How to group values of pandas dataframe and select the latest(by date) from each group?
For example, given a dataframe sorted by date:
id product date
0 220 6647 2014-09-01
1 220 6647 2014-09-03
2 220 6647 2014-10-16
3 826 3380 2014-11-11
4 826 3380 2014-12-09
5 826 3380 2015-05-19
6 901 4555 2014-09-01
7 901 4555 2014-10-05
8 901 4555 2014-11-01
grouping by id or product, and selecting the earliest gives:
id product date
2 220 6647 2014-10-16
5 826 3380 2015-05-19
8 901 4555 2014-11-01
use idxmax in groupby and slice df with loc
df.loc[df.groupby('id').date.idxmax()]
id product date
2 220 6647 2014-10-16
5 826 3380 2015-05-19
8 901 4555 2014-11-01
You can also use tail with groupby to get the last n values of the group:
df.sort_values('date').groupby('id').tail(1)
id product date
2 220 6647 2014-10-16
8 901 4555 2014-11-01
5 826 3380 2015-05-19
To use .tail() as an aggregation method and keep your grouping intact:
df.sort_values('date').groupby('id').apply(lambda x: x.tail(1))
id product date
id
220 2 220 6647 2014-10-16
826 5 826 3380 2015-05-19
901 8 901 4555 2014-11-01
I had a similar problem and ended up using drop_duplicates rather than groupby.
It seems to run significatively faster on large datasets when compared with other methods suggested above.
df.sort_values(by="date").drop_duplicates(subset=["id"], keep="last")
id product date
2 220 6647 2014-10-16
8 901 4555 2014-11-01
5 826 3380 2015-05-19
Given a dataframe sorted by date, you can obtain what you ask for in a number of ways:
Like this:
df.groupby(['id','product']).last()
like this:
df.groupby(['id','product']).nth(-1)
or like this:
df.groupby(['id','product']).max()
If you don't want id and product to appear as index use groupby(['id', 'product'], as_index=False).
Alternatively use:
df.groupby(['id','product']).tail(1)
来源:https://stackoverflow.com/questions/41525911/group-by-pandas-dataframe-and-select-latest-in-each-group