I have a date string formatted like this: "2017-05-31T06:44:13Z".
I need to check whether this date is within a one year span from today's date.
Which is the best method to do it: convert it into a timestamp and check, or convert into a date format?
Convert the timestamp to a datetime object so it can be compared with other datetime objects using <, >, =.
from datetime import datetime
from dateutil.relativedelta import relativedelta
# NOTE this format basically ignores the timezone. This may or may not be what you want
date_to_check = datetime.strptime('2017-05-31T06:44:13Z', '%Y-%m-%dT%H:%M:%SZ')
today = datetime.today()
one_year_from_now = today + relativedelta(years=1)
if today <= date_to_check <= one_year_from_now:
# do whatever
Use the datetime package together with timedelta:
import datetime
then = datetime.datetime.strptime("2017-05-31T06:44:13Z".replace('T',' ')[:-1],'%Y-%m-%d %H:%M:%S')
now = datetime.datetime.now()
d = datetime.timedelta(days = 365)
and simply check if now-d > then.
Related
I have a date object like
Date = '202011'
This is yyyymm format.
I want to get the same month but n years prior. For example if n = 2, then I should get '201811'
Is there any function available to achieve this?
You can use the datetime module and dateutil library for this:
import datetime
from dateutil.relativedelta import relativedelta
fmt = '%Y%m'
date_string = '202011'
n = 2
# Parse `date_string` into a date object.
date = datetime.datetime.strptime(date_string, fmt).date()
# 2020-11-01
# Subtract `n` years.
new_date = date + relativedelta(years=-n)
# Output in the same format.
print(new_date.strftime(fmt)) # -> 201811
Related questions:
Python date string to date object
How do I calculate the date six months from the current date using the datetime Python module?
Just parse it into a Python datetime object and use its replace() method.
from datetime import datetime
years_ago = 2
date = datetime.strptime('202011','%Y%m')
date = date.replace(year=date.year-years_ago)
# This is the modified date object
print(date)
# Formatted back in your format
print(date.strftime('%Y%m'))
This solution does not require any external dependency
Here's a similar solution to the others, but only using the standard library, and as a function.
def subtract_years(date_string: str, diff: int) -> str:
dt = datetime.strptime(date_string, "%Y%m")
new_dt = dt.replace(year=dt.year - diff)
return new_dt.strftime("%Y%m")
# ❯ subtract_years("202011", 2)
# '201811'
I'm trying to make a reminder app, but I can't find out how to show if the entered date is the current date. To give an example:
I added a reminder on December 19, 2021 at 17:45, and I want to know if this date is before today's date.
You can use datetime.datetime.strptime() to convert string to datetime object. Then you can calculate time delta with to check how much time is left.
import datetime
reminder = datetime.datetime.strptime('12/19/2021, 19:11:14',
"%m/%d/%Y, %H:%M:%S")
print(datetime.datetime.now()) # 2021-12-12 17:24:01.573420
time_delta = reminder - datetime.datetime.now()
print(time_delta.seconds) # 6432
print(time_delta.days) # 7
If the current date is after the reminder date, days will be negative.
You can use the python's builtin datetime module.
You can import it like this:
from datetime import datetime
To convert the reminder string to a date time object, you can use the strptime() function.
reminder_date = "December 19, 2021 at 17:45"
reminder = datetime.strptime(reminder_date, "%B %d, %Y at %H:%M")
Next, get the current datetime object.
current = datetime.now()
Finally, you can compare them using python's builtin comparison operators as follows.
if reminder < current:
print("passed")
else:
print("future")
Kindly help below my query:
I got an estimated time from API server like below:
2019-09-25T20:11:23+08:00
it seems like iso 8601 standard with timezone.
I would like to know how to calculate how many days, hours, minutes and seconds left from above value to the current time.
import datetime
Receved_time_frim_API = "2019-09-25T20:11:23+08:00"
Current_time = datetime.datetime.now()
left_days =
left_hour =
left_min =
left_sec =
Your time string contains timezone info. According to https://stackoverflow.com/a/13182163/12112986 it's easy to convert it to datetime object in python 3.7
import datetime
received = datetime.datetime.fromisoformat(Receved_time_frim_API)
In previous versions there is no easy oneliner to convert string with timezone to datetime object. If you're using earlier python version, you can try something crude, like
>>> date, timezone = Receved_time_frim_API.split("+")
>>> tz_hours, tz_minutes = timezone.split(":")
>>> date = datetime.datetime.strptime(date, "%Y-%m-%dT%H:%M:%S")
>>> date -= datetime.timedelta(hours=int(tz_hours))
>>> date -= datetime.timedelta(minutes=int(tz_minutes))
Note that this will work only in case of positive timezones
To substract two datetime objects use
td = date - Current_time
left_days = td.days
left_hour = td.seconds // 3600
left_min = (td.seconds//60)%60
left_sec = td.seconds % 60
Okay first you need to parse the Receved_time_frim_API into datetime format:
from dateutil import parser
Receved_time_frim_API = parser.parse("2019-09-25T20:11:23+08:00")
But you can't just substract this from your Current_time, because datetime.now() is not aware of a timezone:
from datetime import timezone
Current_time = datetime.datetime.now().replace(tzinfo=timezone.utc)
print (Current_time-Receved_time_frim_API)
The result is a datetime.timedelta
I have the following string format (Python 3.6):
'2018-11-19T10:04:57.426872'
I get it as a parameter to my script.
I want to get the date as 'YYYY-MM-DD' and time as 'HH:MM'
I tried to convert it with:
from datetime import datetime
if __name__ == '__main__':
start_timestamp = sys.argv[1]
start_date = datetime.strptime(sys.argv[1], '%Y-%m-%d')
start_time = datetime.strptime(sys.argv[1], '%H:%M')
But this gives:
ValueError: unconverted data remains: T10:04:57.426872
In the above example I want to see:
start_date = '2018-11-19'
start_time = '10:04'
Since the date seems to be in ISO-Format, a simple
start = datetime.datetime.fromisoformat(text)
will parse it correctly. From there you can get your date and time with
start_date = start.strftime("%Y-%m-%d")
start_time = start.strftime("%H:%M")
Edit:
For Python < 3.7, you can use this format:
start = datetime.datetime.strptime(text, "%Y-%m-%dT%H:%M:%S.%f")
For the "duplicate" datetime confusion: I used import datetime. If you use from datetime import datetime, you can get rid of the additional datetime.
Try this:We have one of the best package for parsing dates called dateutil.
from dateutil import parser
date1='2018-11-19T10:04:57.426872'
print 'Start_date:',parser.parse(date1).strftime("%Y-%m-%d")
print 'Start_time:',parser.parse(date1).strftime("%H:%M")
Result:Start_date:2018-11-19
Start_time:10:04
You need to parse the entire string into one datetime object and then extract your required values from that.
dt = datetime.datetime.strptime('2018-11-19T10:04:57.426872', '%Y-%m-%dT%H:%M:%S.%f')
d = dt.date()
t = dt.time()
print(d.strftime('%Y-%m-%d'))
print(t.strftime('%H:%M'))
Which outputs:
2018-11-19
10:04
As an input to an API request I need to get yesterday's date as a string in the format YYYY-MM-DD. I have a working version which is:
yesterday = datetime.date.fromordinal(datetime.date.today().toordinal()-1)
report_date = str(yesterday.year) + \
('-' if len(str(yesterday.month)) == 2 else '-0') + str(yesterday.month) + \
('-' if len(str(yesterday.day)) == 2 else '-0') + str(yesterday.day)
There must be a more elegant way to do this, interested for educational purposes as much as anything else!
You Just need to subtract one day from today's date. In Python datetime.timedelta object lets you create specific spans of time as a timedelta object.
datetime.timedelta(1) gives you the duration of "one day" and is subtractable from a datetime object. After you subtracted the objects you can use datetime.strftime in order to convert the result --which is a date object-- to string format based on your format of choice:
>>> from datetime import datetime, timedelta
>>> yesterday = datetime.now() - timedelta(1)
>>> type(yesterday)
>>> datetime.datetime
>>> datetime.strftime(yesterday, '%Y-%m-%d')
'2015-05-26'
Note that instead of calling the datetime.strftime function, you can also directly use strftime method of datetime objects:
>>> (datetime.now() - timedelta(1)).strftime('%Y-%m-%d')
'2015-05-26'
As a function:
from datetime import datetime, timedelta
def yesterday(frmt='%Y-%m-%d', string=True):
yesterday = datetime.now() - timedelta(1)
if string:
return yesterday.strftime(frmt)
return yesterday
example:
In [10]: yesterday()
Out[10]: '2022-05-13'
In [11]: yesterday(string=False)
Out[11]: datetime.datetime(2022, 5, 13, 12, 34, 31, 701270)
An alternative answer that uses today() method to calculate current date and then subtracts one using timedelta(). Rest of the steps remain the same.
https://docs.python.org/3.7/library/datetime.html#timedelta-objects
from datetime import date, timedelta
today = date.today()
yesterday = today - timedelta(days = 1)
print(today)
print(yesterday)
Output:
2019-06-14
2019-06-13
>>> import datetime
>>> datetime.date.fromordinal(datetime.date.today().toordinal()-1).strftime("%F")
'2015-05-26'
Calling .isoformat() on a date object will give you YYYY-MM-DD
from datetime import date, timedelta
(date.today() - timedelta(1)).isoformat()
I'm trying to use only import datetime based on this answer.
import datetime
oneday = datetime.timedelta(days=1)
yesterday = datetime.date.today() - oneday