How can I convert an Excel date (in a number format) to a proper date in Python?
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1Please clarify: give an example of "Excel data (in a number format)"Eli Bendersky– Eli Bendersky2009-07-10 08:51:11 +00:00Commented Jul 10, 2009 at 8:51
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5Internally, Excel stores dates as floating numbers, and you can distinguish from "normal" numbers only by the format of the cell.rob– rob2009-07-10 08:56:12 +00:00Commented Jul 10, 2009 at 8:56
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1@Roberto Liffredo, yes I know that Excel stored dates as floating numbers, I need to convert them to a proper date and that is why I am asking this question. @eliben, please see Roberto's commentGrzenio– Grzenio2009-07-10 09:26:31 +00:00Commented Jul 10, 2009 at 9:26
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You might also want to check the answer in the duplicate question.j-i-l– j-i-l2015-03-16 16:41:01 +00:00Commented Mar 16, 2015 at 16:41
14 Answers
You can use xlrd.
From its documentation, you can read that dates are always stored as numbers; however, you can use xldate_as_tuple to convert it to a python date.
Note: the version on the PyPI seems more up-to-date than the one available on xlrd's website.
Comments
Here's the bare-knuckle no-seat-belts use-at-own-risk version:
import datetime
def minimalist_xldate_as_datetime(xldate, datemode):
# datemode: 0 for 1900-based, 1 for 1904-based
return (
datetime.datetime(1899, 12, 30)
+ datetime.timedelta(days=xldate + 1462 * datemode)
)
3 Comments
datetimes for Excel dates before 1 Mar 1900. This is due to a bug in Excel that makes it (incorrectly) think that 1900 was a leap year. See Microsoft KB articleAfter testing and a few days wait for feedback, I'll svn-commit the following whole new function in xlrd's xldate module ... note that it won't be available to the diehards still running Python 2.1 or 2.2.
##
# Convert an Excel number (presumed to represent a date, a datetime or a time) into
# a Python datetime.datetime
# @param xldate The Excel number
# @param datemode 0: 1900-based, 1: 1904-based.
# <br>WARNING: when using this function to
# interpret the contents of a workbook, you should pass in the Book.datemode
# attribute of that workbook. Whether
# the workbook has ever been anywhere near a Macintosh is irrelevant.
# @return a datetime.datetime object, to the nearest_second.
# <br>Special case: if 0.0 <= xldate < 1.0, it is assumed to represent a time;
# a datetime.time object will be returned.
# <br>Note: 1904-01-01 is not regarded as a valid date in the datemode 1 system; its "serial number"
# is zero.
# @throws XLDateNegative xldate < 0.00
# @throws XLDateAmbiguous The 1900 leap-year problem (datemode == 0 and 1.0 <= xldate < 61.0)
# @throws XLDateTooLarge Gregorian year 10000 or later
# @throws XLDateBadDatemode datemode arg is neither 0 nor 1
# @throws XLDateError Covers the 4 specific errors
def xldate_as_datetime(xldate, datemode):
if datemode not in (0, 1):
raise XLDateBadDatemode(datemode)
if xldate == 0.00:
return datetime.time(0, 0, 0)
if xldate < 0.00:
raise XLDateNegative(xldate)
xldays = int(xldate)
frac = xldate - xldays
seconds = int(round(frac * 86400.0))
assert 0 <= seconds <= 86400
if seconds == 86400:
seconds = 0
xldays += 1
if xldays >= _XLDAYS_TOO_LARGE[datemode]:
raise XLDateTooLarge(xldate)
if xldays == 0:
# second = seconds % 60; minutes = seconds // 60
minutes, second = divmod(seconds, 60)
# minute = minutes % 60; hour = minutes // 60
hour, minute = divmod(minutes, 60)
return datetime.time(hour, minute, second)
if xldays < 61 and datemode == 0:
raise XLDateAmbiguous(xldate)
return (
datetime.datetime.fromordinal(xldays + 693594 + 1462 * datemode)
+ datetime.timedelta(seconds=seconds)
)
3 Comments
xldate_as_datetime function was added to the xldate module as of xlrd version 0.9.3 (released to PyPI in April 2014).xldate_as_datetime is a much cleaner option than xldate_as_tuple in my opinionPlease refer to this link: Reading date as a string not float from excel using python xlrd
it worked for me:
in shot this the link has:
import datetime, xlrd
book = xlrd.open_workbook("myfile.xls")
sh = book.sheet_by_index(0)
a1 = sh.cell_value(rowx=0, colx=0)
a1_as_datetime = datetime.datetime(*xlrd.xldate_as_tuple(a1, book.datemode))
print 'datetime: %s' % a1_as_datetime
4 Comments
strftime for my formatting a1_as_datetime = datetime.datetime(*xlrd.xldate_as_tuple(a1, book.datemode)).strftime('%Y-%m-%d %H:%M:%S')Expected situation
# Wrong output from cell_values()
42884.0
# Expected output
2017-5-29
Example: Let cell_values(2,2) from sheet number 0 will be the date targeted
Get the required variables as the following
workbook = xlrd.open_workbook("target.xlsx")
sheet = workbook.sheet_by_index(0)
wrongValue = sheet.cell_value(2,2)
And make use of xldate_as_tuple
year, month, day, hour, minutes, seconds = xlrd.xldate_as_tuple(wrongValue, workbook.datemode)
print("{0} - {1} - {2}".format(year, month, day))
That's my solution
Comments
Incase you're using pandas and your read_excel reads in Date formatted as Excel numbers improperly and need to recover the real dates behind...
The lambda function applied on the column uses xlrd to recover the date back
import xlrd
df['possible_intdate'] = df['possible_intdate'].apply(lambda s: xlrd.xldate.xldate_as_datetime(s, 0))
>> df['possible_intdate']
dtype('<M8[ns]')
Comments
Since there's a chance that your excel files are coming from different computers/people; there's a chance that the formatting is messy; so be extra cautious.
I just imported data from 50 odd excels where the dates were entered in DD/MM/YYYY or DD-MM-YYYY, but most of the Excel files stored them as MM/DD/YYYY (Probably because the PCs were setup with en-us instead of en-gb or en-in).
Even more irritating was the fact that dates above 13/MM/YYYY were in DD/MM/YYYY format still. So there was variations within the Excel files.
The most reliable solution I figured out was to manually set the Date column on each excel file to to be Plain Text -- then use this code to parse it:
if date_str_from_excel:
try:
return datetime.strptime(date_str_from_excel, '%d/%m/%Y')
except ValueError:
print("Unable to parse date")
Comments
excel stores dates and times as a number representing the number of days since 1900-Jan-0, if you want to get the dates in date format using python, just subtract 2 days from the days column, as shown below:
Date = sheet.cell(1,0).value-2 //in python
at column 1 in my excel, i have my date and above command giving me date values minus 2 days, which is same as date present in my excel sheet
Comments
This is a revised version from @hounded. My code handles both date and time, something like 43705.591795706
import math
import datetime
def xldate_to_datetime(xldatetime): #something like 43705.6158241088
tempDate = datetime.datetime(1899, 12, 31)
(days, portion) = math.modf(xldatetime)
deltaDays = datetime.timedelta(days=days)
#changing the variable name in the edit
secs = int(24 * 60 * 60 * portion)
detlaSeconds = datetime.timedelta(seconds=secs)
TheTime = (tempDate + deltaDays + detlaSeconds )
return TheTime.strftime("%Y-%m-%d %H:%M:%S")
xldate_to_datetime(43705.6158241088)
# 2019-08-29 14:46:47
1 Comment
A combination of peoples post gave me the date and the time for excel conversion. I did return it as a string
def xldate_to_datetime(xldate):
tempDate = datetime.datetime(1900, 1, 1)
deltaDays = datetime.timedelta(days=int(xldate))
secs = (int((xldate%1)*86400)-60)
detlaSeconds = datetime.timedelta(seconds=secs)
TheTime = (tempDate + deltaDays + detlaSeconds )
return TheTime.strftime("%Y-%m-%d %H:%M:%S")
1 Comment
When converting an excel file to CSV the date/time cell looks like this:
foo, 3/16/2016 10:38, bar,
To convert the datetime text value to datetime python object do this:
from datetime import datetime
date_object = datetime.strptime('3/16/2016 10:38', '%m/%d/%Y %H:%M') # excel format (CSV file)
print date_object will return 2005-06-01 13:33:00