groupby datediff in pandas - python-2.7

Question

I am trying to get the difference between the min date and max date a product is sold in terms of month in a new column.But I am having an unusual return when apply function in groupby.
Any help is much appreciated.
So my steps are :
data :
pch_date day product qty unit_price total_price year_month
421 2013-01-07 tuesday p3 13 4.58 59.54 1
141 2015-09-13 monday p8 3 3.77 11.31 9
249 2015-02-02 monday p5 3 1.80 5.40 2
826 2015-10-09 tuesday p5 6 1.80 10.80 10
427 2014-04-18 friday p7 6 4.21 25.26 4
function definition :
def diff_date(x):
max_date = x.max()
min_date = x.min()
diff_month = (max_date.year - min_date.year)*12 + max_date.month +1
return diff_month
When trying for test:
print diff_date(prod_df['pch_date'])
49 which is correct
But Problem:
print prod_df[['product','pch_date']].groupby(['product']).agg({'pch_date': diff_date}).reset_index()[:5]
Results coming with a extra date:
product pch_date
0 p1 1970-01-01 00:00:00.000000049
1 p10 1970-01-01 00:00:00.000000048
2 p11 1970-01-01 00:00:00.000000045
3 p12 1970-01-01 00:00:00.000000049
4 p13 1970-01-01 00:00:00.000000045
How to get the difference in integer ?

Answer 1

You can use Groupby.apply instead which returns integers and not datetime objects.
df.groupby(['product'])['pch_date'].apply(diff_date).reset_index()
As a workaround for not letting the integer values getting converted to their DatetimeIndex values, you can change the last line of your function to str(diff_month) and you can continue using Groupby.agg as shown:
df.groupby(['product'])['pch_date'].agg({'pch_date': diff_date}).reset_index()