I have strings like:
first = '2018-09-16 15:00:00'
second = '1900-01-01 09:45:55.500597'
I want to compare them.
All methods I found like Convert string date to timestamp in Python requires to know the format of the string.
I don't know the format of the strings (see differences between first and second) all I know is that they can be converted to timestamps.
How can I convert them in order to compare them?
Edit:
The "largest" string that I can get is:
1900-01-01 09:45:55.500597
but I can also get:
1900-01-01
1900-01-01 09:45
1900-01-01 09:45:55
etc..
It's always YYYY-MM-DD HH-MM....
You can use pandas.to_datetime. It offers a lot of flexibility in the string timestamp format, and you can use it on single strings or list/series/tuples of timestamps.
>>> import pandas as pd
>>> day = pd.to_datetime('1900-01-01')
>>> minute = pd.to_datetime('1900-01-01 09:45')
>>> second = pd.to_datetime('1900-01-01 09:45:55')
>>> subsecond = pd.to_datetime('1900-01-01 09:45:55.500597')
>>> assert subsecond > second
>>> assert minute < second
>>> assert day < minute
You can use the dateutil module (pip install python-dateutil):
>>> from dateutil.parser import parse
>>> parse('2018-09-16 15:00:00')
datetime.datetime(2018, 9, 16, 15, 0)
>>> parse('1900-01-01 09:45:55.500597')
datetime.datetime(1900, 1, 1, 9, 45, 55, 500597)
From the list of its features:
Generic parsing of dates in almost any string format;
Once you have the datetime objects, you can compare them directly, there's no need to calculate the timestamps.
Related
I have a list of dates that python recognizes as str. I need to be able to work this as dates, to be able to add and subtract days. The dates are of the form "29-Jun-2017"
How can I have a date list instead of a String list?
datetime.datetime.strptime converts string to datetime object:
>>> datetime.datetime.strptime('29-Jun-2017', '%d-%b-%Y')
datetime.datetime(2017, 6, 29, 0, 0)
Then, datetime.datetime.date() will return date object:
>>> datetime.datetime.strptime('29-Jun-2017', '%d-%b-%Y').date()
datetime.date(2017, 6, 29)
See strftime() and strptime() Behavior for other format codes.
The arrow module makes this type of calculation with dates especially easy.
Here I create a short list of date strings which I convert to arrow dates. Having displayed these dates I use the replace method to subtract two days from each date, creating a new list. Then I format this result for display.
>>> import arrow
>>> data = [ '01-Jan-2017', '31-Dec-2017' ]
>>> data_as_dates = [ arrow.get(_, 'DD-MMM-YYYY') for _ in data]
>>> data_as_dates
[<Arrow [2017-01-01T00:00:00+00:00]>, <Arrow [2017-12-31T00:00:00+00:00]>]
>>> two_days_earlier = [ _.replace(days=-2) for _ in data_as_dates ]
>>> [_.format('YYYY MMMM DD') for _ in two_days_earlier]
['2016 December 30', '2017 December 29']
The dates contained in arrow objects are available as conventional Python dates suitable for use with datetime, for instance.
I want to get datetimes from timestamps like the following :3/1/2014 9:55 with datetime.strptime, or something equivalent.
The month, day of month, and hour is not zero padded, but there doesn't seem to be a formatting directive listed here that is able to parse this automatically.
What's the best approach to do so? Thanks!
strptime is able to parse non-padded values. The fact that they are noted as being padded in the formatting codes table applies to strftime's output. So you can just use
datetime.strptime(datestr, "%m/%d/%Y %H:%M")
strptime isdo not require 0-padded values. See example below
datetime.strptime("3/1/2014 9:55", "%m/%d/%Y %H:%M")
output: datetime.datetime(2014, 3, 1, 9, 55)
The non-pattern way is use dateutil.parse module, it lets to parse the common date formats, even if you don't know what it is using currently
Ex:
>>> import dateutil.parser
>>>
>>> utc_time = '2014-08-13T00:00:00'
>>> verbose_time = '13-Aug-2014'
>>> some_locale = '3/1/2014 9:55'
>>> dateutil.parser.parse(utc_time)
datetime.datetime(2014, 8, 13, 0, 0)
>>> dateutil.parser.parse(verbose_time)
datetime.datetime(2014, 8, 13, 0, 0)
>>> dateutil.parser.parse(some_locale)
datetime.datetime(2014, 3, 1, 9, 55)
Just in case this answer helps someone else -- I came here thinking I had a problem with zero padding, but it was actually to do with 12:00 vs 00:00 and the %I formatter.
The %I formatter is meant to match 12-hour-clock hours, optionally zero-padded. But depending on your data source, you might get data that says that midnight or midday is actually zero, eg:
>>> datetime.strptime('2015/01/01 0:12am', "%Y/%m/%d %I:%M%p")
ValueError: time data '2015/01/01 0:12am' does not match format '%Y/%m/%d %I:%M'
What strptime actually wanted was a 12, not a zero:
>>> datetime.strptime('2015/01/01 12:12am', "%Y/%m/%d %I:%M%p")
datetime.datetime(2015, 1, 1, 0, 12)
But we don't always control our data sources! My solution for this edge case was to catch the exception, try parsing it with a %H, with a quick check that we are in the edge case we think we are in.
def get_datetime(string):
try:
timestamp = datetime.strptime(string, "%m/%d/%Y %I:%M%p")
except ValueError:
# someone used zero for midnight?
timestamp = datetime.strptime(string, "%m/%d/%Y %H:%M%p")
assert string.lower().endswith('am')
assert timestamp.hour == 0
return timestamp
You can see the strftime document here，but in fact they aren't all working well in all platforms，for instance，%-d，%-m don't work on win7 by python 2.7，so you can accomplish like this
>>> date_str = '{d.year}-{d.month}-{d.day}'.format(d=datetime.datetime.now())
>>> print(date_str)
2016-5-23
A function returns date and time in unicode format.
u'2014-03-06T04:38:51Z'
I wish to convert this to date and time format and subtract it with current datetime to get the number of days in between.
Thanks in advance
Check string is unicode
>>> import types
>>> type(u'2014-03-06T04:38:51Z') is types.UnicodeType
True
Converting strings to datetime:
>>> import datetime
>>> datetime.datetime.strptime(u'2014-03-06T04:38:51Z', '%Y-%m-%dT%H:%M:%SZ')
datetime.datetime(2014, 3, 6, 4, 38, 51)
Subtract from today to
>>> import datetime
>>> today = datetime.datetime.today()
>>> yourdate = datetime.datetime.strptime(u'2014-03-06T04:38:51Z', '%Y-%m-%dT%H:%M:%SZ')
>>> difference = today - yourdate
print str(difference)
First you have to convert your string to a datetime.datetime object.
import datetime
then = datetime.datetime.strptime(u'2014-03-06T04:38:51Z', "%Y-%m-%dT%H:%M:%SZ")
then represents itself as datetime.datetime(2014, 3, 6, 4, 38, 51), which looks about right. Then you have to get today's date as a datetime.datetime.
now = datetime.datetime.now()
Finally subtract it from your date (or vice versa - the question didn't make it clear).delta is a datetime.timedelta object that stores increments in days, seconds and microseconds. The latter two are always positive, the first can be negative.
for delta in (now-then, then-now):
print(delta, "::", delta.days, delta.seconds, delta.microseconds)
This prints out:
-1 day, 20:18:14.250142 :: -1 73094 250142
3:41:45.749858 :: 0 13305 749858
Best try it with a few examples to convince yourself it's correct.
What are these date-time formats? I need to convert them to the same format, to check if they are the same. These are just two coming from a separate data source, so I need to find a way to make them the same format. Any ideas?
2013-07-12T07:00:00Z
2013-07-10T11:00:00.000Z
Thanks in advance
That extra .000 is micro seconds.
This will convert a date string of a format to datetime object.
import datetime
d1 = datetime.datetime.strptime("2013-07-12T07:00:00Z","%Y-%m-%dT%H:%M:%SZ")
d2 = datetime.datetime.strptime("2013-07-10T11:00:00.000Z","%Y-%m-%dT%H:%M:%S.%fZ")
Then convert them into any format depending on your requirement, by using:
new_format = "%Y-%m-%d"
d1.strftime(new_format)
perhaps use .isoformat()
string in ISO 8601 format, YYYY-MM-DDTHH:MM:SS[.mmmmmm][+HH:MM]
>>> import datetime
>>> datetime.datetime.utcnow().isoformat() + "Z"
'2013-07-11T22:26:51.564000Z'
>>>
Z specifies "zulu" time or UTC.
You can also add the timezone component by making your datetime object timezone aware by applying the appropriate tzinfo object. With the tzinfo applied the .isoformat() method will include the appropriate utc offset in the output:
>>> d = datetime.datetime.utcnow().replace(tzinfo=datetime.timezone.utc)
>>> d.isoformat()
'2019-11-11T00:52:43.349356+00:00'
You can remove the microseconds by change the microseconds value to 0:
>>> no_ms = d.replace(microsecond=0)
>>> no_ms.isoformat()
'2019-11-11T00:52:43+00:00'
Also, as of python 3.7 the .fromisoformat() method is available to load an iso formatted datetime string into a python datetime object:
>>> datetime.datetime.fromisoformat('2019-11-11T00:52:43+00:00')
datetime.datetime(2019, 11, 11, 0, 52, 43, tzinfo=datetime.timezone.utc)
http://www.ietf.org/rfc/rfc3339.txt
you can try to trim the string
data = "2019-10-22T00:00:00.000-05:00"
result1 = datetime.datetime.strptime(data[0:19],"%Y-%m-%dT%H:%M:%S")
result2 = datetime.datetime.strptime(data[0:23],"%Y-%m-%dT%H:%M:%S.%f")
result3 = datetime.datetime.strptime(data[0:9], "%Y-%m-%d")
use datetime module.
For a variable
import datetime
def convertDate(d):
new_date = datetime.datetime.strptime(d,"%Y-%m-%dT%H:%M:%S.%fZ")
return new_date.date()
convertDate("2019-12-23T00:00:00.000Z")
you can change the ".date()" to ".year", ".month", ".day" etc...
Output: # is now a datetime object
datetime.date(2019, 12, 23)
For a DataFrame column, use apply()
df['new_column'] = df['date_column'].apply(convertDate)
* Short and best way:
str(datetime.datetime.now()).replace(' ','T')
or
str(datetime.datetime.now()).replace(' ','T') + "Z"
I have the following loop I'm trying to use to replace characters in a unicode string. The data I'm getting for this loop is in the following format: YYYY-MM-DD HH:MM:SS
This data is apparently stored in UTC, so when I grab it and append these times & dates to my list appts_list its 4 hours ahead.
I've gotten as far as slicing the unicode string and doing the math on these characters and getting what would be the correct hour I need, but I'm having a problem getting that back into a string so I can write it to my list appts_list.
I'm getting TypeError when I try to write the integer for the correct hour time_slice_int back into the original string. I decided to try to put the entire string into a list and change them there, but that isn't working either.
Ideally I want an appointment for '2013-06-28 15:30:00' to be entered into my appts_list as '2013-06-28 11:30:00'.
The print statements are there for me to debug as I ran it. They are not necessary for the final version.
Anyone have any suggestions or solutions?
for appt in todays_appts:
time = appt['apptdateourtime_c']
time_slice = time[11:13]
time_slice_int = int(time_slice)
time_slice_int -= 4
print(time_slice_int)
appt_time = list(time)
print(appt_time)
print(appt_time[11:13])
#appt_time[11:13] = time_slice_int
#appts_list.append()
print('AppointmentScheduled')
#print(appt['apptdateourtime_c'])
#print(time)
print('')
You should use the datetime module here:
>>> from datetime import datetime, timedelta
>>> strs = '2013-06-28 15:30:00'
>>> d = datetime.strptime(strs, "%Y-%m-%d %H:%M:%S")
datetime.strptime returns a datetime object:
>>> d
datetime.datetime(2013, 6, 28, 15, 30)
>>> d.hour
15
>>> d.month
6
Now decrease 4 hours from the above datetime object(d) using timedelta and assign the new object to a variable:
>>> d1 = d - timedelta(hours = 4)
Now use datetime.strftime to get a string of required format:
>>> datetime.strftime(d1,"%Y-%m-%d %H:%M:%S")
'2013-06-28 11:30:00'