Generate list of days for each employee

Question

I have the following dataframe in Python:

Name	days	Start Date	End Date
EMP1	15	8/8/22	8/26/22
EMP2	3	6/9/22	6/13/22
EMP3	5	8/22/22	8/26/22
EMP3	5	8/1/22	8/5/22
EMP3	6	6/17/22	6/24/22
EMP4	4.5	7/18/22	7/22/22
EMP5	5	7/18/22	7/22/22
EMP6	5	8/15/22	8/19/22
EMP7	9	8/22/22	9/2/22

I need to generate the list of dates between the start and end date for each row whilst keeping the employee name . Is this simple to do in python? Really struggling to get it to do it by row .

desired output (list each day between start and end date and employees name):

Name	date
EMP1	8/8/22
EMP1	8/9/22
EMP1	8/10/22
EMP1	8/11/22
EMP1	8/12/22
EMP1	8/13/22
EMP1	8/14/22
EMP1	8/15/22

could you explain what mean _I have the following table in Python_? — Danila Ganchar, Nov 22 '22 at 14:55
You say that you're struggling. Could we see what you've attempted so far and how it hasn't worked? — Daniel Walker, Nov 22 '22 at 14:58
[How to make good reproducible pandas examples](https://stackoverflow.com/q/20109391/4046632) — buran, Nov 22 '22 at 15:00
@DanielWalker I can only do a simple calculation for the start - end date and list each date. Not sure how to do it for each row, or employee — mgd6, Nov 22 '22 at 15:01

score 0 · Accepted Answer · answered Nov 22 '22 at 15:19

well, convert the start date ,enddate to date range then explode using that columns :

df['Date'] = df.apply(lambda x: pd.date_range(start=x['Start Date'], end=x['End Date']), axis=1)
output = df.explode('Date').drop(columns = ['days','Start Date','End Date'])

output :

>>
    Name       Date
0   EMP1 2022-08-08
0   EMP1 2022-08-09
0   EMP1 2022-08-10
0   EMP1 2022-08-11
0   EMP1 2022-08-12
..   ...        ...
8   EMP7 2022-08-29
8   EMP7 2022-08-30
8   EMP7 2022-08-31
8   EMP7 2022-09-01
8   EMP7 2022-09-02

[69 rows x 2 columns]

Generate list of days for each employee

1 Answers1