I parse a date (format: YYYY-MM-DD HH:MM:SS) from a data file which contains multiple lines of dates.
The problem is that the data contains leap seconds so i'm not able to use datetime. How can I take into account the leap seconds (0-60), so that at the end I would have the same result if I would have used datetime.strptime from the string with the format above (thus, date+time), please?
I have already tried with combine using date for the date and time for the time string. Is it the right way or are there some others?
Thanks in advance.
Just use time.strptime():
#!/usr/bin/env python
import datetime as DT
import time
from calendar import timegm
utc_time_string = '2012-06-30 23:59:60'
utc_time_tuple = time.strptime(utc_time_string, "%Y-%m-%d %H:%M:%S")[:6]
utc_dt = DT.datetime(1970, 1, 1) + DT.timedelta(seconds=timegm(utc_time_tuple))
# -> datetime.datetime(2012, 7, 1, 0, 0)
If the input time is not in UTC then you could handle the leap second in the time_tuple manually e.g., the datetime module may raise ValueError if you pass the leap second directly or it may silently truncate 60 to 59 if it encounters a leap second in indirect (internal) calls.
Related
This question already has answers here:
How to convert integer timestamp into a datetime
(3 answers)
Closed last year.
I am trying to figure out a time format used by someone else. In addition to the date, I have time with an example being the following:
1641859200000
I cant seem to figure out what time or date time format this is. It cannot be HHMMSS, because in this example the second is 85, which is not possible. Any idea what format this is, and how I can convert it using Python to HH:MM:SS ?
Thank you :)
You have a timestamp in seconds from Epoch since January 1, 1970 (midnight UTC/GMT). To convert to a datetime, use:
from datetime import datetime
print(datetime.fromtimestamp(1641859200000 / 1000))
# Output
2022-01-11 01:00:00
Note: you have to divide by 1000 because this timestamp contains milliseconds and Epoch should be in seconds.
This is a Unix-Timestamp.
You can convert in a human-readable format like this:
from datetime import datetime
timestamp = 1641859200000/1000
dt = datetime.fromtimestamp(timestamp)
print(dt)
Edit: didn't check the actual timestamp, whyever, this has to be divided by 1000 as done in the othe answer.
This probably is a unix timestamp https://en.wikipedia.org/wiki/Unix_time. The factor 1000 stems from a millisecond representation I think. Depends on, from where you got this stamp.
You can convert it using:
>>> datetime.datetime.fromtimestamp(1641859200000/1000)
datetime.datetime(2022, 1, 11, 1, 0)
Take a look at dateparser https://dateparser.readthedocs.io/.
It will help you figure out what the date and time is based on the range of input date strings:
pip install dateparser
>>> import dateparser
>>> dateparser.parse('1641859200000')
datetime.datetime(2022, 1, 11, 1, 0)
Your timestamp is miliseconds since Unix epoch in this case but if you ever run into similar problem dateparser could help you.
Regarding the second part of the question. Convertion to HH:MM:SS format
>> dt = datetime.datetime(2022, 1, 11, 1, 0)
>> dt.strftime("%H:%M:%S")
'01:00:00'
Additional info: Available Format Codes
I have measurements taken from 1st January 1993. They were recorded in second elapsed from that date. I would like to have them in date time.
I know in MatLab the function would be
time = datenum(1993,01,01,00,00, time)
However, I struggle to find an equivalent function in Python.
I have tried:
datetime.fromordinal(time) doesn't work because 'module object has no attribute fromordinal'?
datetime.datetime(time) doesn't work (I have a matrix because there are many scans done)
https://docs.python.org/3/library/datetime.html
You will first have to create a datetime object for Jan 1st 1993 and then add the number of seconds to that date. The code below should help you get started.
from datetime import datetime, timedelta
original_date = datetime.strptime('01-01-1993', '%d-%m-%Y')
original_date + timedelta(seconds= 10000)
output: datetime.datetime(1993, 1, 1, 2, 46, 40)
Let us say you have list of timevalues in seconds starting from 1993-01-01 00:00.
Easiest would be:
datevec=[datetime.datetime(1993,1,1,0)+datetime.timedelta(seconds=val) for val in timevector]
It is like UNIX time, but with a different start.
You can compute the offset once:
>>> import datetime as dt
>>> dt.datetime(1993,1,1).timestamp()
725842800.0
and use it in your program:
OFFSET = 725842800.0
mydate = dt.datetime.fromtimestamp(OFFSET + seconds_from_1993)
I have a file in which date field in YYYYMMDD format. I need to pick that date and product to join with another file by subtracting 12 months to get other information.
File1
20180131,Apple
20180228,Orange
20180331,Grapes
File2
20170131,Apple,45
20170131,Orange,20
20170228,Orange,35
20170331,Apple,25
Output
20180131,Apple,45
20180228,Orange,35
20180331,Grapes,null
How to subtract 12 months or 1 year from given date(yyyymmdd) and get the answer in the same format.
You can use strptime from the standard library to parse the date, but unfortunately there is no calendar support in the standard library so you have to use the dateutil library to subtract the year.
import datetime
from dateutil.relativedelta import relativedelta
d = datetime.datetime.strptime('20180131', '%Y%m%d').date()
print((d - relativedelta(years=1)).strftime('%Y%m%d'))
This will print 20170131.
Note that if the input is e.g. 20160229, this will print out 20150228. I'm not sure exactly what the semantics of relativedelta are so be sure to read the docs if this is important to you.
Why not timedelta?
The datetime.timedelta is not appropriate and does not work correctly. A timedelta represents a duration in time, or it represents a number of days, but counter-intuitively, a year is not a duration but instead a calendrical concept. A year is either 365 or 366 days, depending on which year.
It becomes pretty obvious that timedelta will not work once you find the correct test cases:
In [1]: from datetime import date, timedelta
In [2]: date(2018, 1, 1) - timedelta(365)
Out[2]: datetime.date(2017, 1, 1)
In [3]: date(2017, 1, 1) - timedelta(365)
Out[3]: datetime.date(2016, 1, 2)
All that I have is the number 223 (which is the number of days from Jan 01,2012), and the time at which the event occurred at (for example: 09, 55, 56.38 = (hh, mm, ss)).
In Excel I can get a serial number - in the format 10-Aug-12 09:55:56 - with these four numbers.
In Python I've been having some troubles doing the same. Does anyone have any idea about what commands I could be using?
The timedelta class in Python's datetime module lets you create a time difference, in days (or other units), which you can add to/subtract from datetime objects to get a new date.
(Note: the datetime module contains a datetime object. Yup.)
So, you could construct a datetime object using 1st Jan 2012 and the hour, minute and second values you've got, and then add a timedelta of 223 days to it.
To get that datetime as a string in your desired format, the strftime method on the datetime object is your friend.
Putting it all on one line:
from datetime import datetime, timedelta
serial_number = (datetime(2012, 1, 1, 9, 55, 56) + timedelta(223)).strftime('%d-%h-%y %H:%M:%S')
I was working on code to generate the time for an entire day with 30 second intervals. I tried using DT.datetime and DT.time but I always end up with either a datetime value or a timedelta value like (0,2970). Can someone please tell me how to do this.
So I need a list that has data like:
[00:00:00]
[00:00:01]
[00:00:02]
till [23:59:59] and needed to compare it against a datetime value like 6/23/2011 6:38:00 AM.
Thanks!
Is there a reason you want to use datetimes instead of just 3 for loops counting up? Similarly, do you want to do something fancy or do you want to just compare against the time? If you don't need to account for leap seconds or anything like that, just do it the easy way.
import datetime
now = datetime.datetime.now()
for h in xrange(24):
for m in xrange(60):
for s in xrange(60):
time_string = '%02d:%02d:%02d' % (h,m,s)
if time_string == now.strftime('%H:%m:%S'):
print 'you found it! %s' % time_string
Can you give any more info about why you are doing this? It seems like you would be much much better off parsing the datetimes or using strftime to get what you need instead of looping through 60*60*24 times.
There's a great answer on how to get a list of incremental values for seconds for a 24-hour day. I reused a part of it.
Note 1. I'm not sure how you're thinking of comparing time with datetime. Assuming that you're just going to compare the time part and extracting that.
Note 2. The time.strptime call expects a 12-hour AM/PM-based time, as in your example. Its result is then passed to time.strftime that returns a 24-hour-based time.
Here's what I think you're looking for:
my_time = '6/23/2011 6:38:00 AM' # time you defined
from datetime import datetime, timedelta
from time import strftime, strptime
now = datetime(2013, 1, 1, 0, 0, 0)
last = datetime(2013, 1, 1, 23, 59, 59)
delta = timedelta(seconds=1)
times = []
while now <= last:
times.append(now.strftime('%H:%M:%S'))
now += delta
twenty_four_hour_based_time = strftime('%H:%M:%S', strptime(my_time, '%m/%d/%Y %I:%M:%S %p'))
twenty_four_hour_based_time in times # returns True