<p>如果我能够处理<code>merge_asof</code>,这将是一个简单的<code>max_date</code>案例,因此我做了很多工作:</p>
<pre><code>max_date = pd.to_datetime('2018-04-01')
# set_index for easy extraction by id
df_emails.set_index('CustID', inplace=True)
# we want this later in the final output
df_emails['NextDateSentOrEndOfData'] = df_emails.groupby('CustID').shift(-1).fillna(max_date)
# cuts function for groupby
def cuts(df):
custID = df.CustID.iloc[0]
bins=list(df_emails.loc[[custID], 'DateSent']) + [max_date]
return pd.cut(df.TripDate, bins=bins, right=False)
# bin the dates:
s = df_trips.groupby('CustID', as_index=False, group_keys=False).apply(cuts)
# aggregate the info:
new_df = (df_trips.groupby([df_trips.CustID, s])
.TotalSpend.agg(['sum', 'size'])
.reset_index()
)
# get the right limit:
new_df['NextDateSentOrEndOfData'] = new_df.TripDate.apply(lambda x: x.right)
# drop the unnecessary info
new_df.drop('TripDate', axis=1, inplace=True)
# merge:
df_emails.reset_index().merge(new_df,
on=['CustID','NextDateSentOrEndOfData'],
how='left'
)
</code></pre>
<p>输出:</p>
<pre><code> CustID DateSent NextDateSentOrEndOfData sum size
0 2 2018-01-20 2018-02-19 125.0 2.0
1 2 2018-02-19 2018-03-31 250.0 1.0
2 2 2018-03-31 2018-04-01 NaN NaN
3 4 2018-01-10 2018-02-26 NaN NaN
4 4 2018-02-26 2018-04-01 200.0 2.0
5 5 2018-02-01 2018-02-07 NaN NaN
6 5 2018-02-07 2018-04-01 NaN NaN
</code></pre>