How to convert date to timestamp using Python? - python

I need to convert this result to timestamp:
>>> print (datetime.date(2010, 1, 12) + datetime.timedelta(days = 3))
2010-01-15
I need to compare the value with this timestamp:
>>> datetime.datetime.now()
2011-10-24 10:43:43.371294
How can I achieve this?

I need to convert this result to timestamp
import time
mydate = datetime.date(2010, 1, 12) + datetime.timedelta(days = 3)
time.mktime(mydate.timetuple())
I need to compare the value with this timestamp:
a = datetime.datetime(2010, 1, 12) + datetime.timedelta(days = 3)
b = datetime.datetime.now()
a < b
a > b
a == b

oneDate = datetime.date(2010, 1, 12) + datetime.timedelta(days = 3)
now = datetime.datetime.now()
The first is date, the second is datetime. So if you just want to compare the dates (day, month, year), convert the second to date:
oneDate < now.date()
returns True

datetime.datetime.now() will return an instance of datetime.datetime which has a date() method returning an instance of datetime.date. You can then compare that to the result of datetime.date(2010, 1, 12) + datetime.timedelta(days = 3)

Related

Iterating over date range in python and setting the start and end date

I know this question is a repeated one. But what I am trying to do is, I want to iterate through a date range and for each iteration i need to set the fromDate and toDate.
for ex:
If I give the date range as startDate = '2022-10-31'
and endDate = '2022-11-04'
and for each iteration fromDate = '2022-10-31' and toDate = '2022-11-01'
next iteration fromDate = '2022-11-01' and endDate = '2022-11-02' and so on.
I did some research and got to know how to iterate through dateRange.
sample code:
import datetime
start_date = datetime.date(2022, 10, 31)
end_date = datetime.date(2022, 11, 04)
dates_2011_2013 = [ start_date + datetime.timedelta(n) for n in range(int ((end_date - start_date).days))]
This just prints the incremented dates in the date Range. Am new to Python language. Any help is appreciated.
Thank you.
You can change the code slightly to,
import datetime
start_date = datetime.date(2022, 10, 31)
end_date = datetime.date(2022, 11, 4)
dates_2011_2013 = [ (start_date + datetime.timedelta(n), start_date + datetime.timedelta(n+1)) for n in range(int ((end_date - start_date).days))]
[(datetime.date(2022, 10, 31), datetime.date(2022, 11, 1)),
(datetime.date(2022, 11, 1), datetime.date(2022, 11, 2)),
(datetime.date(2022, 11, 2), datetime.date(2022, 11, 3)),
(datetime.date(2022, 11, 3), datetime.date(2022, 11, 4))]
First item of the tuple is start date and the second item is end date.
I want to iterate through a date range
another option is using while:
while start_date < end_date:
print(start_date, start_date + datetime.timedelta(1))
start_date += datetime.timedelta(1)
'''
2022-10-31 2022-11-01
2022-11-01 2022-11-02
2022-11-02 2022-11-03
2022-11-03 2022-11-04

Exclude specific dates from date range - Python

I tried to exclude specific dates from a range of date and not getting valid output and below is the code. Could you please help to fix it.
from datetime import timedelta, date
sdate = date(2020, 7, 1)
edate = date(2020, 7, 7)
delta = edate - sdate
toRemoveDate = [date(2020, 7, 6), date(2020, 7, 2)]
for i in range(delta.days + 1):
day = sdate + timedelta(days=i)
print(day)
for j in range(len(toRemoveDate)):
if day != toRemoveDate[j]:
print(day)
Thanks,
Anand
try this,
from datetime import timedelta, date
sdate = date(2020, 7, 1)
edate = date(2020, 7, 7)
delta = edate - sdate
toRemoveDate = [date(2020, 7, 6), date(2020, 7, 2)]
for i in range(delta.days + 1):
day = sdate + timedelta(days=i)
if day not in toRemoveDate: # <-- check if day not in list
print(day)
Try .drop()
from datetime import timedelta, date
import pandas as pd
sdate = date(2020, 7, 1)
edate = date(2020, 7, 7)
range = pd.date_range(start=str(sdate), end=str(edate))
toRemoveDate = [date(2020, 7, 6), date(2020, 7, 2)]
days = range.drop(toRemoveDate)
print(days)

Convert two datetime objects to monthly datetime range

I have two datetime objects, let's say start_date and end_date. What I want to do is to make a list of datetime objects.
>>> from datetime import datetime
>>> from dateutil.rrule import rrule, MONTHLY
>>> start_date = datetime(2018, 9, 6, 0,)
>>> end_date = datetime(2018, 11, 26, 23, 59, 59)
>>> list(rrule(MONTHLY, dtstart=start_date, until=end_date))
[datetime.datetime(2018, 9, 6, 0, 0), datetime.datetime(2018, 10, 6, 0, 0), datetime.datetime(2018, 11, 6, 0, 0)]
I can do this with rrule(), it's moving from date to date but I want it to go month-wise and also include the end_date,
[
datetime.datetime(2018,9,6,0,0),
datetime.datetime(2018,9,30,23,59),
datetime.datetime(2018,10,1,0,0),
datetime.datetime(2018,10,31,23,59),
datetime.datetime(2018,11,1,0,0),
datetime.datetime(2018,11,26,23,59)
]
I would prefer to do this without using pandas or numpy.
Try this
import datetime
def last_day_of_month(date):
if date.month == 12:
return date.replace(day=31)
return date.replace(month=date.month+1, day=1) - datetime.timedelta(days=1)
def desired_output(start_date,end_date):
curr_date = start_date
desired_list = []
desired_list.append(start_date)
while curr_date<=end_date:
if last_day_of_month(curr_date)>end_date:
desired_list.append(end_date)
break
desired_list.append(last_day_of_month(curr_date))
curr_date = last_day_of_month(curr_date) + datetime.timedelta(1)
if curr_date<end_date:
desired_list.append(curr_date)
else:
break
return desired_list
print(desired_output(datetime.datetime(2018,9,6,0,0), datetime.datetime(2018,12,26,0,0)))
here's how you can get a list of dates between two ranges. you can modify it to add certain dates.
from datetime import datetime
from datetime import timedelta
import calendar
dates = []
start_date = datetime(2018, 9, 6, 0,)
end_date = datetime(2018, 10, 10, 0,)
index_date = start_date
while True:
index_date = index_date
dates.append(index_date)
dayy = calendar.monthrange(index_date.year,index_date.month)
index_date = index_date.replace(day= dayy[1])
if index_date > end_date:
break
dates.append(index_date)
index_date = index_date + timedelta(days=1)
if index_date > end_date:
break
print dates

Adding <type 'datetime.date'> and <type 'datetime.time'>

I have 2017-07-21 as a'datetime.date' object and
15:30:00 as a datetime.time' object
How do I combine the two to get
2017-07-21-15:30:00?
Use datetime.combine from datetime.
import datetime
d = datetime.date(2017, 07, 21)
t = datetime.time(15, 30, 0)
dt = datetime.datetime.combine(d, t)
classmethod datetime.combine(date, time, tzinfo=self.tzinfo)
Return a new datetime object whose date components are equal to the given date object’s, and whose time components are equal to the given time object’s.
source
>>> from datetime import datetime, date, time
>>> d = date(2017, 7, 21)
>>> t = time(15, 30, 0)
>>> newDate = datetime.combine(d, t)
>>> newDate
datetime.datetime(2017, 7, 21, 15, 30)
>>> newDate.strftime(%Y-%m-%d-%H:%M:%S)
'2017-07-21-15:30:00'
You can use the datetime.combine method as above. More about info here

Calculate the end of the previous quarter

I'm looking for an elegant and pythonic way to get the date of the end of the previous quarter.
Something like this:
def previous_quarter(reference_date):
...
>>> previous_quarter(datetime.date(2013, 5, 31))
datetime.date(2013, 3, 31)
>>> previous_quarter(datetime.date(2013, 2, 1))
datetime.date(2012, 12, 31)
>>> previous_quarter(datetime.date(2013, 3, 31))
datetime.date(2012, 12, 31)
>>> previous_quarter(datetime.date(2013, 11, 1))
datetime.date(2013, 9, 30)
Edit: Have I tried anything?
Yes, this seems to work:
def previous_quarter(ref_date):
current_date = ref_date - timedelta(days=1)
while current_date.month % 3:
current_date -= timedelta(days=1)
return current_date
But it seems unnecessarily iterative.
You can do it the "hard way" by just looking at the month you receive:
def previous_quarter(ref):
if ref.month < 4:
return datetime.date(ref.year - 1, 12, 31)
elif ref.month < 7:
return datetime.date(ref.year, 3, 31)
elif ref.month < 10:
return datetime.date(ref.year, 6, 30)
return datetime.date(ref.year, 9, 30)
Using dateutil:
import datetime as DT
import dateutil.rrule as rrule
def previous_quarter(date):
date = DT.datetime(date.year, date.month, date.day)
rr = rrule.rrule(
rrule.DAILY,
bymonth=(3,6,9,12), # the month must be one of these
bymonthday=-1, # the day has to be the last of the month
dtstart = date-DT.timedelta(days=100))
result = rr.before(date, inc=False) # inc=False ensures result < date
return result.date()
print(previous_quarter(DT.date(2013, 5, 31)))
# 2013-03-31
print(previous_quarter(DT.date(2013, 2, 1)))
# 2012-12-31
print(previous_quarter(DT.date(2013, 3, 31)))
# 2012-12-31
print(previous_quarter(DT.date(2013, 11, 1)))
# 2013-09-30
Exploit the data pattern involved and turn the problem into a table-lookup - your classic space-time tradeff:
from datetime import date
PQTBL = (((12,31,-1),)*3 + ((3,31,0),)*3 + ((6,30,0),)*3 + ((9,30,0),)*3)
def previous_quarter(ref):
entry = PQTBL[ref.month-1]
return date(ref.year+entry[2], entry[0], entry[1])
Find the first day and month of the quarter, then use relativedelta to subtract a day.
from dateutil.relativedelta import relativedelta
def previous_quarter(ref):
first_month_of_quarter = ((ref.month - 1) // 3) * 3 + 1
return ref.replace(month=first_month_of_quarter, day=1) - relativedelta(days=1)
It's almost certain you would be happier using pandas (a python library), it has many functions for "business time" data.
http://pandas.pydata.org/pandas-docs/dev/timeseries.html
Reworked Justin Ethier's code for a "next quarter" version. Also added timezone via pytz and strftime formatting. #justin-ethier
import pytz
from datetime import datetime, timedelta
import datetime as dt
def nextQuarter():
ref = datetime.now(pytz.timezone('America/New_York'))
if ref.month < 4:
next = dt.datetime(ref.year, 3, 31, 23, 59, 59).strftime('%m-%d-%Y %H:%M:%S')
elif ref.month < 7:
next = dt.datetime(ref.year, 6, 30, 23, 59, 59).strftime('%m-%d-%Y %H:%M:%S')
elif ref.month < 10:
next = dt.datetime(ref.year, 9, 30, 23, 59, 59).strftime('%m-%d-%Y %H:%M:%S')
else:
next = dt.datetime(ref.year + 1, 12, 31, 23, 59, 59).strftime('%m-%d-%Y %H:%M:%S')
return next
next = nextQuarter()
import datetime
def previous_quarter(ref):
quarter = (ref.month - 1) // 3
prev_quarter = (quarter - 1) % 4
return datetime.datetime(ref.year if quarter>0 else ref.year-1, prev_quarter*3+1, 1)
Solution using only python's datetime library -
import datetime
def get_quarter_end(dt):
'''
given a datetime object, find the end of the quarter
'''
quarter_of_month = int((dt.month-1)/3 + 1)
#======================================================
# find the first day of the next quarter
#======================================================
# if in last quarter then go to the next year
year = dt.year + 1 if quarter_of_month==4 else dt.year
# if in last quarter then month is january (or 1)
month = 1 if quarter_of_month==4 else (quarter_of_month*3) + 1
first_of_next_quarter = datetime.datetime(year = year,
month = month,
day = 1
)
# last day of quarter for dt will be minus 1 day of first of next quarter
quarter_end_dt = first_of_next_quarter - datetime.timedelta(days=1)
return quarter_end_dt
if __name__=='__main__':
dt = datetime.datetime.strptime('2016-07-15', '%Y-%m-%d')
target_dt = get_quarter_end(dt)
and if you want to retreive the last fours quarter you can do this
if ref.month < 4:
list1 = [datetime.date(ref.year - 1, 12, 31),
datetime.date(ref.year - 1, 9, 30),
datetime.date(ref.year - 1, 6, 30),
datetime.date(ref.year - 1, 3, 31)]
list1 = [i.strftime('%Y%m%d') for i in list1]
return list1
elif ref.month < 7:
return [datetime.date(ref.year, 3, 31),
datetime.date(ref.year - 1, 12, 31),
datetime.date(ref.year - 1, 9, 30),
datetime.date(ref.year - 1, 6, 30)]
elif ref.month < 10:
return [datetime.date(ref.year, 6, 30),
datetime.date(ref.year, 3, 31),
datetime.date(ref.year - 1, 12, 31),
datetime.date(ref.year - 1, 9, 30)]
return [datetime.date(ref.year, 9, 30),
datetime.date(ref.year, 6, 30),
datetime.date(ref.year, 3, 30),
datetime.date(ref.year - 1, 12, 31)]

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