I want to compare two date and times with each other and then use that information for other stuff. Like if delta > 23 hours do this, elif delta > 11 hours, do that etc. I thought this would be a reasonable way to write it, but python won't accept it! It says:
ValueError: 'h' is a bad directive in format '%m/%d/%Y%h:%m:%s'
Isn't h the standard way to write hours in Python? :o
My dates are written in this format: "12/28/13 16:49:19", "m/d/y h:m:s" if that's any help!
from datetime import datetime
date_format = "%m/%d/%Y%h:%m:%s"
then=(dictionary["date"])
now= time.strftime("%c")
a = datetime.strptime(str(then), date_format)
b = datetime.strptime(str(now), date_format)
delta = b - a
print(delta.hour)
The 24 hour format is %H, a capital H, not a lowercase. The same for the minutes and seconds. You'll need a space as well, and you have a year format without the century, so use lower-case y:
date_format = "%m/%d/%y %H:%M:%S"
See the strftime() and strptime() behaviour documentation. %h doesn't exist as a format character.
And instead of using time.strftime('%c') to represent now, then parse that again, use:
b = datetime.now()
datetime.timedelta objects do not have an hours attribute; calculate the hours from the total seconds in the delta:
delta = b - a
print(delta.total_seconds() // 60)
or compare the delta object against another timedelta():
if delta > timedelta(hours=23):
Demo:
>>> from datetime import datetime
>>> date_format = "%m/%d/%y %H:%M:%S"
>>> datetime.strptime('12/28/13 16:49:19', date_format)
datetime.datetime(2013, 12, 28, 16, 49, 19)
Related
My expertise lack when it comes to understanding this time format. I am guessing the ,XXX is XXX/1000 of a second?
Nevertheless I am trying to add a text files that contains time stamp like these and sum up the total.
Below is an example,
00:03:33,950
00:03:34,590
This is what I have so far but I'm not sure how to add up the last part
Hours = s.split(":")[0]
Minutes = s.split(":")[1]
Seconds = (s.split(":")[2]).split(",")[0]
Total_seconds = (Hours * 3600) + (Minutes * 60) + (Seconds)
Total_Time = str(datetime.timedelta(seconds=Total_seconds))
Reed this documentation about time.strftime() format
For example
from time import gmtime, strftime
strftime("%a, %d %b %Y %H:%M:%S +0000", gmtime())
--'Thu, 28 Jun 2001 14:17:15 +0000'--
Actually, you're halfway there.
All you have to do is to to convert your strs into int and pass them as parameters to the appropriate timedelta keywords.
from datetime import timedelta
Hours = int(s.split(":")[0])
Minutes = int(s.split(":")[1])
Seconds = int((s.split(":")[2]).split(",")[0])
Milliseconds = int((s.split(":")[2]).split(",")[1])
duration = timedelta(hours=Hours, minutes=Minutes, seconds=Seconds, milliseconds=Milliseconds)
After adding all the durations you need, str() the final timedelta object.
>>> durations_1 = timedelta(hours=2,milliseconds=750)
>>> durations_2 = timedelta(milliseconds=251)
>>> durations_sum = durations_1 + durations_2
>>> str(durations_sum)
'2:00:01.001000'
>>> str(durations_sum).replace('.',',')
'2:00:01,001000'
Seems like this should be so simple but for the life of me, I can't find the answer. I pull two datetimes/timestamps from the database:
2015-08-10 19:33:27.653
2015-08-10 19:31:28.209
How do I subtract the first from the second, preferably the result being in milliseconds? And yes, I have the date in there, too, because I need it to work at around midnight, as well.
Parse your strings as datetime.datetime objects and subtract them:
from datetime import datetime
d1 = datetime.strptime("2015-08-10 19:33:27.653", "%Y-%m-%d %H:%M:%S.%f")
d2 = datetime.strptime("2015-08-10 19:31:28.209", "%Y-%m-%d %H:%M:%S.%f")
print(d1 - d2)
Gives me:
0:01:59.444000
Also check out timedelta documentation for all possible operations.
you can do subtraction on 2 datetime objects to get the difference
>>> import time
>>> import datetime
>>>
>>> earlier = datetime.datetime.now()
>>> time.sleep(10)
>>> now = datetime.datetime.now()
>>>
>>> diff = now - earlier
>>> diff.seconds
10
convert your strings to datetime objects with time.strptime
datetime.strptime("2015-08-10 19:33:27.653", "%Y-%m-%d %H:%M:%S.%f")
timedelta.seconds does not represent the total number of seconds in the timedelta, but the total number of seconds modulus 60.
Call the function timedelta.total_seconds() instead of accessing the timedelta.seconds property.
For python 3.4, first you'd need to convert the strings representing times into datetime objects, then the datetime module has helpful tools work with dates and times.
from datetime import datetime
def to_datetime_object(date_string, date_format):
s = datetime.strptime(date_string, date_format)
return s
time_1 = '2015-08-10 19:33:27'
time_2 = '2015-08-10 19:31:28'
date_format = "%Y-%m-%d %H:%M:%S"
time_1_datetime_object = to_datetime_object(time_1, date_format)
time_2_datetime_object = to_datetime_object(time_2, date_format)
diff_time = time_1_datetime_object - time_2_datetime_object
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
I need to calculate difference between time (and if it exceed 24 hours then days)
Like:
from datetime import datetime
from time import strftime
s1 = '24:11:2014:14:28:42'
s2 = datetime.now().strftime("%d:%m:%Y:%H:%M:%S")
FMT = '%d:%m:%Y:%H:%M:%S'
timedelta = datetime.now.strftime(s2,FMT) - datetime.now.strftime(s1,FMT)
print (timedelta)
But this is not detecting more than 24 hours, If found this code which can detect the days:
from datetime import datetime
date_format = "%d/%m/%Y %H%M%S"
a = datetime.strptime('22/10/2014 090000', date_format)
b = datetime.strptime('25/11/2014 100000', date_format)
delta = b - a
print (delta.days)
What I want is something like this in return: "2 days 03:35:00 HH:MM:SS" in return"
The timedelta you are getting from b - a already has all the information you need, have a look at https://docs.python.org/2/library/datetime.html#datetime.timedelta.
I'm working on a simple program to tell an individual how long they have been alive.
I know how to get the current date, and get their birthday. The only problem is I have no way of subtracting the two, I know a way of subtracting two dates, but unfortunately it does not include hours, minutes, or seconds.
I am looking for a method that can subtract two dates and return the difference down to the second, not merely the day.
from datetime import datetime
birthday = datetime(1988, 2, 19, 12, 0, 0)
diff = datetime.now() - birthday
print diff
# 8954 days, 7:03:45.765329
Use UTC time otherwise age in seconds can go backwards during DST transition:
from datetime import datetime
born = datetime(1981, 12, 2) # provide UTC time
age = datetime.utcnow() - born
print(age.total_seconds())
You also can't use local time if your program runs on a computer that is in a different place (timezone) from where a person was born or if the time rules had changed in this place since birthday. It might introduce several hours error.
If you want to take into account leap seconds then the task becomes almost impossible.
When substracting two datetime objects you will get a new datetime.timedelta object.
from datetime import datetime
x = datetime.now()
y = datetime.now()
delta = y - x
It will give you the time difference with resolution to microsencods.
For more information take a look at the official documentation.
Create a datetime.datetime from your date:
datetime.datetime.combine(birthdate, datetime.time())
Now you can subtract it from datetime.datetime.now().
>>> from datetime import date, datetime, time
>>> bday = date(1973, 4, 1)
>>> datetime.now() - datetime.combine(bday, time())
datetime.timedelta(14392, 4021, 789383)
>>> print datetime.now() - datetime.combine(bday, time())
14392 days, 1:08:13.593813
import datetime
born = datetime.date(2002, 10, 31)
today = datetime.date.today()
age = today - born
print(age.total_seconds())
Output: 463363200.0
Since DateTime.DateTime is an immutable type method like these always produce a new object the difference of two DateTime object produces a DateTime.timedelta type:
from datetime import date,datetime,time,timedelta
dt=datetime.now()
print(dt)
dt2=datetime(1997,7,7,22,30)
print(dt2)
delta=dt-dt2
print(delta)
print(int(delta.days)//365)
print(abs(12-(dt2.month-dt.month)))
print(abs(dt.day))
The output timedelta(8747,23:48:42.94) or what ever will be days when u test the code indicates that the time delta encodes an offset of 8747 days and 23hour and 48 minute ...
The Output
2021-06-19 22:27:36.383761
1997-07-07 22:30:00
8747 days, 23:57:36.383761
23 Year
11 Month
19 Day