Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

Pandas - Convert dataframe with start and end date to daily data

I have one record per ID with start date and end date

id  age state   start_date  end_date
123 18  CA     2/17/2019    5/4/2019
223 24  AZ     1/17/2019    3/4/2019

I want to create a record for each day between the start and end day, so I can join daily activity data to it. The target output would look something like this

id  age state   start_date
123 18  CA      2/17/2019
123 18  CA      2/18/2019
123 18  CA      2/19/2019
123 18  CA      2/20/2019
123 18  CA      2/21/2019
            …
123 18  CA      5/2/2019
123 18  CA      5/3/2019
123 18  CA      5/4/2019

And of course do this for all ids and their respective start dates in the dataset. Any help is much appreciated - thanks!

like image 769
L Xandor Avatar asked Aug 02 '19 22:08

L Xandor


2 Answers

melt, GroupBy, resample & ffill

First we melt (unpivot) your two date columns to one. Then we resample on day basis:

melt = df.melt(id_vars=['id', 'age', 'state'], value_name='date').drop('variable', axis=1)
melt['date'] = pd.to_datetime(melt['date'])

melt = melt.groupby('id').apply(lambda x: x.set_index('date').resample('d').first())\
           .ffill()\
           .reset_index(level=1)\
           .reset_index(drop=True)

Output

          date     id   age state
0   2019-02-17  123.0  18.0    CA
1   2019-02-18  123.0  18.0    CA
2   2019-02-19  123.0  18.0    CA
3   2019-02-20  123.0  18.0    CA
4   2019-02-21  123.0  18.0    CA
..         ...    ...   ...   ...
119 2019-02-28  223.0  24.0    AZ
120 2019-03-01  223.0  24.0    AZ
121 2019-03-02  223.0  24.0    AZ
122 2019-03-03  223.0  24.0    AZ
123 2019-03-04  223.0  24.0    AZ

[124 rows x 4 columns]

Edit:

I had to revisit this problem in a project, and looks like using DataFrame.apply with pd.date_range and DataFrame.explode is almost 3x faster:

df["date"] = df.apply(
    lambda x: pd.date_range(x["start_date"], x["end_date"]), axis=1
)
df = (
    df.explode("date", ignore_index=True)
    .drop(columns=["start_date", "end_date"])
)

Output

      id  age state       date
0    123   18    CA 2019-02-17
1    123   18    CA 2019-02-18
2    123   18    CA 2019-02-19
3    123   18    CA 2019-02-20
4    123   18    CA 2019-02-21
..   ...  ...   ...        ...
119  223   24    AZ 2019-02-28
120  223   24    AZ 2019-03-01
121  223   24    AZ 2019-03-02
122  223   24    AZ 2019-03-03
123  223   24    AZ 2019-03-04

[124 rows x 4 columns]
like image 137
Erfan Avatar answered Oct 30 '22 13:10

Erfan


Use listcomp and pd.date_range on values of columns start_date and end_date to create list of date for each rec. Next, construct a new dataframe from result of listcomp and join back to the other 3 columns of df. Finally, set_index, stack and reset_index back

a = [pd.date_range(*r, freq='D') for r in df[['start_date', 'end_date']].values]
df[['id', 'age', 'state']].join(pd.DataFrame(a)).set_index(['id', 'age', 'state']) \
                          .stack().droplevel(-1).reset_index()

Out[187]:
      id  age state          0
0    123   18    CA 2019-02-17
1    123   18    CA 2019-02-18
2    123   18    CA 2019-02-19
3    123   18    CA 2019-02-20
4    123   18    CA 2019-02-21
5    123   18    CA 2019-02-22
6    123   18    CA 2019-02-23
7    123   18    CA 2019-02-24
8    123   18    CA 2019-02-25
9    123   18    CA 2019-02-26
10   123   18    CA 2019-02-27
11   123   18    CA 2019-02-28
12   123   18    CA 2019-03-01
13   123   18    CA 2019-03-02
14   123   18    CA 2019-03-03
15   123   18    CA 2019-03-04
16   123   18    CA 2019-03-05
17   123   18    CA 2019-03-06
18   123   18    CA 2019-03-07
19   123   18    CA 2019-03-08
20   123   18    CA 2019-03-09
21   123   18    CA 2019-03-10
22   123   18    CA 2019-03-11
23   123   18    CA 2019-03-12
24   123   18    CA 2019-03-13
25   123   18    CA 2019-03-14
26   123   18    CA 2019-03-15
27   123   18    CA 2019-03-16
28   123   18    CA 2019-03-17
29   123   18    CA 2019-03-18
..   ...  ...   ...        ...
94   223   24    AZ 2019-02-03
95   223   24    AZ 2019-02-04
96   223   24    AZ 2019-02-05
97   223   24    AZ 2019-02-06
98   223   24    AZ 2019-02-07
99   223   24    AZ 2019-02-08
100  223   24    AZ 2019-02-09
101  223   24    AZ 2019-02-10
102  223   24    AZ 2019-02-11
103  223   24    AZ 2019-02-12
104  223   24    AZ 2019-02-13
105  223   24    AZ 2019-02-14
106  223   24    AZ 2019-02-15
107  223   24    AZ 2019-02-16
108  223   24    AZ 2019-02-17
109  223   24    AZ 2019-02-18
110  223   24    AZ 2019-02-19
111  223   24    AZ 2019-02-20
112  223   24    AZ 2019-02-21
113  223   24    AZ 2019-02-22
114  223   24    AZ 2019-02-23
115  223   24    AZ 2019-02-24
116  223   24    AZ 2019-02-25
117  223   24    AZ 2019-02-26
118  223   24    AZ 2019-02-27
119  223   24    AZ 2019-02-28
120  223   24    AZ 2019-03-01
121  223   24    AZ 2019-03-02
122  223   24    AZ 2019-03-03
123  223   24    AZ 2019-03-04

[124 rows x 4 columns]
like image 31
Andy L. Avatar answered Oct 30 '22 14:10

Andy L.