Is there a way to guess datetime format of a given string in python?
e.g. desired usage would be:
> guess_format('2020-02-24 07:22')
'%Y-%m-%d %H:%M'
There's dateutil project which automates datetime string conversion to valid Datetime objects:
> from dateutil.parser import parse
> parse('2020-02-24 07:22')
datetime.datetime(2020, 2, 24, 7, 22)
but can it produce valid formatting strings?
The pydateinfer package provides the possibility to infer the datetime format string of a given date string.
Example:
>>> import dateinfer
>>> dateinfer.infer(['Mon Jan 13 09:52:52 MST 2014', 'Tue Jan 21 15:30:00 EST 2014'])
'%a %b %d %H:%M:%S %Z %Y'
Related
Please, How do I converts strings like this:
2021-01-15 14:22:56.692234+00:00 into datetime value as this :
Jan. 15, 2021, 3:07 p.m.
You can use datetime module to do the parsing.
from datetime import datetime
datetime.strptime('2019-01-04T16:41:24+0200', "%Y-%m-%dT%H:%M:%S%z")
Use the datetime module, specifically, strptime to convert a string into a datetime object, then strftime to print it out in your desired format.
from datetime import datetime
parsed = datetime.strptime(
'2021-01-15 14:22:56.692234+00:00',
'%Y-%m-%d %H:%M:%S.%f%z'
)
formatted = parsed.strftime('%b. %d, %Y, %-I:%M %p')
print(formatted)
This will give:
'Jan. 15, 2021, 2:22 PM'
There are no way to reproduce exactly p.m. but to replace it manually:
formatted = formatted.replace('AM', 'a.m.').replace('PM', 'p.m.')
print(formatted)
Which gives:
'Jan. 15, 2021, 2:22 p.m.'
I am trying to convert a string into date format in Python.
I am using following statement
datetime_obj = datetime.datetime.strptime("Sun Aug 19 16:24:31 PDT 2018", "%a %b %d %H:%M:%S %Z %Y")
However, I get an error -
ValueError: time data 'Sun Aug 19 16:24:31 PDT 2018' does not match format '%a %b %d %H:%M:%S %Z %Y'
If I remove the timezone from the date string and the format string, the code works perfect. Which leads me to believe that the issue is related to the timezone but I am not sure what actions should be taken.
I am in eastern timezone and the time zone in the string is in Pacific timezone.
Appreciate any help on this.
As mentioned in this answer you can use python-dateutil for this:
>>> from dateutil import parser
>>> datetime_obj = parser.parse("Sun Aug 19 16:24:31 PDT 2018")
datetime.datetime(2018, 8, 19, 16, 24, 31)
Standard datetime module behaves very strangely with parsing timezones, as I see reading this answer in question related to similar problem.
I want to convert my date into DateTime object for MySQL.
My string format is: Mon Aug 27 04:47:45 +0000 2018
Expected Output: 'YYYY-M-D H:mm:ss'
from datetime import datetime
t = datetime.strptime('Mon Aug 27 04:47:45 +0000 2008', '%a %b %d %H:%M:%S % z %Y')
t.strftime('%Y-%m-%d %H:%M:%S')
Refer section 8.1.8
here
If you are using python 3, this solution would work -
from datetime import datetime
x = 'Mon Aug 27 04:47:45 +0000 2018'
x = datetime.strftime(datetime.strptime(x, '%a %b %d %I:%M:%S %z %Y'), '%Y-%m-%d %H:%M:%S')
# OP '2018-08-27 04:47:45'
But for python 2, you might get a ValueError: 'z' is a bad directive.... In that case, you'll either have to use something like pytz or dateutil. The table that you need to look for all these conversions can be found here
Edit: You can't have Expected Output: 'YYYY-M-D H:mm:ss' if you convert your datetime string to datetime object. Datetime object has it's own format. Above gives you a string of the format that you want
from datetime import datetime
date_as_dt_object = datetime.strptime(dt, '%a %b %d %H:%M:%S %z %Y')
You can use date_as_dt_object in a raw query or an ORM. If used in a raw query pass it as a string like:
query = "select * from table where date >" + str(date_as_dt_object)
Check out this list for Python's strftime directives.
http://strftime.org/
I work with api in python3 in this api return date like this
'Jun 29, 2018 12:44:14 AM'
but i need just hours, minute ad second like this
12:44:14
are there a fonction that can format this
It looks like the output is a string. So, you can use string slicing:
x = 'Jun 29, 2018 12:44:18 AM'
time = x[-11:-3]
It's best to use negative indexing here because the day may be single-digit or double-digit, so a solution like time = x[13:21] won't work every time.
If you're inclined, you may wish to use strptime() and strftime() to take your string, convert it into a datetime object, and then convert that into a string in HH:MM:SS format. (You may wish to consult the datetime module documentation for this approach).
Use the datetime module. Use .strptime() to convert string to datetime object and then .strftime() to convert to your required string output. Your sample datetime string is represented as '%b %d, %Y %H:%M:%S %p'
Ex:
import datetime
s = 'Jun 29, 2018 12:44:14 AM'
print( datetime.datetime.strptime(s, '%b %d, %Y %H:%M:%S %p').strftime("%H:%M:%S") )
Output:
12:44:14
I have a date in excel which is given in: dd mmm yy format i.e.,
29 Jun 18
How do I convert this string into a date object?
I get the error:
time data '13 Jul 18' does not match format '%d %m %Y'
when I try
datetime.strptime(input, '%d %m %Y')
What should the correct date format be?
Since the year in your excell is only two digits (i.e., 18 and not 2018) you need to use %y instead of %Y in your format string:
datetime.strptime(input, '%d %b %y')
For example:
datetime.strptime( '13 Jul 18', '%d %b %y')
Results with:
datetime.datetime(2018, 7, 13, 0, 0)
See this page for more information about date/time string format.
You can use python datetime module or you can use dateutil parser to parse the string date to valid datetime object. I'd go with dateutil parser as I don't have to define string format. Here is an example
from dateutil.parser import parse
dt = parse("Thu Sep 25 10:36:28 BRST 2003")
Remember to install dateutil by pip install python-dateutil
You would have to import datetime and import xlrd
Use xlrd to open the excel workbook as
book = xlrd.open_workbook("Excel.xlsx")
sheet = book.sheet_by_name("Worksheet")
Use this to convert
obj = datetime.datetime(*xlrd.xldate_as_tuple(sheet.cell(row,column).value, book.datemode))
from datetime import datetime
datetime.strptime("29 Jun 18", "%d %b %y").date()
Here you get a datetime.date object, I don't know if that's good enough for you. I recommend you to visit the documentation on the module
You can make use of strptime which follows the pattern:
datetime.strptime(date_string, format)
example:
from datetime import datetime
dt = datetime.strptime('19 Jul 2017', '%d %b %y')
Hope this helps :)
Your format is incorrect: %b is Locale's short month and %y is two digit year
import time
time.strptime('13 Jul 18', '%d %b %y')
time.struct_time(tm_year=2018, tm_mon=7, tm_mday=13, tm_hour=0, tm_min=0, tm_sec=0, tm_wday=4, tm_yday=194, tm_isdst=-1)