Calculate a list of dates in Pandas column

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I have a Pandas dataframe including two dates column in datetime. And I want to generate a list of dates in this date rage as a new column, so I can explode the entry into multiple rows later.

I tried the following list comprehension.

orders_df['list_of_dates'] = [orders_df['start_date'] + timedelta(days=n) for n in range(orders_df['date_difference'])]

But the following message was received

TypeError: 'Series' object cannot be interpreted as an integer

Any thoughts on the solution would be much appreciated.

pandas timedelta dataframe list python
1 Answers

Use nested list comprehension with range:

from datetime import timedelta

rng = pd.date_range('2017-04-03', periods=5)
orders_df = pd.DataFrame({'start_date': rng, 'date_difference': 2})  

orders_df['list_of_dates'] = [[d + timedelta(days=x) for x in range(n)] 
                                      for d, n 
                                      in zip(orders_df['start_date'],
                                             orders_df['date_difference'])]

print (orders_df)
  start_date  date_difference                               list_of_dates
0 2017-04-03                2  [2017-04-03 00:00:00, 2017-04-04 00:00:00]
1 2017-04-04                2  [2017-04-04 00:00:00, 2017-04-05 00:00:00]
2 2017-04-05                2  [2017-04-05 00:00:00, 2017-04-06 00:00:00]
3 2017-04-06                2  [2017-04-06 00:00:00, 2017-04-07 00:00:00]
4 2017-04-07                2  [2017-04-07 00:00:00, 2017-04-08 00:00:00]

If need also new column is possible use Index.repeat with GroupBy.cumcount for counter Series converted to timedeltas by to_timedelta:

df = orders_df.loc[orders_df.index.repeat(orders_df['date_difference'])]
g = df.groupby(level=0).cumcount()
df['new'] = df['start_date'] + pd.to_timedelta(g, unit='d')
df = df.reset_index(drop=True)
print (df)
  start_date  date_difference        new
0 2017-04-03                2 2017-04-03
1 2017-04-03                2 2017-04-04
2 2017-04-04                2 2017-04-04
3 2017-04-04                2 2017-04-05
4 2017-04-05                2 2017-04-05
5 2017-04-05                2 2017-04-06
6 2017-04-06                2 2017-04-06
7 2017-04-06                2 2017-04-07
8 2017-04-07                2 2017-04-07
9 2017-04-07                2 2017-04-08
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