I have looked around to see if I can find a simple method in Python to find out if a date has passed.
For example:-
If the date is 01/05/2015, and the date; 30/04/2015 was in-putted into Python, it would return True, to say the date has passed.
This needs to be as simple and efficient as possible.
Thanks for any help.
you may use datetime, first parse String to date, then you can compare
import datetime
d1 = datetime.datetime.strptime('05/01/2015', "%d/%m/%Y").date()
d2 = datetime.datetime.strptime('30/04/2015', "%d/%m/%Y").date()
d2>d1
from datetime import datetime
present = datetime.now()
print datetime(2015,04,30) < present #should return true
Sourced some material from this question/answer: How to compare two dates?
Just compare them?
>>> t1 = datetime.datetime.now()
>>> t2 = datetime.datetime.now()
>>> t1>t2
False
>>> t1<t2
True
You can create a simple function which does this:
def has_expired(date):
import datetime
return date < datetime.datetime.now()
Related
I need to get the current date with numbers, like that: 14:45:35:233 08.05.2016. I didn't find a function for that in the time module, is there any short way to do that?
It sounds like you want parse from specific date time format to another one. Maybe it'd be what you looking for, take a look:
>>> import datetime
>>> strdate = '14:45:35:233 08.05.2016'
>>> dt = datetime.datetime.strptime(strdate, '%H:%M:%S:233 %d.%m.%Y')
>>> dt.strftime('%Y-%m-%d')
'2016-05-08'
Use strftime() and get whatever format you need.
Use strftime to format time.
datetime.datetime.now() will give current time.
To get the time in "Hours:Minutes:Seconds:Microseconds Date.Month.Year " Format use strftime("%H:%M:%S:%f %d.%m.%Y")
import datetime
a=datetime.datetime.now()
a.strftime("%H:%M:%S:%f %d.%m.%Y")
Output
'16:50:54:238874 08.05.2016'
Use this :
>>> import datetime
>>> import time
>>> ts = time.time()
>>> datetime.datetime.fromtimestamp(ts).strftime('%H.%M.%S %d.%m.%Y')
Output :
'18.20.06 08.05.2016'
I have a string output from another program that shows the date as
16/05/03 # (YY/MM/DD)
and I wish to change it to
03/05/16 #(DD/MM/YY)
and here is how the date is supplied
(date = info[4].replace('"', '')
i have tried
dates = str(date)[::-1]
but that gave me an output of
40/50/61
not quite what I wanted
any ideas using a minimal code as possible?
>>> '/'.join('16/05/03'.split('/')[::-1])
'03/05/16'
or
>>> '/'.join(reversed('16/05/03'.split('/')))
'03/05/16'
or using datetime library:
>> from datetime import datetime
>>> datetime.strftime(datetime.strptime('16/05/03', '%y/%m/%d'), '%d/%m/%y')
'03/05/16'
Using datetime give you alot more control with changing the format to suite what you want.
import datetime
d = datetime.strptime('16/05/03', '%y/%m/%d')
print d.strftime('%d/%m/%y')
I have a datetime object and I'm trying to individually get a string with the date, and one with the time. I'd like the values for theDate and theTime to be strings.
theDate = myDatetime.date()
theTime = myDatetime.time()
Something along those lines. I tried str(datetime.date) but it gave me a reference in memory, any other ideas? Thanks in advance for any help.
Use the datetime.strftime() method on the datetime object:
theDate = myDatetime.strftime('%Y-%m-%d')
theTime = myDatetime.strftime('%H:%M:%S')
Alternatively, turn your date and time objects into strings for their default string representations:
theDate = str(myDatetime.date())
theTime = str(myDatetime.time())
Demo:
>>> import datetime
>>> myDatetime = datetime.datetime.now()
>>> myDatetime.strftime('%Y-%m-%d')
'2013-06-19'
>>> myDatetime.strftime('%H:%M:%S')
'16:49:44'
>>> str(myDatetime.date())
'2013-06-19'
>>> str(myDatetime.time())
'16:49:44.447010'
The default string format for datetime.time objects includes the microsecond component.
I have two different dates and I want to know the difference in days between them. The format of the date is YYYY-MM-DD.
I have a function that can ADD or SUBTRACT a given number to a date:
def addonDays(a, x):
ret = time.strftime("%Y-%m-%d",time.localtime(time.mktime(time.strptime(a,"%Y-%m-%d"))+x*3600*24+3600))
return ret
where A is the date and x the number of days I want to add. And the result is another date.
I need a function where I can give two dates and the result would be an int with date difference in days.
Use - to get the difference between two datetime objects and take the days member.
from datetime import datetime
def days_between(d1, d2):
d1 = datetime.strptime(d1, "%Y-%m-%d")
d2 = datetime.strptime(d2, "%Y-%m-%d")
return abs((d2 - d1).days)
Another short solution:
from datetime import date
def diff_dates(date1, date2):
return abs(date2-date1).days
def main():
d1 = date(2013,1,1)
d2 = date(2013,9,13)
result1 = diff_dates(d2, d1)
print '{} days between {} and {}'.format(result1, d1, d2)
print ("Happy programmer's day!")
main()
You can use the third-party library dateutil, which is an extension for the built-in datetime.
Parsing dates with the parser module is very straightforward:
from dateutil import parser
date1 = parser.parse('2019-08-01')
date2 = parser.parse('2019-08-20')
diff = date2 - date1
print(diff)
print(diff.days)
Answer based on the one from this deleted duplicate
I tried the code posted by larsmans above but, there are a couple of problems:
1) The code as is will throw the error as mentioned by mauguerra
2) If you change the code to the following:
...
d1 = d1.strftime("%Y-%m-%d")
d2 = d2.strftime("%Y-%m-%d")
return abs((d2 - d1).days)
This will convert your datetime objects to strings but, two things
1) Trying to do d2 - d1 will fail as you cannot use the minus operator on strings and
2) If you read the first line of the above answer it stated, you want to use the - operator on two datetime objects but, you just converted them to strings
What I found is that you literally only need the following:
import datetime
end_date = datetime.datetime.utcnow()
start_date = end_date - datetime.timedelta(days=8)
difference_in_days = abs((end_date - start_date).days)
print difference_in_days
Try this:
data=pd.read_csv('C:\Users\Desktop\Data Exploration.csv')
data.head(5)
first=data['1st Gift']
last=data['Last Gift']
maxi=data['Largest Gift']
l_1=np.mean(first)-3*np.std(first)
u_1=np.mean(first)+3*np.std(first)
m=np.abs(data['1st Gift']-np.mean(data['1st Gift']))>3*np.std(data['1st Gift'])
pd.value_counts(m)
l=first[m]
data.loc[:,'1st Gift'][m==True]=np.mean(data['1st Gift'])+3*np.std(data['1st Gift'])
data['1st Gift'].head()
m=np.abs(data['Last Gift']-np.mean(data['Last Gift']))>3*np.std(data['Last Gift'])
pd.value_counts(m)
l=last[m]
data.loc[:,'Last Gift'][m==True]=np.mean(data['Last Gift'])+3*np.std(data['Last Gift'])
data['Last Gift'].head()
I tried a couple of codes, but end up using something as simple as (in Python 3):
from datetime import datetime
df['difference_in_datetime'] = abs(df['end_datetime'] - df['start_datetime'])
If your start_datetime and end_datetime columns are in datetime64[ns] format, datetime understands it and return the difference in days + timestamp, which is in timedelta64[ns] format.
If you want to see only the difference in days, you can separate only the date portion of the start_datetime and end_datetime by using (also works for the time portion):
df['start_date'] = df['start_datetime'].dt.date
df['end_date'] = df['end_datetime'].dt.date
And then run:
df['difference_in_days'] = abs(df['end_date'] - df['start_date'])
pd.date_range('2019-01-01', '2019-02-01').shape[0]
I'm adding UTC time strings to Bitbucket API responses that currently only contain Amsterdam (!) time strings. For consistency with the UTC time strings returned elsewhere, the desired format is 2011-11-03 11:07:04 (followed by +00:00, but that's not germane).
What's the best way to create such a string (without a microsecond component) from a datetime instance with a microsecond component?
>>> import datetime
>>> print unicode(datetime.datetime.now())
2011-11-03 11:13:39.278026
I'll add the best option that's occurred to me as a possible answer, but there may well be a more elegant solution.
Edit: I should mention that I'm not actually printing the current time – I used datetime.now to provide a quick example. So the solution should not assume that any datetime instances it receives will include microsecond components.
If you want to format a datetime object in a specific format that is different from the standard format, it's best to explicitly specify that format:
>>> import datetime
>>> datetime.datetime.now().strftime("%Y-%m-%d %H:%M:%S")
'2011-11-03 18:21:26'
See the documentation of datetime.strftime() for an explanation of the % directives.
Starting from Python 3.6, the isoformat() method is flexible enough to also produce this format:
datetime.datetime.now().isoformat(sep=" ", timespec="seconds")
>>> import datetime
>>> now = datetime.datetime.now()
>>> print unicode(now.replace(microsecond=0))
2011-11-03 11:19:07
In Python 3.6:
from datetime import datetime
datetime.now().isoformat(' ', 'seconds')
'2017-01-11 14:41:33'
https://docs.python.org/3.6/library/datetime.html#datetime.datetime.isoformat
This is the way I do it. ISO format:
import datetime
datetime.datetime.now().replace(microsecond=0).isoformat()
# Returns: '2017-01-23T14:58:07'
You can replace the 'T' if you don't want ISO format:
datetime.datetime.now().replace(microsecond=0).isoformat(' ')
# Returns: '2017-01-23 15:05:27'
Yet another option:
>>> import time
>>> time.strftime("%Y-%m-%d %H:%M:%S")
'2011-11-03 11:31:28'
By default this uses local time, if you need UTC you can use the following:
>>> time.strftime("%Y-%m-%d %H:%M:%S", time.gmtime())
'2011-11-03 18:32:20'
Keep the first 19 characters that you wanted via slicing:
>>> str(datetime.datetime.now())[:19]
'2011-11-03 14:37:50'
I usually do:
import datetime
now = datetime.datetime.now()
now = now.replace(microsecond=0) # To print now without microsecond.
# To print now:
print(now)
output:
2019-01-13 14:40:28
Since not all datetime.datetime instances have a microsecond component (i.e. when it is zero), you can partition the string on a "." and take only the first item, which will always work:
unicode(datetime.datetime.now()).partition('.')[0]
As of Python 3.6+, the best way of doing this is by the new timespec argument for isoformat.
isoformat(timespec='seconds', sep=' ')
Usage:
>>> datetime.now().isoformat(timespec='seconds')
'2020-10-16T18:38:21'
>>> datetime.now().isoformat(timespec='seconds', sep=' ')
'2020-10-16 18:38:35'
We can try something like below
import datetime
date_generated = datetime.datetime.now()
date_generated.replace(microsecond=0).isoformat(' ').partition('+')[0]
>>> from datetime import datetime
>>> dt = datetime.now().strftime("%Y-%m-%d %X")
>>> print(dt)
'2021-02-05 04:10:24'
f-string formatting
>>> import datetime
>>> print(f'{datetime.datetime.now():%Y-%m-%d %H:%M:%S}')
2021-12-01 22:10:07
This I use because I can understand and hence remember it better (and date time format also can be customized based on your choice) :-
import datetime
moment = datetime.datetime.now()
print("{}/{}/{} {}:{}:{}".format(moment.day, moment.month, moment.year,
moment.hour, moment.minute, moment.second))
I found this to be the simplest way.
>>> t = datetime.datetime.now()
>>> t
datetime.datetime(2018, 11, 30, 17, 21, 26, 606191)
>>> t = str(t).split('.')
>>> t
['2018-11-30 17:21:26', '606191']
>>> t = t[0]
>>> t
'2018-11-30 17:21:26'
>>>
You can also use the following method
import datetime as _dt
ts = _dt.datetime.now().timestamp()
print("TimeStamp without microseconds: ", int(ts)) #TimeStamp without microseconds: 1629275829
dt = _dt.datetime.now()
print("Date & Time without microseconds: ", str(dt)[0:-7]) #Date & Time without microseconds: 2021-08-18 13:07:09
Current TimeStamp without microsecond component:
timestamp = list(str(datetime.timestamp(datetime.now())).split('.'))[0]