Pandas groupby聚合截断最早的日期而不是最早的日期

Question

I'm trying to aggregate from the end of a date range instead of from the beginning. Despite the fact that I would think that adding closed='right' to the grouper would solve the issue, it doesn't. Please let me know how I can achieve my desired output shown at the bottom, thanks.

import pandas as pd
df = pd.DataFrame(columns=['date','number'])
df['date'] = pd.date_range('1/1/2000', periods=8, freq='T')
df['number'] = pd.Series(range(8))
df

    date                number
0   2000-01-01 00:00:00 0
1   2000-01-01 00:01:00 1
2   2000-01-01 00:02:00 2
3   2000-01-01 00:03:00 3
4   2000-01-01 00:04:00 4
5   2000-01-01 00:05:00 5
6   2000-01-01 00:06:00 6
7   2000-01-01 00:07:00 7

With the groupby and aggregation of the date I get the following. Since I have 8 dates and I'm grouping by periods of 3 it must choose whether to truncate the earliest date group or the oldest date group, and it chooses the oldest date group (the oldest date group has a count of 2):

df.groupby(pd.Grouper(key='date', freq='3T')).agg('count')

date                number
2000-01-01 00:00:00 3
2000-01-01 00:03:00 3
2000-01-01 00:06:00 2

My desired output is to instead truncate the earliest date group:

date                number
2000-01-01 00:00:00 2
2000-01-01 00:02:00 3
2000-01-01 00:05:00 3

Please let me know how this can be achieved, I'm hopeful there's just a parameter that can be set that I've overlooked. Note that this is similar to this question, but my question is specific to the date truncation.

EDIT: To reframe the question (thanks Alexdor) the default behavior in pandas is to bin by period [0, 3), [3, 6), [6, 9) but instead I'd like to bin by (-1, 2], (2, 5], (5, 8]

Answer 1

It seems like the grouper function build up the bins starting from the oldest time in the series that you pass to it. I couldn't see a way to make it build up the bins from the newest time, but it's fairly easy to construct the bins from scratch.

freq = '3min'

minTime = df.date.min()
maxTime = df.date.max()
deltaT = pd.Timedelta(freq)
minTime -= deltaT - (maxTime - minTime) % deltaT # adjust min time to start of first bin
r = pd.date_range(start=minTime, end=maxTime, freq=freq)

df.groupby(pd.cut(df["date"], r)).agg('count')

Gives

date                                     date number        
(1999-12-31 23:58:00, 2000-01-01 00:01:00]  2   2
(2000-01-01 00:01:00, 2000-01-01 00:04:00]  3   3
(2000-01-01 00:04:00, 2000-01-01 00:07:00]  3   3

Answer 2

This is one hack, which let's you group by a constant group size, counting bottom up.

from itertools import chain

def grouper(x, k=3):
    n = len(df.index)
    return list(chain.from_iterable([[0]*int(n//k)] + [[i]*k for i in range(1, int(n/k)+1)]))

df['grouper'] = grouper(df, 3)

res = df.groupby('grouper', as_index=False)\
        .agg({'date': 'first', 'number': 'count'})\
        .drop('grouper', 1)

#                  date  number
# 0 2000-01-01 00:00:00       2
# 1 2000-01-01 00:02:00       3
# 2 2000-01-01 00:05:00       3