groupby pandas之后过滤行

Question

我在熊猫里有一张桌子:

import pandas as pd

df = pd.DataFrame({
    'LeafID':[1,1,2,1,3,3,1,6,3,5,1],
    'pidx':[10,10,300,10,30,40,20,10,30,45,20],
    'pidy':[20,20,400,20,15,20,12,43,54,112,23],
    'count':[10,20,30,40,80,10,20,50,30,10,70],
    'score':[10,10,10,22,22,3,4,5,9,0,1]
})

LeafID  count       pidx     pidy   score
0   1       10           10        20     10
1   1       20           10        20     10
2   2       30          300       400     10
3   1       40           10        20     22
4   3       80           30        15     22
5   3       10           40        20      3
6   1       20           20        12      4
7   6       50           10        43      5
8   3       30           20        54      9
9   5       10           45       112      0
10  1       70           20        23      1

我想做一个groupby然后过滤出现pidx大于2的行.

也就是说,过滤行为pidx10和20.

我尝试使用,df.groupby('pidx').count()但它没有帮助我.同样对于那些行,我必须做0.4*计数+ 0.6*得分.

期望的输出是:

LeafID    count       pidx     pidy    final_score
   1       10           10        20
   1       20           10        20
   1       40           10        20
   6       50           10        43
   1       20           20        12
   3       30           20        54
   1       70           20        23

Answer 1

这是在执行groupby之后直接应用过滤器.在您提供的数据中,pidx的值仅为20,因此会被过滤掉.

df.groupby('pidx').filter(lambda x: len(x) > 2)

   LeafID  count  pidx  pidy
0       1     10    10    20
1       1     20    10    20
3       1     40    10    20
7       6     50    10    43

Answer 2

您可以value_counts与boolean indexing和一起使用isin：

df = pd.DataFrame({
    'LeafID':[1,1,2,1,3,3,1,6,3,5,1],
    'pidx':[10,10,300,10,30,40,20,10,30,45,20],
    'pidy':[20,20,400,20,15,20,12,43,54,112,23],
    'count':[10,20,30,40,80,10,20,50,30,10,70],
    'score':[10,10,10,22,22,3,4,5,9,0,1]
})
print (df)
    LeafID  count  pidx  pidy  score
0        1     10    10    20     10
1        1     20    10    20     10
2        2     30   300   400     10
3        1     40    10    20     22
4        3     80    30    15     22
5        3     10    40    20      3
6        1     20    20    12      4
7        6     50    10    43      5
8        3     30    30    54      9
9        5     10    45   112      0
10       1     70    20    23      1

s = df.pidx.value_counts()
idx = s[s>2].index
print (df[df.pidx.isin(idx)])
   LeafID  count  pidx  pidy  score
0       1     10    10    20     10
1       1     20    10    20     10
3       1     40    10    20     22
7       6     50    10    43      5

时间：

np.random.seed(123)
N = 1000000


L1 = list('abcdefghijklmnopqrstu')
L2 = list('efghijklmnopqrstuvwxyz')
df = pd.DataFrame({'LeafId':np.random.randint(1000, size=N),
                   'pidx': np.random.randint(10000, size=N),
                   'pidy': np.random.choice(L2, N),
                   'count':np.random.randint(1000, size=N)})
print (df)


print (df.groupby('pidx').filter(lambda x: len(x) > 120))

def jez(df):
    s = df.pidx.value_counts()
    return df[df.pidx.isin(s[s>120].index)]

print (jez(df))

In [55]: %timeit (df.groupby('pidx').filter(lambda x: len(x) > 120))
1 loop, best of 3: 1.17 s per loop

In [56]: %timeit (jez(df))
10 loops, best of 3: 141 ms per loop

In [62]: %timeit (df[df.groupby('pidx').pidx.transform('size') > 120])
10 loops, best of 3: 102 ms per loop

In [63]: %timeit (df[df.groupby('pidx').pidx.transform(len) > 120])
1 loop, best of 3: 685 ms per loop

In [64]: %timeit (df[df.groupby('pidx').pidx.transform('count') > 120])
10 loops, best of 3: 104 ms per loop

因为final_score您可以使用：

df['final_score'] = df['count'].mul(.4).add(df.score.mul(.6))