How to accurately minus X month on a date in Python?

Question

So this code what I want:

import datetime
d = datetime.date.today()

three_months_ago = d - timedelta(months=3)

However, as we know, the 'months' param does not exist in timedelta.

I admit I can program like this to achieve the goal:

if d.month > 3:
    three_months_ago = datetime.date(d.year, d.month-3, d.day)
else:
    three_months_ago = datetime.date(d.year-1, d.month-3+12, d.day)

But this seems really stupid...

Can you guys tell me how to realize this smartly?

You could use [`dateparser`](https://dateparser.readthedocs.org/en/latest/). Have a look at this [answer](http://stackoverflow.com/a/35627268/2932244). — JRodDynamite, Mar 10 '16 at 06:11

score 15 · Accepted Answer · answered Mar 10 '16 at 06:25

15

This could help:

>>>from dateutil.relativedelta import relativedelta
>>>import datetime
>>>datetime.date.today()
datetime.date(2016, 3, 10)
>>>datetime.date.today() - relativedelta(months=3)
datetime.date(2015, 12, 10)

You can use relativedelta() to add or subtract weeks and years too.

answered Mar 10 '16 at 06:25

Girish Jadhav

174
4

Nice solution because don't have to work with pandas etc – igorkf Jun 25 '20 at 18:13

score 0 · Answer 2 · answered Mar 10 '16 at 06:14

0

Numpy's timedelta has months support, ie:

np.timedelta64(3, 'M')

answered Mar 10 '16 at 06:14

xvan

4,554
1
22
37

How to accurately minus X month on a date in Python?

2 Answers2