CODEWITHSUNDEEP
pandasdate-difference
user1700890
user1700890

Reputation: 7730

Pandas date difference in one column

Here is my dataframe:

import pandas as pd
df_manual = pd.DataFrame({'A': ['one', 'one', 'two', 'two', 'one'] ,
                   'B': ['Ar', 'Br', 'Cr', 'Ar','Ar'] ,
                   'C': ['12/15/2011', '11/11/2001', '08/7/2015', '07/3/1999','03/03/2000' ]})

I would like to create column which would contain date difference for column see (with prior grouping). Here is what I wrote:

df_manual['C']=pd.to_datetime(df_manual['C'])
df_manual['diff'] =
df_manual.groupby(['A'])['C'].transform(lambda x: x.diff())

But the result I get is not day difference. Resulting difference between 2001-11-11 and 2000-03-03 is a date 1971-09-11, while I need number of days in between.

Any idea how to achieve it?

Upvotes: 2

Views: 2632

Answers (1)

root
root

Reputation: 33773

Use apply instead of transform:

df_manual['diff'] = df_manual.groupby(['A'])['C'].apply(lambda x: x.diff())

The resulting output:

     A   B          C       diff
0  one  Ar 2011-12-15        NaT
1  one  Br 2001-11-11 -3686 days
2  two  Cr 2015-08-07        NaT
3  two  Ar 1999-07-03 -5879 days
4  one  Ar 2000-03-03  -618 days

If you want df_manual['diff'] to be an integer instead of a timedelta, use the dt.days accessor:

df_manual['diff'] = df_manual.groupby(['A'])['C'].apply(lambda x: x.diff()).dt.days

Upvotes: 4

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