Convert dataframe with start and end date to daily data

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 row: pd.date_range(row["start_date"], row["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]

Original answer:

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]