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问题:
I want to group my dataframe by two columns and then sort the aggregated results within the groups.
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
I would now like to sort the count column in descending order within each of the groups. And then take only the top three rows. To get something like:
count job source market A 5 D 4 B 3 sales E 7 C 6 B 4
回答1:
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(level=0, 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.order(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
回答2:
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
回答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
回答4:
If you don't need to sum a column, then use @tvashtar's answer. If you do need to sum, then you can use @joris' answer or this one which is very similar to it.
df.groupby(['job']).apply(lambda x: (x.groupby('source') .sum() .sort_values('count', ascending=False)) .head(3))
