将属性.years与apply和axis=1一起用于按行处理：

df = pd.DataFrame({'start':['2015-10-02','2014-11-05'],
                   'end':['2018-01-02','2018-10-05']})

df['start'] = pd.to_datetime(df['start'])
df['end'] = pd.to_datetime(df['end'])

from dateutil.relativedelta import relativedelta

df['y'] = df.apply(lambda x: relativedelta(x['end'], x['start']).years, axis=1)

或者使用list comprehension：
x一个一个一个一个x一个一个二个x
编辑：

df = pd.DataFrame({'start':['2015-10-02','2014-11-05'],
                   'end':['2018-01-02',np.nan]})

df['start'] = pd.to_datetime(df['start'])
df['end'] = pd.to_datetime(df['end'])

from dateutil.relativedelta import relativedelta

m = df[['start','end']].notnull().all(axis=1)
df.loc[m, 'y'] = df[m].apply(lambda x: relativedelta(x['end'], x['start']).years, axis=1)
print (df)
       start        end    y
0 2015-10-02 2018-01-02  2.0
1 2014-11-05        NaT  NaN

可以按年单位划分timedelta系列，如有必要，还可以四舍五入：

# data from jezrael

df['years'] = (df['end'] - df['start']) / np.timedelta64(1, 'Y')
df['years_floor'] = df['years'].round()

print(df)

       start        end     years  years_floor
0 2015-10-02 2018-01-02  2.253297          2.0
1 2014-11-05        NaT       NaN          NaN

你可以通过

(df['end'] - df['start'])/pd.Timedelta(1, 'Y')

并且如果需要的话对结果进行舍入。
在Pandasv0.23.4和以后你可以做

(df['end'] - df['start'])//pd.Timedelta(1, 'Y')

直接得到全年的差价。
更新：在panda v0.25和更高版本中，不支持pd.Timedelta(1, 'Y')，因为年不是一致的度量单位（有时是365d，有时是366d）。如果365d近似值可以接受，您可以改为使用pd.Timedelta(1, 'Y')：

(df['end'] - df['start'])/pd.Timedelta(365, 'D')

检查此答案calculate the difference between two datetime.date() dates in years and months

from dateutil import relativedelta as rdelta
from datetime import date
d1 = date(2001,5,1)
d2 = date(2012,1,1)
rd = rdelta.relativedelta(d2,d1)
rd
relativedelta(years=+10, months=+8)