add_dates []

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with pandas; multiple dates

Given

In [1]: 
import pandas as pd
df = pd.DataFrame({'begin': ['20201107', '20201121'], 'offset': [5, -6]})
df
Out[1]: 
      begin  offset
0  20201107       5
1  20201121      -6

In [2]: 
df.dtypes
Out[2]: 
begin     object
offset     int64
dtype: object

we want

      begin  offset       end
0  20201107       5  20201112
1  20201121      -6  20201115

where end date = begin date + offset

Solution:

In [3]: 
fmt='%Y%m%d'
from datetime import datetime
df['end'] = (df['begin'].apply(lambda x: datetime.strptime(x, fmt)) +
             pd.to_timedelta(df['offset'], 'd')).dt.strftime(fmt)
df
Out[3]: 
      begin  offset       end
0  20201107       5  20201112
1  20201121      -6  20201115

Intermediate steps:

In [4]: 
df['begin'].apply(lambda x: datetime.strptime(x, fmt))
Out[4]: 
0   2020-11-07
1   2020-11-21
Name: begin, dtype: datetime64[ns]

In [5]: 
df['begin'].apply(lambda x: datetime.strptime(x, fmt)) + pd.to_timedelta(df['offset'], 'd')
Out[5]: 
0   2020-11-12
1   2020-11-15
dtype: datetime64[ns]

In [6]: 
(df['begin'].apply(lambda x: datetime.strptime(x, fmt)) +
 pd.to_timedelta(df['offset'], 'd')).dt.strftime(fmt)
Out[6]: 
0    20201112
1    20201115
dtype: object

without pandas; single date

For a single date, we can do this without pandas also

In [1]: 
from datetime import datetime, timedelta
print(datetime.now())
2020-11-21 23:54:10.971599

In [2]: 
print(datetime.now() + timedelta(days=10))
2020-12-01 23:54:32.393776

Table of Contents

Add dates

with pandas; multiple dates

without pandas; single date