Calculate the same month from previous year-CodePudding

I have a date object like

Date = '202011'

This is yyyymm format.

I want to get the same month but n years prior. For example if n = 2, then I should get '201811'

Is there any function available to achieve this?

CodePudding user response：

Just parse it into a Python datetime object and use its replace() method.

from datetime import datetime

years_ago = 2
date = datetime.strptime('202011','%Y%m')
date = date.replace(year=date.year-years_ago)

# This is the modified date object
print(date)

# Formatted back in your format
print(date.strftime('%Y%m'))

This solution does not require any external dependency

CodePudding user response：

You can use the datetime module and dateutil library for this:

import datetime
from dateutil.relativedelta import relativedelta

fmt = '%Y%m'
date_string = '202011'
n = 2

# Parse `date_string` into a date object.
date = datetime.datetime.strptime(date_string, fmt).date()
# 2020-11-01

# Subtract `n` years.
new_date = date   relativedelta(years=-n)

# Output in the same format.
print(new_date.strftime(fmt))  # -> 201811

Related questions:

CodePudding user response：

Here's a similar solution to the others, but only using the standard library, and as a function.

def subtract_years(date_string: str, diff: int) -> str:
    dt = datetime.strptime(date_string, "%Y%m")
    new_dt = dt.replace(year=dt.year - diff)
    return new_dt.strftime("%Y%m")

# ❯ subtract_years("202011", 2)
# '201811'