How to get date after subtracting days in pandas

Question

I have a dataframe:

In [15]: df
Out[15]: 
        date  day
0 2015-10-10   23
1 2015-12-19    9
2 2016-03-05   34
3 2016-09-17   23
4 2016-04-30    2

I want to subtract the number of days from the date and create a new column.

In [16]: df.dtypes
Out[16]: 
date    datetime64[ns]
day              int64

Desired output something like:

In [15]: df
Out[15]: 
        date  day date1
0 2015-10-10   23 2015-09-17
1 2015-12-19    9 2015-12-10
2 2016-03-05   34 2016-01-29
3 2016-09-17   23 2016-08-25
4 2016-04-30    2 2016-04-28

I tried but this does not work:

df['date1']=df['date']+pd.Timedelta(df['date'].dt.day-df['day'])

it throws error :

TypeError: unsupported type for timedelta days component: Series

Answer 1

You can use to_timedelta:

df['date1'] = df['date'] -  pd.to_timedelta(df['day'], unit='d')

print (df)
        date  day      date1
0 2015-10-10   23 2015-09-17
1 2015-12-19    9 2015-12-10
2 2016-03-05   34 2016-01-31
3 2016-09-17   23 2016-08-25
4 2016-04-30    2 2016-04-28

If need Timedelta use apply, but it is slower:

df['date1'] = df['date'] -  df.day.apply(lambda x: pd.Timedelta(x, unit='D'))

print (df)
        date  day      date1
0 2015-10-10   23 2015-09-17
1 2015-12-19    9 2015-12-10
2 2016-03-05   34 2016-01-31
3 2016-09-17   23 2016-08-25
4 2016-04-30    2 2016-04-28

Timings:

#[5000 rows x 2 columns]
df = pd.concat([df]*1000).reset_index(drop=True)

In [252]: %timeit df['date'] -  df.day.apply(lambda x: pd.Timedelta(x, unit='D'))
10 loops, best of 3: 45.3 ms per loop

In [253]: %timeit df['date'] -  pd.to_timedelta(df['day'], unit='d')
1000 loops, best of 3: 1.71 ms per loop

Answer 2

import dateutil.relativedelta
def calculate diff(v):
    return v['date'] - dateutil.relativedelta.relativedelta(day=v['day'])
df['date1']=df.apply(calculate_diff, axis=1)

given that v['date'] is datetime object