Python 如何过滤组s最大和最小行使用`transform`'

使用DataFrame.transform，其中2个掩码乘以Series.eq，按位OR—|链接:

g = df.groupby('decile')['score']
out = df[df['score'].eq(g.transform('min')) | df['score'].eq(g.transform('max'))]
print (out)
    id     score  decile
1    2  0.715189       6
2    3  0.602763       5
3    4  0.544883       4
5    6  0.645894       5
6    7  0.437587       2
9   10  0.383442       1
10  11  0.791725       7
11  12  0.528895       3
12  13  0.568045       4
13  14  0.925597       8
15  16  0.087129       0
16  17  0.020218       0
17  18  0.832620       7
19  20  0.870012       8
20  21  0.978618       9
22  23  0.461479       3
23  24  0.780529       6
24  25  0.118274       1
27  28  0.944669       9
29  30  0.414662       2

Details:

print (df.assign(min=df['score'].eq(g.transform('min')),
                 max=df['score'].eq(g.transform('max')),
                 both = lambda x: x['min'] | x['max']))

    id     score  decile    min    max   both
0    1  0.548814       4  False  False  False
1    2  0.715189       6   True  False   True
2    3  0.602763       5   True  False   True
3    4  0.544883       4   True  False   True
4    5  0.423655       2  False  False  False
5    6  0.645894       5  False   True   True
6    7  0.437587       2  False   True   True
7    8  0.891773       8  False  False  False
8    9  0.963663       9  False  False  False
9   10  0.383442       1  False   True   True
10  11  0.791725       7   True  False   True
11  12  0.528895       3  False   True   True
12  13  0.568045       4  False   True   True
13  14  0.925597       8  False   True   True
14  15  0.071036       0  False  False  False
15  16  0.087129       0  False   True   True
16  17  0.020218       0   True  False   True
17  18  0.832620       7  False   True   True
18  19  0.778157       6  False  False  False
19  20  0.870012       8   True  False   True
20  21  0.978618       9  False   True   True
21  22  0.799159       7  False  False  False
22  23  0.461479       3   True  False   True
23  24  0.780529       6  False   True   True
24  25  0.118274       1   True  False   True
25  26  0.639921       5  False  False  False
26  27  0.143353       1  False  False  False
27  28  0.944669       9   True  False   True
28  29  0.521848       3  False  False  False
29  30  0.414662       2   True  False   True

lambda函数的另一个 idea ，但解决方案很慢:

out = df[df.groupby('decile')['score'].transform(lambda x: x.eq(x.min()) | x.eq(x.max()))]

np.random.seed(0)
df = pd.DataFrame({'id' : range(1,30001),
                  'score' : np.random.uniform(size = 30000)})

df['decile'] = pd.qcut(df['score'], 5000, labels=False)


In [44]: %%timeit
    ...: g = df.groupby('decile')['score']
    ...: out = df[df['score'].eq(g.transform('min')) | df['score'].eq(g.transform('max'))]
    ...: 
3.27 ms ± 239 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

In [45]: %%timeit
    ...: df[df.groupby('decile')['score'].transform(lambda x: x.eq(x.min()) | x.eq(x.max()))]
    ...: 
1.45 s ± 12.2 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)