Pandas convert numeric year to text value till day resolution

I have dataframe like as below

cust_id,purchase_date
   1,10/01/1998
   1,10/12/1999
   2,13/05/2016
   3,14/02/2018
   3,15/03/2019

I would like to do the below

a) display the output in text format as 5 years and 9 months instead of 5.93244 etc.

I tried the below

from datetime import timedelta
df['purchase_date'] = pd.to_datetime(df['purchase_date'])
gb = df_new.groupby(['unique_key'])
df_cust_age = gb['purchase_date'].agg(min_date=np.min, max_date=np.max).reset_index()
df_cust_age['diff_in_days'] = df_cust_age['max_date'] - df_cust_age['min_date']
df_cust_age['years_diff'] = df_cust_age['diff_in_days']/timedelta(days=365)

but the above code gives the output in decimal numbers.

I expect my output to be like as below

cust_id,years_diff
  1, 1 years and 11 months and 0 day
  2, 0 years
  3, 1 year and 1 month and 1 day

>Solution :

If possible create 'default' month with 30 days use this custom function:

#https://stackoverflow.com/a/13756038/2901002

def td_format(td_object):
    seconds = int(td_object.total_seconds())
    periods = [
        ('year',        60*60*24*365),
        ('month',       60*60*24*30),
        ('day',         60*60*24),
        ('hour',        60*60),
        ('minute',      60),
        ('second',      1)
    ]

    strings=[]
    for period_name, period_seconds in periods:
        if seconds > period_seconds:
            period_value , seconds = divmod(seconds, period_seconds)
            has_s = 's' if period_value > 1 else ''
            strings.append("%s %s%s" % (period_value, period_name, has_s))

    return ", ".join(strings) if len(strings) > 0 else '0 year'


df_cust_age['years_diff'] = df_cust_age['diff_in_days'].apply(td_format)
print (df_cust_age)
   cust_id   min_date   max_date diff_in_days       years_diff
0        1 1998-10-01 1999-10-12     376 days  1 year, 11 days
1        2 2016-05-13 2016-05-13       0 days           0 year
2        3 2018-02-14 2019-03-15     394 days  1 year, 29 days