pandas groupby sort within groups
You could also just do it in one go, by doing the sort first and using head to take the first 3 of each group.
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
What you want to do is actually again a groupby (on the result of the first groupby): sort and take the first three elements per group.
Starting from the result of the first groupby:
In [60]: df_agg = df.groupby(['job','source']).agg({'count':sum})
We group by the first level of the index:
In [63]: g = df_agg['count'].groupby('job', group_keys=False)
Then we want to sort ('order') each group and take the first three elements:
In [64]: res = g.apply(lambda x: x.sort_values(ascending=False).head(3))
However, for this, there is a shortcut function to do this, nlargest
:
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
So in one go, this looks like:
df_agg['count'].groupby('job', group_keys=False).nlargest(3)
Here's other example of taking top 3 on sorted order, and sorting within the groups:
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