Convert Excel style date with pandas

Question

I have to parse an xml file which gives me datetimes in Excel style; for example: 42580.3333333333.

Does Pandas provide a way to convert that number into a regular datetime object?

Answer 1

OK I think the easiest thing is to construct a TimedeltaIndex from the floats and add this to the scalar datetime for 1900,1,1:

In [85]:
import datetime as dt
import pandas as pd
df = pd.DataFrame({'date':[42580.3333333333, 10023]})
df

Out[85]:
           date
0  42580.333333
1  10023.000000

In [86]:
df['real_date'] = pd.TimedeltaIndex(df['date'], unit='d') + dt.datetime(1900,1,1)
df

Out[86]:
           date                  real_date
0  42580.333333 2016-07-31 07:59:59.971200
1  10023.000000 1927-06-12 00:00:00.000000

OK it seems that excel is a bit weird with it's dates thanks @ayhan:

In [89]:
df['real_date'] = pd.TimedeltaIndex(df['date'], unit='d') + dt.datetime(1899, 12, 30)
df

Out[89]:
           date                  real_date
0  42580.333333 2016-07-29 07:59:59.971200
1  10023.000000 1927-06-10 00:00:00.000000

See related: How to convert a python datetime.datetime to excel serial date number

Answer 2

you can directly parse with pd.to_datetime, with keywords unit='D' and origin='1899-12-30':

import pandas as pd

df = pd.DataFrame({'xldate': [42580.3333333333]})

df['date'] = pd.to_datetime(df['xldate'], unit='D', origin='1899-12-30')

df['date']
Out[2]: 
0   2016-07-29 07:59:59.999971200
Name: date, dtype: datetime64[ns]

further reading:

• What is story behind December 30, 1899 as base date?
• an answer from Martijn Pieters how to handle excel ordinal value < 60 correctly

Answer 3

You can use the 3rd party xlrd library before passing to pd.to_datetime:

import xlrd

def read_date(date):
    return xlrd.xldate.xldate_as_datetime(date, 0)

df = pd.DataFrame({'date':[42580.3333333333, 10023]})

df['new'] = pd.to_datetime(df['date'].apply(read_date), errors='coerce')

print(df)

           date                 new
0  42580.333333 2016-07-29 08:00:00
1  10023.000000 1927-06-10 00:00:00

Answer 4

Another option to solve it with "apply lambda" for pd.Series:

xl_date = 42580

df['DATE SERIES'] = df['DATE SERIES'].apply(lambda x: datetime.fromordinal(datetime(1900, 1, 1).toordinal() + xldate - 2))