Python Pandas : Return the consecutive missing weekdays dates and assign rate next to missing dates in a dataframe

Question

Dates       rates
7/26/2019   1.04
7/30/2019   1.0116
7/31/2019   1.005
8/1/2019    1.035
8/2/2019    1.01
8/6/2019    0.9886
8/12/2019   0.965

df = df.merge(
    p         


        

Answer 1

you can use reindex with bdate_range to create all the missing values in rates for business days only:

new_df = df.set_index('Dates')\
           .reindex( pd.bdate_range(df.Dates.min(), df.Dates.max(), name='Dates'), 
                     method='bfill')\
           .reset_index() 
print (new_df)
        Dates   rates
0  2019-07-26  1.0400
1  2019-07-29  1.0116
2  2019-07-30  1.0116
3  2019-07-31  1.0050
4  2019-08-01  1.0350
5  2019-08-02  1.0100
6  2019-08-05  0.9886
7  2019-08-06  0.9886
8  2019-08-07  0.9650
9  2019-08-08  0.9650
10 2019-08-09  0.9650
11 2019-08-12  0.9650

Answer 2

You could create a Series of all business days then outer merge and bfill the missing values. This will retain any non-business days in your initial DataFrame (if any) and will also use their values in the filling.

import pandas as pd
#df['Dates'] = pd.to_datetime(df['Dates'])

s = pd.Series(pd.date_range(df['Dates'].min(), df['Dates'].max(), freq='D'),
              name='Dates')
s = s[s.dt.dayofweek.lt(5)]

df = df.merge(s, how='outer').sort_values('Dates').bfill()

---

        Dates   rates
0  2019-07-26  1.0400
7  2019-07-29  1.0116
1  2019-07-30  1.0116
2  2019-07-31  1.0050
3  2019-08-01  1.0350
4  2019-08-02  1.0100
8  2019-08-05  0.9886
5  2019-08-06  0.9886
9  2019-08-07  0.9650
10 2019-08-08  0.9650
11 2019-08-09  0.9650
6  2019-08-12  0.9650