np.take(['others','PC','Ln'], idx)
In [211]: newvals
Out[211]:
array(['PC', 'PC', 'Ln', 'Ln', 'Ln', 'others'],
dtype='|S6')
# Finally assign those into col2
In [212]: df.col2 = newvals
In [213]: df
Out[213]:
col1 col2
0 1 PC
1 2 PC
2 3 Ln
3 4 Ln
4 5 Ln
5 6 others
</code></pre>
<p>运行时测试-</p>
<pre><code>In [251]: df=pd.DataFrame({'col1':[1,2,3,4,5,6], 'col2':list('AASOSP')})
In [252]: df = pd.concat([df]*10000).reset_index(drop=True)
In [253]: lis1
Out[253]: ['A']
In [254]: lis2
Out[254]: ['S', 'O']
In [255]: def in1d_based(df, lis1, lis2):
...: a = df.col2.values
...: return np.take(['others','PC','Ln'], np.in1d(a,lis1) + 2*np.in1d(a,lis2))
...:
# @jezrael's soln
In [256]: %timeit np.where(df.col2.isin(lis1),'PC', np.where(df.col2.isin(lis2),'Ln','others'))
100 loops, best of 3: 3.78 ms per loop
In [257]: %timeit in1d_based(df, lis1, lis2)
1000 loops, best of 3: 1.89 ms per loop
</code></pre>