Python/Pandas — convert day and hour columns into index of hour

Question

I have a Data Frame that looks like this:

df
         Date    Hr    CO2_resp
0      5/1/02   600    0.000889
1      5/2/02   600    0.000984
2      5/4/02   900    0.000912

How would I go about creating a column Ind that represents a number index of hours elapsed since midnight 5/1/02? Such that the column would read

df
         Date    Hr   Ind      CO2_resp
0      5/1/02   600     6      0.000889
1      5/2/02   600    30      0.000984
2      5/4/02   800    80      0.000912

Thanks.

Answer 1

Assuming that the Date is a string, and Hr is an integer, you could apply a function to parse the Date , get the hours (days * 24) from the timedelta with your reference date, and add the hours.

Something like this -

df.apply(lambda x: 
     (datetime.datetime.strptime(x['Date'], '%m/%d/%y')
      - datetime.datetime.strptime('5/1/02', '%m/%d/%y')).days
     * 24 + x['Hr'] / 100,
     axis=1)

Answer 2

You can use to_datetime with to_timedelta . Then convert timedelta to hours by np.timedelta64(1, 'h') and last if type of output is always int , cast by astype :

#convert column Date to datetime
df['Date'] = pd.to_datetime(df.Date)

df['Ind'] = ((df.Date 
              - pd.to_datetime('2002-05-01') 
              + pd.to_timedelta(df.Hr / 100, unit='h')) / np.timedelta64(1, 'h')).astype(int)
print (df)
        Date   Hr  CO2_resp  ind
0 2002-05-01  600  0.000889    6
1 2002-05-02  600  0.000984   30
2 2002-05-04  900  0.000912   81

If not dividing by 100 column Hr , output is different:

df['Ind'] = ((df.Date 
              - pd.to_datetime('2002-05-01') 
              + pd.to_timedelta(df.Hr,unit='h')) / np.timedelta64(1, 'h')).astype(int)
print (df)
        Date   Hr  CO2_resp  Ind
0 2002-05-01  600  0.000889  600
1 2002-05-02  600  0.000984  624
2 2002-05-04  900  0.000912  972