我想按两列对我的数据框进行分组,然后对这些组中的聚合结果进行排序。
In [167]: df
Out[167]:
count job source
0 2 sales A
1 4 sales B
2 6 sales C
3 3 sales D
4 7 sales E
5 5 market A
6 3 market B
7 2 market C
8 4 market D
9 1 market E
In [168]: df.groupby(['job','source']).agg({'count':sum})
Out[168]:
count
job source
market A 5
B 3
C 2
D 4
E 1
sales A 2
B 4
C 6
D 3
E 7
我现在想在每个组中以降序对“count”列进行排序,然后只取前三行。得到类似的东西:
count
job source
market A 5
D 4
B 3
sales E 7
C 6
B 4
你也可以一次性完成,先做排序,然后用head取每组的前3个。
In[34]: df.sort_values(['job','count'],ascending=False).groupby('job').head(3)
Out[35]:
count job source
4 7 sales E
2 6 sales C
1 4 sales B
5 5 market A
8 4 market D
6 3 market B
你要做的实际上又是一个groupby(根据第一个groupby的结果):排序并获取每个组的前三个元素。
从第一个groupby的结果开始:
In [60]: df_agg = df.groupby(['job','source']).agg({'count':sum})
我们按索引的第一级分组:
In [63]: g = df_agg['count'].groupby('job', group_keys=False)
然后我们要对每个组进行排序('order')并取前三个元素:
In [64]: res = g.apply(lambda x: x.sort_values(ascending=False).head(3))
但是,为此,有一个快捷函数可以执行此操作,nmax
:
In [65]: g.nlargest(3)
Out[65]:
job source
market A 5
D 4
B 3
sales E 7
C 6
B 4
dtype: int64
所以一口气,这看起来像:
df_agg['count'].groupby('job', group_keys=False).nlargest(3)
这是另一个按排序顺序取前3名并在组内排序的示例:
In [43]: import pandas as pd
In [44]: df = pd.DataFrame({"name":["Foo", "Foo", "Baar", "Foo", "Baar", "Foo", "Baar", "Baar"], "count_1":[5,10,12,15,20,25,30,35], "count_2" :[100,150,100,25,250,300,400,500]})
In [45]: df
Out[45]:
count_1 count_2 name
0 5 100 Foo
1 10 150 Foo
2 12 100 Baar
3 15 25 Foo
4 20 250 Baar
5 25 300 Foo
6 30 400 Baar
7 35 500 Baar
### Top 3 on sorted order:
In [46]: df.groupby(["name"])["count_1"].nlargest(3)
Out[46]:
name
Baar 7 35
6 30
4 20
Foo 5 25
3 15
1 10
dtype: int64
### Sorting within groups based on column "count_1":
In [48]: df.groupby(["name"]).apply(lambda x: x.sort_values(["count_1"], ascending = False)).reset_index(drop=True)
Out[48]:
count_1 count_2 name
0 35 500 Baar
1 30 400 Baar
2 20 250 Baar
3 12 100 Baar
4 25 300 Foo
5 15 25 Foo
6 10 150 Foo
7 5 100 Foo