How to convert a given ordinal number (from Excel) to a date

Question

I have a Value 38142 I need to convert it into date format using python. if use this number in excel and do a right click and format cell at that time the value will be converted to 04/06/2004 and I need the same result using python. How can I achieve this

Answer 1

The offset in Excel is the number of days since 1900/01/01, with 1 being the first of January 1900, so add the number of days as a timedelta to 1899/12/31:

from datetime import datetime, timedelta

def from_excel_ordinal(ordinal: float, _epoch0=datetime(1899, 12, 31)) -> datetime:
    if ordinal >= 60:
        ordinal -= 1  # Excel leap year bug, 1900 is not a leap year!
    return (_epoch0 + timedelta(days=ordinal)).replace(microsecond=0)

You have to adjust the ordinal by one day for any date after 1900/02/28; Excel has inherited a leap year bug from Lotus 1-2-3 and treats 1900 as a leap year. The code above returns datetime(1900, 2, 28, 0, 0) for both 59 and 60 to correct for this, with fractional values in the range [59.0 - 61.0) all being a time between 00:00:00.0 and 23:59:59.999999 on that day.

The above also supports serials with a fraction to represent time, but since Excel doesn't support microseconds those are dropped.

Answer 2

from datetime import datetime, timedelta

def from_excel_ordinal(ordinal, epoch=datetime(1900, 1, 1)):
    # Adapted from above, thanks to @Martijn Pieters 

    if ordinal > 59:
        ordinal -= 1  # Excel leap year bug, 1900 is not a leap year!
    inDays = int(ordinal)
    frac = ordinal - inDays
    inSecs = int(round(frac * 86400.0))

    return epoch + timedelta(days=inDays - 1, seconds=inSecs) # epoch is day 1

excelDT = 42548.75001           # Float representation of 27/06/2016  6:00:01 PM in Excel format  
pyDT = from_excel_ordinal(excelDT)

The above answer is fine for just a date value, but here I extend the above solution to include time and return a datetime values as well.

Answer 3

I would recomment the following:

import pandas as pd

def convert_excel_time(excel_time):

    return pd.to_datetime('1900-01-01') + pd.to_timedelta(excel_time,'D')

Or

import datetime

def xldate_to_datetime(xldate):
    temp = datetime.datetime(1900, 1, 1)
    delta = datetime.timedelta(days=xldate)
    return temp+delta

Is taken from https://gist.github.com/oag335/9959241

Answer 4

I came to this question when trying to do the same above, but for entire columns within a df. I made this function, which did it for me:

import pandas as pd    
from datetime import datetime, timedelta
import copy as cp

def xlDateConv(df, *cols):      
    tempDt = []
    fin = cp.deepcopy(df)
    for col in [*cols]:
        for i in range(len(fin[col])):
            tempDate = datetime(1900, 1, 1)
            delta = timedelta(float(fin[col][i]))
            tempDt.append(pd.to_datetime(tempDate+delta))

        fin[col] = tempDt
        tempDt = []
    return fin

Note that you need to type each column, quoted (as string), as one parameter, which can most likely be improved (list of columns as input, for instance). Also, it returns a copy of the original df (doesn't change the original).

Btw, partly inspired by this (https://gist.github.com/oag335/9959241).

Answer 5

If you are working with Pandas this could be useful

    import xlrd
    import datetime as dt
    
    def from_excel_datetime(x):
        return dt.datetime(*xlrd.xldate_as_tuple(x, datemode=0))
    
    df['date'] = df.excel_date.map(from_excel_datetime)

If the date seems to be 4 years delayed, maybe you can try with datemode 1.

:param datemode: 0: 1900-based, 1: 1904-based.