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I have following data in one of my columns:

df['DOB']

0 01-01-84
1 31-07-85
2 24-08-85
3 30-12-93
4 09-12-77
5 08-09-90
6 01-06-88
7 04-10-89
8 15-11-91
9 01-06-68
Name: DOB, dtype: object

I want to convert this to a datatype column.
I tried following:

print(pd.to_datetime(df1['Date.of.Birth']))
0 1984-01-01
1 1985-07-31
2 1985-08-24
3 1993-12-30
4 1977-09-12
5 1990-08-09
6 1988-01-06
7 1989-04-10
8 1991-11-15
9 2068-01-06
Name: DOB, dtype: datetime64[ns]

What should be my code change to get the date as 1968-01-06 instead of 2068-01-06? Thanks in advance

In this specific case, I would use this:

pd.to_datetime(df['DOB'].str[:-2] + '19' + df['DOB'].str[-2:])

Note that this will break if you have DOBs after 1999!

Output:

0 1984-01-01
1 1985-07-31
2 1985-08-24
3 1993-12-30
4 1977-09-12
5 1990-08-09
6 1988-01-06
7 1989-04-10
8 1991-11-15
9 1968-01-06
dtype: datetime64[ns]

You can first convert to datetimes and if years are above or equal 2020 then subtract 100 years created by DateOffset:

df['DOB'] = pd.to_datetime(df['DOB'], format='%d-%m-%y')
df.loc[df['DOB'].dt.year >= 2020, 'DOB'] -= pd.DateOffset(years=100)
#same like
#mask = df['DOB'].dt.year >= 2020
#df.loc[mask, 'DOB'] = df.loc[mask, 'DOB'] - pd.DateOffset(years=100)
print (df)
 DOB
0 1984-01-01
1 1985-07-31
2 1985-08-24
3 1993-12-30
4 1977-12-09
5 1990-09-08
6 1988-06-01
7 1989-10-04
8 1991-11-15
9 1968-06-01

Or you can add 19 or 20 to years by Series.str.replace and set valuies by numpy.where with condition.

Notice: Solution working also for years 00 for 2000, up to 2020.

s1 = df['DOB'].str.replace(r'-(d+)$', r'-191')
s2 = df['DOB'].str.replace(r'-(d+)$', r'-201')
mask = df['DOB'].str[-2:].astype(int) <= 20
df['DOB'] = pd.to_datetime(np.where(mask, s2, s1))

print (df)
 DOB
0 1984-01-01
1 1985-07-31
2 1985-08-24
3 1993-12-30
4 1977-09-12
5 1990-08-09
6 1988-01-06
7 1989-04-10
8 1991-11-15
9 1968-01-06

If all years are below 2000:

s1 = df['DOB'].str.replace(r'-(d+)$', r'-191')
df['DOB'] = pd.to_datetime(s1, format='%d-%m-%Y')
print (df)
 DOB
0 1984-01-01
1 1985-07-31
2 1985-08-24
3 1993-12-30
4 1977-12-09
5 1990-09-08
6 1988-06-01
7 1989-10-04
8 1991-11-15
9 1968-06-01

Another solution is to treat the DOB as a date, and take it back to the previous century only if it is in the future (i.e. after "now"). Example:

from datetime import datetime, date

df=pd.DataFrame.from_dict('DOB':['01-06-68','01-06-08'])
df['DOB'] = df['DOB'].apply(lambda x: datetime.strptime(x,'%d-%m-%y'))
df['DOB'] = df['DOB'].apply(lambda x: x if x<datetime.now() else date(x.year-100,x.month,x.day))

In general (in case of uncertainty), it would be better to explicitly specify the year:

pd.to_datetime(data['Date.of.Birth'].apply(lambda x: '-'.join(x.split('-')[:-1] + ['19' + x.split('-')[2]])))

I ran this with the following data frame:

 0 1
0 0 01-01-84
1 1 31-07-85
2 2 24-08-85
3 3 30-12-93
4 4 09-12-77
5 5 08-09-90
6 6 01-06-88
7 7 04-10-89
8 8 15-11-91
9 9 01-06-68


pd.to_datetime(data[1].apply(lambda x: '-'.join(x.split('-')[:-1] + ['19' + x.split('-')[2]])))


0 1984-01-01
1 1985-07-31
2 1985-08-24
3 1993-12-30
4 1977-09-12
5 1990-08-09
6 1988-01-06
7 1989-04-10
8 1991-11-15
9 1968-01-06
Name: 1, dtype: datetime64[ns]

You can use the code below if there are only 19 and 20 as starts, like:

df['DOB'] = pd.to_datetime(df['DOB'].str.replace('20([^20]*)$', '19'))

And if there are no 20s anywhere else:

df['DOB'] = pd.to_datetime(df['DOB'].str.replace('20', '19'))

And now:

print(df['DOB'])

Is:

0 1984-01-01
1 1985-07-31
2 1985-08-24
3 1993-12-30
4 1977-09-12
5 1990-08-09
6 1988-01-06
7 1989-04-10
8 1991-11-15
9 1968-01-06
dtype: datetime64[ns]