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Converting timezone to new format and inserting into list in python [duplicate]

I'm adding UTC time strings to Bitbucket API responses that currently only contain Amsterdam (!) time strings. For consistency with the UTC time strings returned elsewhere, the desired format is 2011-11-03 11:07:04 (followed by +00:00, but that's not germane).
What's the best way to create such a string (without a microsecond component) from a datetime instance with a microsecond component?
>>> import datetime
>>> print unicode(datetime.datetime.now())
2011-11-03 11:13:39.278026
I'll add the best option that's occurred to me as a possible answer, but there may well be a more elegant solution.
Edit: I should mention that I'm not actually printing the current time – I used datetime.now to provide a quick example. So the solution should not assume that any datetime instances it receives will include microsecond components.

If you want to format a datetime object in a specific format that is different from the standard format, it's best to explicitly specify that format:
>>> import datetime
>>> datetime.datetime.now().strftime("%Y-%m-%d %H:%M:%S")
'2011-11-03 18:21:26'
See the documentation of datetime.strftime() for an explanation of the % directives.
Starting from Python 3.6, the isoformat() method is flexible enough to also produce this format:
datetime.datetime.now().isoformat(sep=" ", timespec="seconds")

>>> import datetime
>>> now = datetime.datetime.now()
>>> print unicode(now.replace(microsecond=0))
2011-11-03 11:19:07

In Python 3.6:
from datetime import datetime
datetime.now().isoformat(' ', 'seconds')
'2017-01-11 14:41:33'
https://docs.python.org/3.6/library/datetime.html#datetime.datetime.isoformat

This is the way I do it. ISO format:
import datetime
datetime.datetime.now().replace(microsecond=0).isoformat()
# Returns: '2017-01-23T14:58:07'
You can replace the 'T' if you don't want ISO format:
datetime.datetime.now().replace(microsecond=0).isoformat(' ')
# Returns: '2017-01-23 15:05:27'

Yet another option:
>>> import time
>>> time.strftime("%Y-%m-%d %H:%M:%S")
'2011-11-03 11:31:28'
By default this uses local time, if you need UTC you can use the following:
>>> time.strftime("%Y-%m-%d %H:%M:%S", time.gmtime())
'2011-11-03 18:32:20'

Keep the first 19 characters that you wanted via slicing:
>>> str(datetime.datetime.now())[:19]
'2011-11-03 14:37:50'

I usually do:
import datetime
now = datetime.datetime.now()
now = now.replace(microsecond=0) # To print now without microsecond.
# To print now:
print(now)
output:
2019-01-13 14:40:28

Since not all datetime.datetime instances have a microsecond component (i.e. when it is zero), you can partition the string on a "." and take only the first item, which will always work:
unicode(datetime.datetime.now()).partition('.')[0]

As of Python 3.6+, the best way of doing this is by the new timespec argument for isoformat.
isoformat(timespec='seconds', sep=' ')
Usage:
>>> datetime.now().isoformat(timespec='seconds')
'2020-10-16T18:38:21'
>>> datetime.now().isoformat(timespec='seconds', sep=' ')
'2020-10-16 18:38:35'

We can try something like below
import datetime
date_generated = datetime.datetime.now()
date_generated.replace(microsecond=0).isoformat(' ').partition('+')[0]

>>> from datetime import datetime
>>> dt = datetime.now().strftime("%Y-%m-%d %X")
>>> print(dt)
'2021-02-05 04:10:24'

f-string formatting
>>> import datetime
>>> print(f'{datetime.datetime.now():%Y-%m-%d %H:%M:%S}')
2021-12-01 22:10:07

This I use because I can understand and hence remember it better (and date time format also can be customized based on your choice) :-
import datetime
moment = datetime.datetime.now()
print("{}/{}/{} {}:{}:{}".format(moment.day, moment.month, moment.year,
moment.hour, moment.minute, moment.second))

I found this to be the simplest way.
>>> t = datetime.datetime.now()
>>> t
datetime.datetime(2018, 11, 30, 17, 21, 26, 606191)
>>> t = str(t).split('.')
>>> t
['2018-11-30 17:21:26', '606191']
>>> t = t[0]
>>> t
'2018-11-30 17:21:26'
>>>

You can also use the following method
import datetime as _dt
ts = _dt.datetime.now().timestamp()
print("TimeStamp without microseconds: ", int(ts)) #TimeStamp without microseconds: 1629275829
dt = _dt.datetime.now()
print("Date & Time without microseconds: ", str(dt)[0:-7]) #Date & Time without microseconds: 2021-08-18 13:07:09

Current TimeStamp without microsecond component:
timestamp = list(str(datetime.timestamp(datetime.now())).split('.'))[0]

How to get timedelta in seconds from a string? Example: +414 00:45:41.004000

I have data of timedeltas which looks like this:
time_delta = '+414 00:45:41.004000'
So, these values are strings and they are of the format ddd hh:mm:ss.f. I now want to get this deltas to seconds. I tried to use .total_seconds() but it did not work.
How could I achieve what I am trying to do?

If you always assume the same input format, you can build a function as below (result to be checked with a simple case) :
import datetime as dt
def parseTimeDelta(time_delta_str):
splitted = time_delta_str.split(' ')
day_part = int(splitted[0][1:])
time_part = dt.datetime.strptime(splitted[1], "%H:%M:%S.%f")
delta = dt.timedelta(days=day_part, hours=time_part.hour, minutes=time_part.minute, seconds=time_part.second,microseconds=time_part.microsecond)
return delta.total_seconds()
time_delta = '+414 00:45:41.004000'
parseTimeDelta(time_delta)

can do this with pandas library
import pandas as pd
# Create the Timedelta object
td = pd.Timedelta('3 days 06:05:01.000000111')
print(td)
print(td.seconds)

Unfortunately we can't create a timedelta with a formatted string directly, but we can get a similar effect with regex then unpack parsed values into a timedelta.
import re
import datetime
# Create parser for your time format with named groups that match timedelta kwargs
time_parser = re.compile(r"\+(?P<days>\d+)\s+(?P<hours>\d{2}):(?P<minutes>\d{2}):(?P<seconds>\d{2})\.(?P<microseconds>\d+)")
# Get the values from your example string
regex_match = time_parser.match("+414 00:45:41.004000")
time_dict = regex_match.groupdict()
# Convert the time values to integers from strings
timedelta_kwargs = {k: int(v) for k, v in time_dict.items()}
# Make a time delta object
delta = datetime.timedelta(**timedelta_kwargs)
# Get total seconds
delta_in_seconds = delta.total_seconds()
Organise that into some functions and you'll get the functionality you're looking for with standard python packages.

Convert text to Time format in python

I have time values in text as:
a="060453"
b="135309"
I want to convert the above into "HH:MM:SS" format and also get the difference in the same format.
13:53:09 - 06:04:53 = 07:49:06
Regards

I don't think your math is right but I think this is kind of what you want:
>>> import datetime
>>> print datetime.datetime.strptime("135309", "%H%M%S") - datetime.datetime.strptime("060453", "%H%M%S")
7:48:16
>>>

This should get you the timedelta you need-
from datetime import datetime
a="060453"
b="135309"
a = datetime.strptime(a,'%H%M%S')
b = datetime.strptime(b,'%H%M%S')
print(b-a)

How to convert from mm/dd/yyyy format to yyyy-mm-dd in python 3.0 when given 1/4/2014

So the I generally understand how I would convert from mm/dd/yyyy format to yyyy-mm-dd if the initial date given was something like 01/04/2014. However I am only given 1/4/2014 (without zeroes). Is there a clean and more efficient way of converting this in python 3 rather than writing a bunch of if statements?

>>> from datetime import datetime
>>> date = datetime.strptime("1/4/2014", "%m/%d/%Y")
>>> datetime.strftime(date, "%Y-%m-%d")
'2014-01-04'
Python datetime strftime()
How strptime and strftime work

from datetime import datetime
date = datetime.strptime("1/4/2014", "%m/%d/%Y")
print(datetime.strftime(date, "%Y-%m-%d"))
'2014-01-04'

Abdou's suggestion using the datetime module is best, because you get the benefit of datetime's sanity-checking of the values. If you want to do it using only string manipulations, then notice that you can supply a fill character when invoking a string's rjust method:
>>> "1".rjust(2,"0")
'01'
so:
>>> x = "1/4/2345"
>>> f = x.split("/")
>>> f[2] + "-" + f[0].rjust(2,"0") + "-" + f[1].rjust(2,"0")
'2345-01-04'

Assuming that you don't need to validate the input dates, you don't need any if statements, or date processing functions: you can do it all with the string .split and .format methods.
def us_date_to_iso(us_date):
return '{2}-{0:>02}-{1:>02}'.format(*us_date.split('/'))
# test
for s in ('1/4/2014', '1/11/1999', '31/5/2015', '25/12/2016'):
print(s, us_date_to_iso(s))
output
1/4/2014 2014-01-04
1/11/1999 1999-01-11
31/5/2015 2015-31-05
25/12/2016 2016-25-12

String to time conversion issue in python

This is my code - I want to convert a string into a time in Python - it sort of works:
import datetime
firstTime = ("18:08:14")
firstTime = datetime.datetime.strptime(firstTime, "%H:%M:%S")
print (firstTime)
The problem is, I get '1900-01-01 18:08:14' instead of just '18:08:14'. I know this is a fairly basic thing, but I'm new to Python and any help would be appreciated.

As my comment suggests, use a time class rather than datetime since you don't need the date part:
>>> from datetime import datetime
>>> firsttime = datetime.strptime('18:08:14','%H:%M:%S')
>>> print(firsttime)
1900-01-01 18:08:14
>>> print(firsttime.time())
18:08:14
or simply:
>>> firsttime = datetime.strptime('18:08:14','%H:%M:%S').time()
>>> print(firsttime)
18:08:14

try this..
>>> from datetime import datetime
>>> date=datetime.now()
>>> date.strftime('%H:%M:%S')
'23:55:17'
>>>

Your code seems fine. Just change strptime to strftime.
import datetime
firstTime = ("18:08:14")
firstTime = datetime.datetime.strftime(firstTime, "%H:%M:%S")
print (firstTime)
strptime takes a string and convert to datetime object.
strftime creates formatted string from given date,time,datetime object according to a specific format