CODEWITHSUNDEEP
pythonpython-2.7python-3.xdate
Michail N
Michail N

Reputation: 3845

Convert integer date format to human readable format

I have a date time format where dates are represented as integers from 1/1/1900 . For example: 1 is 1/1/1900 and 42998 is 20/9/2017.

How can I convert this format to a human readable format like dd/mm/yyyy ? I checked datetime documentation but I did not find any way to do this. I want to do this either on python 3 or 2.7.

Thanks for any suggestion.

Upvotes: 2

Views: 1291

Answers (3)

EdChum
EdChum

Reputation: 394091

You can define your dates as offsets from your basetime and construct a datetime:

In[22]:
import datetime as dt
dt.datetime(1900,1,1) + dt.timedelta(42998)

Out[22]: datetime.datetime(2017, 9, 22, 0, 0)

Once it's a datetime object you can convert this to a str via strftime using whatever format you desire:

In[24]:
(dt.datetime(1900,1,1) + dt.timedelta(42998-1)).strftime('%d/%m/%Y')

Out[24]: '21/09/2017'

So you can define a user func to do this:

In[27]:
def dtToStr(val):
    base = dt.datetime(1900,1,1)
    return (base + dt.timedelta(val-1)).strftime('%d/%m/%Y')
dtToStr(42998)

Out[27]: '21/09/2017'

Upvotes: 2

iurii_n
iurii_n

Reputation: 1370

import datetime

base_date = datetime.datetime(1900, 1, 1)
convert = lambda x: base_date + datetime.timedelta(days=x-1)
>>> convert(42998)
datetime.datetime(2017, 9, 21, 0, 0)

Upvotes: 1

p-robot
p-robot

Reputation: 4904

You can use a datetime object to do that. 

import datetime
d = datetime.datetime(1900, 1, 1, 0, 0)
d + datetime.timedelta(days = 42998)
>> datetime.datetime(2017, 9, 22, 0, 0)

Upvotes: 0

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