I want to covert a string item to the datetime format that I require.
The string item is this : '3/27/2013 2:54:00 PM'
My code is as follows: map(datetime.datetime.strptime,'3/27/2013 2:54:00 PM', '%m/%d/%Y %I:%M:%S %p')
I get the following error : stray % in format '%'
Any help is appreciated.
Not sure why you use map, you can just use strptime on that string, worked for me
import datetime
dt = datetime.datetime.strptime('3/27/2013 2:54:00 PM', '%m/%d/%Y %I:%M:%S %p')
print(dt)
Related
I am very new in python working on dates. I have 2 type of dates (because getting in array dynamically). Something like this ['2022-02-17 08:29:36.345374' , '2021-03-18 08:29:36']
I am trying to get these in this format 17/02/2022 08:29 PM.
Until now trying something like this:
def update_date(datetime):
date_formating = datetime.strptime(date_time, '%m/%d/%Y %I:%M %p')
return date_formating
but getting errors like:
ValueError: time data '2022-02-17 08:29:36.345374' does not match format '%m/%d/%Y %I:%M %p'
I need a solution which can turn both formats I am getting into desired format. Any help would be highly appreciated.
Try a list comprehension with datetime.fromisoformat(date_string):
>>> from datetime import datetime
>>> date_strs = ['2022-02-17 08:29:36.345374', '2021-03-18 08:29:36']
>>> [datetime.fromisoformat(s).strftime('%d/%m/%Y %I:%M %p') for s in date_strs]
['17/02/2022 08:29 AM', '18/03/2021 08:29 AM']
The times in the given list come in two different formats
with microseconds
with seconds (and no microseconds)
This code highlights the differences and would work:
import datetime as dt
t = ['2022-02-17 08:29:36.345374' , '2021-03-18 08:29:36']
# format with microseconds
d1 = dt.datetime.strptime(t[0], '%Y-%m-%d %H:%M:%S.%f')
# format with seconds
d2 = dt.datetime.strptime(t[1], '%Y-%m-%d %H:%M:%S')
# format user wants
x = d1.strftime('%m/%d/%Y %I:%M %p')
y = d2.strftime('%m/%d/%Y %I:%M %p')
print(x)
print(y)
The result:
02/17/2022 08:29 AM
03/18/2021 08:29 AM
I am trying to parse the string '10/23/2019 6:02:05 PM EST' into a datetime with time zone using Python 3.7.
Code:
from datetime import datetime
timestamp = datetime.strptime(date_str, '%m/%d/%Y %I:%M:%S %p %Z')
Error:
ValueError: time data '10/23/2019 6:02:05 PM EST' does not match format '%m/%d/%Y %I:%M:%S %p %Z'
When I create a datetime and output it using the same formatting I get the correct output. The only difference is that there is a 0 in front of the hour, but adding 0 in front of the 6 in my date string results in the same error.
My current solution is to parse the datetime without the timezone and then localize it, but this is not ideal.
date_lst = date.split()
date_str = ' '.join(date_lst[0:3])
timestamp = datetime.strptime(date_str, '%m/%d/%Y %I:%M:%S %p')
new_tz = pytz.timezone(date_lst[3])
timestamp_tz = new_tz.localize(timestamp)```
It is reasonable to expect that parsing a string with a timezone included would produce a timezone aware datetime object.
Try it
>>timestamp = datetime.strptime('10/23/2019 6:02:05 PM EST', '%m/%d/%Y %I:%M:%S %p EST')
>>2019-10-23 06:02:05
You can try this.
I have a DATETIME format like this: '17-09-2019 06:22:00 PM'. I'm trying to use this format to convert '2019-09-17 06:22:00 PM'.
datetime.strptime(startingDate, '%y-%m-%d %I:%M:%S %p')
But I get an error. Anyone know how to do this?
Your input time format is wrong, it should match the string you are putting in.
"%Y-%m-%d %I:%M:%S %p" becomes "%d-%m-%Y %I:%M:%S %p"
Notice that the order of the date has changed.
import datetime
startingDate = "17-09-2019 06:22:00 PM"
dt = datetime.datetime.strptime(startingDate, "%d-%m-%Y %I:%M:%S %p")
datetime.datetime.strftime(dt, "%Y-%m-%d %I:%M:%S %p")
# "2019-09-17 06:22:00 PM"
Try this:
from datetime import datetime
startingDate = '17-09-2019 06:22:00 PM'
newDate = datetime.strptime(startingDate, '%d-%m-%Y %I:%M:%S %p').strftime("%Y-%m-%d %I:%M:%S %p")
print(newDate)
https://repl.it/repls/JuniorPushyLoopfission
I am trying to convert from string to timestamp using:
from datetime import datetime
date_object = datetime.strptime('09-MAR-15 12.54.45.000000000 AM', '%d-%b-%y %I.%M.%S.%f %p')
I get:
ValueError:
time data '09-MAR-15 12.54.45.000000000 AM' does not match format
'%d-%b-%y %I.%M.%S.%f %p'
The below will work as long as the the part after the decimal point always ends in 000. :-) %f captures microseconds, while I guess your timestamp uses nanoseconds?
date_object = datetime.strptime('09-MAR-15 12.54.45.000000000 AM',
'%d-%b-%y %I.%M.%S.%f000 %p')
You might consider just chopping off those three digits. E.g.
date_object = datetime.strptime(
re.sub(r'\d{3}( .M)$', r'\1', '09-MAR-15 12.54.45.000000000 AM'),
'%d-%b-%y %I.%M.%S.%f %p')
This question already has answers here:
Convert 12-hour date/time to 24-hour date/time
(8 answers)
Convert string "Jun 1 2005 1:33PM" into datetime
(26 answers)
Closed 7 years ago.
I have a string containing time stamp in format
(DD/MM/YYYY HH:MM:SS AM/PM), e.g."12/20/2014 15:25:05 pm"
.
The time here is in 24 Hrs format.
I need to convert into same format but with time in 12-Hrs format.
I am using python version 2.6.
I have gone through time library of python but couldn't come up with any solution.
View Live ideOne use Python datetime,
>>> from datetime import datetime as dt
>>> date_str='12/20/2014 15:25:05 pm'
>>> date_obj = dt.strptime(date_str, '%m/%d/%Y %H:%M:%S %p')
>>> dt.strftime(date_obj, '%m/%d/%Y %I:%M:%S %p')
'12/20/2014 03:25:05 PM'
The trick is to convert your Input date string to Python datetime object and then convert it back to date string
import datetime
#Input Date String
t = "12/20/2014 15:25:05 pm"
#Return a datetime corresponding to date string
dateTimeObject = datetime.datetime.strptime(t, '%m/%d/%Y %H:%M:%S %p')
print dateTimeObject
Output: 2014-12-20 15:25:05
#Return a string representing the date
dateTimeString = datetime.datetime.strftime(dateTimeObject, '%m/%d/%Y %I:%M:%S %p')
print dateTimeString
Output: 12/20/2014 03:25:05 PM
After creating a datetime object using strptime you then call strftime and pass the desired format as a string see the docs:
In [162]:
t = "12/20/2014 15:25:05 pm"
dt.datetime.strftime(dt.datetime.strptime(t, '%m/%d/%Y %H:%M:%S %p'), '%m/%d/%Y %I:%M:%S %p')
Out[162]:
'12/20/2014 03:25:05 PM'
Shortest & simplest solution --
I really appreciate & admire (coz I barely manage to read man pages :P) you going through time documentation, but why use "astonishing" & "cryptic" code when simple code could get the job done
Just extract the hour part as int & replace it by hrs-12 if it is greater than 12
t = "12/20/2014 15:25:05 pm"
hrs = int( t.split()[1][:2] )
if hrs > 12:
t = t.replace( str(hrs), str(hrs-12) )
Output
See explaination & live output here
Using Lambda
If you like one liners, checkout f() below
t = "12/20/2014 15:25:05 pm"
f = lambda tym: tym.replace(str(int(tym.split()[1][:2])), str(int(tym.split()[1][:2])-12)) if int(tym.split()[1][:2]) > 12 else tym
print(f(t))