根据列值选择行

Man*_*wal 3 slice pandas

我有一个像这样的数据框

data = {'ID': [1,2,3,4,5,6,7,8,9],
       'Doc':['Order','Order','Inv','Order','Order','Shp','Order', 'Order','Inv'],
       'Rep':[101,101,101,102,102,102,103,103,103]}
frame = pd.DataFrame(data)


    Doc     ID  Rep
0   Order   1   101
1   Order   2   101
2   Inv     3   101
3   Order   4   102
4   Order   5   102
5   Shp     6   102
6   Order   7   103
7   Order   8   103
8   Inv     9   103
Run Code Online (Sandbox Code Playgroud)

现在我想为 Rep 选择仅 Doc 类型为 Inv 的行。

我想要一个数据框作为

    Doc     ID  Rep
0   Order   1   101
1   Order   2   101
2   Inv     3   101
6   Order   7   103
7   Order   8   103
8   Inv     9   103
Run Code Online (Sandbox Code Playgroud)

所有代表都会有 Doc 类型的订单,所以我试图做这样的事情

frame[frame.Rep == frame.Rep[frame.Doc == 'Inv']] 
Run Code Online (Sandbox Code Playgroud)

但我收到一个错误

ValueError:只能比较标记相同的系列对象

jez*_*ael 6

您可以使用两次boolean indexing- 首先Rep按条件获取所有行,然后按isin以下方式获取所有行:

a = frame.loc[frame['Doc'] == 'Inv', 'Rep']
print (a)
2    101
8    103
Name: Rep, dtype: int64

df = frame[frame['Rep'].isin(a)]
print (df)
     Doc  ID  Rep
0  Order   1  101
1  Order   2  101
2    Inv   3  101
6  Order   7  103
7  Order   8  103
8    Inv   9  103
Run Code Online (Sandbox Code Playgroud)

解决方案query

a = frame.query("Doc == 'Inv'")['Rep']
df = frame.query("Rep in @a")
print (df)
     Doc  ID  Rep
0  Order   1  101
1  Order   2  101
2    Inv   3  101
6  Order   7  103
7  Order   8  103
8    Inv   9  103
Run Code Online (Sandbox Code Playgroud)

时间

np.random.seed(123)
N = 1000000
L = ['Order','Shp','Inv']
frame = pd.DataFrame({'Doc': np.random.choice(L, N,  p=[0.49, 0.5, 0.01]),
                     'ID':np.arange(1,N+1),
                     'Rep':np.random.randint(1000, size=N)})
print (frame.head())

     Doc  ID  Rep
0    Shp   1   95
1  Order   2  147
2  Order   3  282
3    Shp   4   82
4    Shp   5  746

In [204]: %timeit (frame.groupby('Rep').filter(lambda x: 'Inv' in x['Doc'].values))
1 loop, best of 3: 250 ms per loop

In [205]: %timeit (frame[frame['Rep'].isin(frame.loc[frame['Doc'] == 'Inv', 'Rep'])])
100 loops, best of 3: 17.3 ms per loop

In [206]: %%timeit
     ...: a = frame.query("Doc == 'Inv'")['Rep']
     ...: frame.query("Rep in @a")
     ...: 
100 loops, best of 3: 14.5 ms per loop
Run Code Online (Sandbox Code Playgroud)

编辑:

谢谢John Galt的好建议:

df = frame.query("Rep in %s" % frame.query("Doc == 'Inv'")['Rep'].tolist()) 
print (df)
     Doc  ID  Rep
0  Order   1  101
1  Order   2  101
2    Inv   3  101
6  Order   7  103
7  Order   8  103
8    Inv   9  103
Run Code Online (Sandbox Code Playgroud)