I need to parse strings representing 6-digit dates in the format yymmdd where yy ranges from 59 to 05 (1959 to 2005). According to the time module docs, Python's default pivot year is 1969 which won't work for me.
Is there an easy way to override the pivot year, or can you suggest some other solution? I am using Python 2.7. Thanks!
I'd use datetime and parse it out normally. Then I'd use datetime.datetime.replace on the object if it is past your ceiling date -- Adjusting it back 100 yrs.:
import datetime
dd = datetime.datetime.strptime(date,'%y%m%d')
if dd.year > 2005:
dd = dd.replace(year=dd.year-100)
Prepend the century to your date using your own pivot:
year = int(date[0:2])
if 59 <= year <= 99:
date = '19' + date
else
date = '20' + date
and then use strptime with the %Y directive instead of %y.
import datetime
date = '20-Apr-53'
dt = datetime.datetime.strptime( date, '%d-%b-%y' )
if dt.year > 2000:
dt = dt.replace( year=dt.year-100 )
^2053 ^1953
print dt.strftime( '%Y-%m-%d' )
You can also perform the following:
today=datetime.datetime.today().strftime("%m/%d/%Y")
today=today[:-4]+today[-2:]
Recently had a similar case, ended up with this basic calculation and logic:
pivotyear = 1969
century = int(str(pivotyear)[:2]) * 100
def year_2to4_digit(year):
return century + year if century + year > pivotyear else (century + 100) + year
If you are dealing with very recent dates as well as very old dates and want to use the current date as a pivot (not just the current year), try this code:
import datetime
def parse_date(date_str):
parsed = datetime.datetime.strptime(date_str,'%y%m%d')
current_date = datetime.datetime.now()
if parsed > current_date:
parsed = parsed.replace(year=parsed.year - 100)
return parsed
Related
Is there a simple way to convert numbers like 1030, 0131, 1231, etc to an mm-dd (or even mm-dd-yyyy assuming everything is 2020) format, so that I can perform date calculations with them? For example, I would like to be able to do (1231 minus 0131) = 11 months.
Of course, I could just do the following to change the formatting, but looking to see if there's a more intuitive way!
startDate = startDate[:2] + "-" + startDate[2:] + "-2020"
endDate = endDate[:2] + "-" + endDate[2:] + "-2020"
You can convert to datetime object straight away, using strptime directive "%m%d". This allows you to make timedelta calculations later on. The default year is 1900; if you wish, you can replace by 2020.
from datetime import datetime
startDate, endDate = "0131", "1231"
startDate, endDate = (datetime.strptime(startDate, "%m%d").replace(year=2020),
datetime.strptime(endDate,"%m%d").replace(year=2020))
deltaDays = (endDate-startDate).days
# 335
Let me know if it is not exactly what you want:
from datetime import datetime
s = "1030"
d = '-'.join([s[:2], s[2:]]) + '-2020'
date_object = datetime.strptime(d, '%m-%d-%Y').date()
I have a string as Julian date like "16152" meaning 152'nd day of 2016 or "15234" meaning 234'th day of 2015.
How can I convert these Julian dates to format like 20/05/2016 using Python 3 standard library?
I can get the year 2016 like this: date = 20 + julian[0:1], where julian is the string containing the Julian date, but how can I calculate the rest according to 1th of January?
The .strptime() method supports the day of year format:
>>> import datetime
>>>
>>> datetime.datetime.strptime('16234', '%y%j').date()
datetime.date(2016, 8, 21)
And then you can use strftime() to reformat the date
>>> date = datetime.date(2016, 8, 21)
>>> date.strftime('%d/%m/%Y')
'21/08/2016'
Well, first, create a datetime object (from the module datetime)
from datetime import datetime
from datetime import timedelta
julian = ... # Your julian datetime
date = datetime.strptime("1/1/" + jul[:2], "%m/%d/%y")
# Just initializing the start date, which will be January 1st in the year of the Julian date (2 first chars)
Now add the days from the start date:
daysToAdd = int(julian[2:]) # Taking the days and converting to int
date += timedelta(days = daysToAdd - 1)
Now, you can just print it as is:
print(str(date))
Or you can use strftime() function.
print(date.strftime("%d/%m/%y"))
Read more about strftime format string here
Easy way
Convert from regular date to Julian date
print datetime.datetime.now().strftime("%y%j")
Convert from Julian date to regular date
print datetime.datetime.strptime('19155', '%y%j').strftime("%d-%m-%Y")
I used this for changing a Juian date to xml xsd:datetime
def julianDate2ISO8601(d, offset='+00:00'):
"""
return ISO8601 formated datetime from julian date
optional offset [+|-]hh:mm
"""
d = str(d) # make sure it is a string
# replace leading number with correct century
centuryArray = ['19','20','21']
d = centuryArray[int(d[:1])] + d[1:]
# format string to iso 8601 datetime
return datetime.datetime.strptime(d, '%Y%j').date().strftime(
'%Y-%m-%dT00:00:00') + offset
I must have the current year and month in datetime.
I use this:
datem = datetime.today().strftime("%Y-%m")
datem = datetime.strptime(datem, "%Y-%m")
Is there maybe another way?
Try this solution:
from datetime import datetime
currentSecond= datetime.now().second
currentMinute = datetime.now().minute
currentHour = datetime.now().hour
currentDay = datetime.now().day
currentMonth = datetime.now().month
currentYear = datetime.now().year
Use:
from datetime import datetime
today = datetime.today()
datem = datetime(today.year, today.month, 1)
I assume you want the first of the month.
Use:
from datetime import datetime
current_month = datetime.now().strftime('%m') // 02 //This is 0 padded
current_month_text = datetime.now().strftime('%h') // Feb
current_month_text = datetime.now().strftime('%B') // February
current_day = datetime.now().strftime('%d') // 23 //This is also padded
current_day_text = datetime.now().strftime('%a') // Fri
current_day_full_text = datetime.now().strftime('%A') // Friday
current_weekday_day_of_today = datetime.now().strftime('%w') //5 Where 0 is Sunday and 6 is Saturday.
current_year_full = datetime.now().strftime('%Y') // 2018
current_year_short = datetime.now().strftime('%y') // 18 without century
current_second= datetime.now().strftime('%S') //53
current_minute = datetime.now().strftime('%M') //38
current_hour = datetime.now().strftime('%H') //16 like 4pm
current_hour = datetime.now().strftime('%I') // 04 pm
current_hour_am_pm = datetime.now().strftime('%p') // 4 pm
current_microseconds = datetime.now().strftime('%f') // 623596 Rarely we need.
current_timzone = datetime.now().strftime('%Z') // UTC, EST, CST etc. (empty string if the object is naive).
Reference: 8.1.7. strftime() and strptime() Behavior
Reference: strftime() and strptime() Behavior
The above things are useful for any date parsing, not only now or today. It can be useful for any date parsing.
e.g.
my_date = "23-02-2018 00:00:00"
datetime.strptime(str(my_date),'%d-%m-%Y %H:%M:%S').strftime('%Y-%m-%d %H:%M:%S+00:00')
datetime.strptime(str(my_date),'%d-%m-%Y %H:%M:%S').strftime('%m')
And so on...
>>> from datetime import date
>>> date.today().month
2
>>> date.today().year
2020
>>> date.today().day
13
The question asks to use datetime specifically.
This is a way that uses datetime only:
from datetime import datetime
year = datetime.now().year
month = datetime.now().month
Late answer, but you can also use:
import time
ym = time.strftime("%Y-%m")
You can write the accepted answer as a one-liner using date.replace:
datem = datetime.today().replace(day=1)
You can always use a sub-string method:
import datetime;
today = str(datetime.date.today());
curr_year = int(today[:4]);
curr_month = int(today[5:7]);
This will get you the current month and year in integer format. If you want them to be strings you simply have to remove the " int " precedence while assigning values to the variables curr_year and curr_month.
using the regular expression extract the date and time from the string then convert the string date to a datetimeindex using strptime
import re
s = "Audio was recorded at 21:50:00 02/07/2019 (UTC) by device 243B1F05 at gain setting 2 while battery state was 3.6V."
t = re.search(' (\d{2}:\d{2}:\d{2} \d{2}\/\d{2}\/\d{4}) ', s).group(1)
date=datetime.datetime.strptime(t,'%H:%M:%S %m/%d/%Y')
print(type(date))
Please what's wrong with my code:
import datetime
d = "2013-W26"
r = datetime.datetime.strptime(d, "%Y-W%W")
print(r)
Display "2013-01-01 00:00:00", Thanks.
A week number is not enough to generate a date; you need a day of the week as well. Add a default:
import datetime
d = "2013-W26"
r = datetime.datetime.strptime(d + '-1', "%Y-W%W-%w")
print(r)
The -1 and -%w pattern tells the parser to pick the Monday in that week. This outputs:
2013-07-01 00:00:00
%W uses Monday as the first day of the week. While you can pick your own weekday, you may get unexpected results if you deviate from that.
See the strftime() and strptime() behaviour section in the documentation, footnote 4:
When used with the strptime() method, %U and %W are only used in calculations when the day of the week and the year are specified.
Note, if your week number is a ISO week date, you'll want to use %G-W%V-%u instead! Those directives require Python 3.6 or newer.
In Python 3.8 there is the handy datetime.date.fromisocalendar:
>>> from datetime import date
>>> date.fromisocalendar(2020, 1, 1) # (year, week, day of week)
datetime.date(2019, 12, 30, 0, 0)
In older Python versions (3.7-) the calculation can use the information from datetime.date.isocalendar to figure out the week ISO8601 compliant weeks:
from datetime import date, timedelta
def monday_of_calenderweek(year, week):
first = date(year, 1, 1)
base = 1 if first.isocalendar()[1] == 1 else 8
return first + timedelta(days=base - first.isocalendar()[2] + 7 * (week - 1))
Both works also with datetime.datetime.
To complete the other answers - if you are using ISO week numbers, this string is appropriate (to get the Monday of a given ISO week number):
import datetime
d = '2013-W26'
r = datetime.datetime.strptime(d + '-1', '%G-W%V-%u')
print(r)
%G, %V, %u are ISO equivalents of %Y, %W, %w, so this outputs:
2013-06-24 00:00:00
Availabe in Python 3.6+; from docs.
import datetime
res = datetime.datetime.strptime("2018 W30 w1", "%Y %W w%w")
print res
Adding of 1 as week day will yield exact current week start. Adding of timedelta(days=6) will gives you the week end.
datetime.datetime(2018, 7, 23)
If anyone is looking for a simple function that returns all working days (Mo-Fr) dates from a week number consider this (based on accepted answer)
import datetime
def weeknum_to_dates(weeknum):
return [datetime.datetime.strptime("2021-W"+ str(weeknum) + str(x), "%Y-W%W-%w").strftime('%d.%m.%Y') for x in range(-5,0)]
weeknum_to_dates(37)
Output:
['17.09.2021', '16.09.2021', '15.09.2021', '14.09.2021', '13.09.2021']
In case you have the yearly number of week, just add the number of weeks to the first day of the year.
>>> import datetime
>>> from dateutil.relativedelta import relativedelta
>>> week = 40
>>> year = 2019
>>> date = datetime.date(year,1,1)+relativedelta(weeks=+week)
>>> date
datetime.date(2019, 10, 8)
Another solution which worked for me that accepts series data as opposed to strptime only accepting single string values:
#fw_to_date
import datetime
import pandas as pd
# fw is input in format 'YYYY-WW'
# Add weekday number to string 1 = Monday
fw = fw + '-1'
# dt is output column
# Use %G-%V-%w if input is in ISO format
dt = pd.to_datetime(fw, format='%Y-%W-%w', errors='coerce')
Here's a handy function including the issue with zero-week.
I need to convert dates into Excel serial numbers for a data munging script I am writing. By playing with dates in my OpenOffice Calc workbook, I was able to deduce that '1-Jan 1899 00:00:00' maps to the number zero.
I wrote the following function to convert from a python datetime object into an Excel serial number:
def excel_date(date1):
temp=dt.datetime.strptime('18990101', '%Y%m%d')
delta=date1-temp
total_seconds = delta.days * 86400 + delta.seconds
return total_seconds
However, when I try some sample dates, the numbers are different from those I get when I format the date as a number in Excel (well OpenOffice Calc). For example, testing '2009-03-20' gives 3478032000 in Python, whilst excel renders the serial number as 39892.
What is wrong with the formula above?
*Note: I am using Python 2.6.3, so do not have access to datetime.total_seconds()
It appears that the Excel "serial date" format is actually the number of days since 1900-01-00, with a fractional component that's a fraction of a day, based on http://www.cpearson.com/excel/datetime.htm. (I guess that date should actually be considered 1899-12-31, since there's no such thing as a 0th day of a month)
So, it seems like it should be:
def excel_date(date1):
temp = dt.datetime(1899, 12, 30) # Note, not 31st Dec but 30th!
delta = date1 - temp
return float(delta.days) + (float(delta.seconds) / 86400)
While this is not exactly relevant to the excel serial date format, this was the top hit for exporting python date time to Excel. What I have found particularly useful and simple is to just export using strftime.
import datetime
current_datetime = datetime.datetime.now()
current_datetime.strftime('%x %X')
This will output in the following format '06/25/14 09:59:29' which is accepted by Excel as a valid date/time and allows for sorting in Excel.
if the problem is that we want DATEVALUE() excel serial number for dates, the toordinal() function can be used. Python serial numbers start from Jan1 of year 1 whereas excel starts from 1 Jan 1900 so apply an offset. Also see excel 1900 leap year bug (https://support.microsoft.com/en-us/help/214326/excel-incorrectly-assumes-that-the-year-1900-is-a-leap-year)
def convert_date_to_excel_ordinal(day, month, year) :
offset = 693594
current = date(year,month,day)
n = current.toordinal()
return (n - offset)
With the 3rd party xlrd.xldate module, you can supply a tuple structured as (year, month, day, hour, minute, second) and, if necessary, calculate a day fraction from any microseconds component:
from datetime import datetime
from xlrd import xldate
from operator import attrgetter
def excel_date(input_date):
components = ('year', 'month', 'day', 'hour', 'minute', 'second')
frac = input_date.microsecond / (86400 * 10**6) # divide by microseconds in one day
return xldate.xldate_from_datetime_tuple(attrgetter(*components)(input_date), 0) + frac
res = excel_date(datetime(1900, 3, 1, 12, 0, 0, 5*10**5))
# 61.50000578703704
According to #akgood's answer, when the datetime is before 1/0/1900, the return value is wrong, the corrected return expression may be:
def excel_date(date1):
temp = dt.datetime(1899, 12, 30) # Note, not 31st Dec but 30th!
delta = date1 - temp
return float(delta.days) + (-1.0 if delta.days < 0 else 1.0)*(delta.seconds)) / 86400
This worked when I tested using the csv package to create a spreadsheet:
from datetime import datetime
def excel_date(date1):
return date1.strftime('%x %-I:%M:%S %p')
now = datetime.now()
current_datetime=now.strftime('%x %-I:%M:%S %p')
time_data.append(excel_date(datetime.now()))
...