将年龄函数应用于数据框列

Question

I have a dataframe df with a column date of birth . The head looks like:

    Date of birth
0      1957-04-30
1      1966-11-10
2      1966-11-10
3      NOT KNOWN
4      1958-10-28
5      1958-06-04

I also have a variable called referencePeriodEndDate which is a date in a yyyy-mm-dd format and for example looks like 2017-03-31

I am trying to create a new column called Age which is the age of from the Date of birth upto the referencePeriodEndDate

so the function to apply to each row would looks like:

(`referencePeriodEndDate` - df["Date of birth"]) / 365.25

There is the potential for rows in the Date of birth column to be empty ( null ) or have the entry 'NOT KNOWN' so I need to return the value 'NOT KNOWN' to the effected column in the new Age column.

I have come up with the following but it refuses to work (or return an error)

    df["Age"].apply(lambda row: TimeCalc(df,referencePeriodEndDate) if row.notnull() else "NOT KNOWN")


def TimeCalc(rawDatabase,referencePeriodEndDate):

     Age = ((referencePeriodEndDate - rawDatabase["Date of birth"]) / 365.25)

     return Age

The desired output would look like:

 Date of birth            Age
    30/04/1957    59.91786448
    10/11/1966    50.38740589
    10/11/1966    50.38740589
     NOT KNOWN      NOT KNOWN
    28/10/1958    58.42299795
    04/06/1958    58.82272416

Answer 1

You can using to_datetime

df['Dateofbirth']=pd.to_datetime(df['Dateofbirth'],errors='coerce')
df['Age']=(pd.to_datetime('2017-03-31')-df['Dateofbirth']).dt.days/365.25

df.fillna('unknow')
Out[370]: 
           Dateofbirth      Age
0  1957-04-30 00:00:00  59.9179
1  1966-11-10 00:00:00  50.3874
2  1966-11-10 00:00:00  50.3874
3               unknow   unknow
4  1958-10-28 00:00:00   58.423
5  1958-06-04 00:00:00  58.8227