I have tried multiple ways like splitting each value and the rearranging. and the code seems too lengthy can someone help me to write it in a shorter way.
time = "28/01/2023 13:17:58"
I split it to
lis = [28,01,2023]
lis2 = [13,17,58]
then rearranged it to
t = 28/01/2023 13:17:58
can someone tell me how I can use datetime to simplify this?
date = "28/01/2023 13:17:58"
datetime.datetime.strptime(date, '%d/%m/%Y %H:%M:%S')
Output = datetime.datetime(2023, 1, 28, 13, 17, 58)
from datetime import datetime
time = "28/01/2023 13:17:58"
datetime_obj = datetime.strptime(time, "%d/%m/%Y %H:%M:%S")
print(datetime_obj) # 2023-01-28 13:17:58
Related
As far as I can tell there is no way to distinguish the difference between these two date strings ('2020-10-07', '2020-10-07T00:00:00') once they are parsed by dateutil. I really would like to be able to tell the difference between a standalone date and a date with a timestamp of zero.
import dateutil.parser
import datetime
date_str = '2020-10-07'
time_str = '2020-10-07T00:00:00'
s = dateutil.parser.parse(date_str)
e = dateutil.parser.parse(time_str)
The ultimate goal is to set the time to the beginning of the day in the end of the day when it is a standalone date but leave the date alone when there is a time included. Get close with something like this but it still can't differentiate from this one case. If do you know of any good solution to this that would be really helpful.
if s == e and s.time() == datetime.time.min:
e = datetime.datetime.combine(e, datetime.time.max)
Post is somewhat useful but it's outdated and I'm not even sure that it would work for my use case. Finding if a python datetime has no time information
Here's a function which uses a simple try/except to test if the input can be parsed to a date (i.e. has no time information) or a datetime object (i.e. has time information). If the input format is different from ISO format, you could also implement specific strptime directives.
from datetime import date, time, datetime
def hasTime(s):
"""
Parameters
----------
s : string
ISO 8601 formatted date / datetime string.
Returns
-------
tuple, (bool, datetime.datetime).
boolean will be True if input specifies a time, otherwise False.
"""
try:
return False, datetime.combine(date.fromisoformat(t), time.min)
except ValueError:
return True, datetime.fromisoformat(t)
# do nothing else here; will raise an error if input can't be parsed
for t in ('2020-10-07', '2020-10-07T00:00:00', 'not-a-date'):
print(t, hasTime(t))
# output:
# >>> 2020-10-07 (False, datetime.datetime(2020, 10, 7, 0, 0))
# >>> 2020-10-07T00:00:00 (True, datetime.datetime(2020, 10, 7, 0, 0))
# >>> ValueError: Invalid isoformat string: 'not-a-date'
Looking for help on how to sort a python3 dictonary by a datetime object (as shown below, a value in the dictionary) using the timestamp below.
datetime: "2018-05-08T14:06:54-04:00"
Any help would be appreciated, spent a bit of time on this and know that to create the object I can do:
format = "%Y-%m-%dT%H:%M:%S"
# Make strptime obj from string minus the crap at the end
strpTime = datetime.datetime.strptime(ts[:-6], format)
# Create string of the pieces I want from obj
convertedTime = strpTime.strftime("%B %d %Y, %-I:%m %p")
But I'm unsure how to go about comparing that to the other values where it accounts for both day and time correctly, and cleanly.
Again, any nudges in the right direction would be greatly appreciated!
Thanks ahead of time.
Datetime instances support the usual ordering operators (< etc), so you should order in the datetime domain directly, not with strings.
Use a callable to convert your strings to timezone-aware datetime instances:
from datetime import datetime
def key(s):
fmt = "%Y-%m-%dT%H:%M:%S%z"
s = ''.join(s.rsplit(':', 1)) # remove colon from offset
return datetime.strptime(s, fmt)
This key func can be used to correctly sort values:
>>> data = {'s1': "2018-05-08T14:06:54-04:00", 's2': "2018-05-08T14:05:54-04:00"}
>>> sorted(data.values(), key=key)
['2018-05-08T14:05:54-04:00', '2018-05-08T14:06:54-04:00']
>>> sorted(data.items(), key=lambda item: key(item[1]))
[('s2', '2018-05-08T14:05:54-04:00'), ('s1', '2018-05-08T14:06:54-04:00')]
I have a question very similar to this one and this one but I'm stuck on some rounding issue.
I have a time series from a netCDF file and I'm trying to convert them to a datetime format. The format of the time series is in 'days since 1990-01-01 00:00:00'. Eventually I want output in the format .strftime('%Y%m%d.%H%M'). So for example I read my netCDF file as follows
import netCDF4
nc = netCDF4.Dataset(file_name)
time = np.array(nc['time'][:])
I then have
In [180]: time[0]
Out[180]: 365
In [181]: time[1]
Out[181]: 365.04166666651145
I then did
In [182]: start = datetime.datetime(1990,1,1)
In [183]: delta = datetime.timedelta(time[1])
In [184]: new_time = start + delta
In [185]: print(new_time.strftime('%Y%m%d.%H%M'))
19910101.0059
Is there a a way to "round" to the nearest hour so I get 19910101.0100?
You can round down with datetime.replace(), and round up by adding an hour to the rounded down value using datetime.timedelta(hours=1).
import datetime
def round_to_hour(dt):
dt_start_of_hour = dt.replace(minute=0, second=0, microsecond=0)
dt_half_hour = dt.replace(minute=30, second=0, microsecond=0)
if dt >= dt_half_hour:
# round up
dt = dt_start_of_hour + datetime.timedelta(hours=1)
else:
# round down
dt = dt_start_of_hour
return dt
Note that since we're using replace the values we're not replacing (like the timezone - tzinfo) will be preserved.
I don't think datetime provides a way to round times, you'll have to provide the code to do that yourself. Something like this should work:
def round_to_hour(dt):
round_delta = 60 * 30
round_timestamp = datetime.datetime.fromtimestamp(dt.timestamp() + round_delta)
round_dt = datetime.datetime.fromtimestamp(round_timestamp)
return round_dt.replace(microsecond=0, second=0, minute=0)
I'm new to python. my question is how to pass parameter to date.strftime() or a workaround
Below is the code
from datetime import date
dl_date = date.today()
p = '%d%b%Y' # the format may vary %d%B%Y or %d%m%Y or % d%M%Y etc
file_date_format = "{0}/{1}/{2}".format(str(dl_date.strftime('%r')),str(dl_date.strftime('%r').upper())
, str(dl_date.strftime('%r'))) % (p[:2], p[2:4], p[4:6])
print(file_date_format)
Help is much appreciated.
No need to use percent style string formatting here. Just stick the p slices directly in the strftime calls.
from datetime import date
dl_date = date.today()
p = '%d%b%Y' # the format may vary %d%B%Y or %d%m%Y or % d%M%Y etc
file_date_format = "{0}/{1}/{2}".format(
str(dl_date.strftime(p[:2])),
str(dl_date.strftime(p[2:4]).upper()),
str(dl_date.strftime(p[4:6]))
)
print(file_date_format)
Result:
14/NOV/2014
I take date from QDateTimeEdit and convert it to seconds like this:
import time
from datetime import datetime
date = self.__ui.dateTimeEdit.date().toString("dd/MM/yy")
dateString = str(date)
seconds = time.mktime(datetime.strptime(dateString, "%d/%m/%y").timetuple())
This works well, but since it looks to long to me, my question is: Is it possible to convert self.__ui.dateTimeEdit.date() directly, without those string conversions?
EDIT1
Unfortunately toMSecsSinceEpoch() as falsetru suggested, doesn't work for me.
AttributeError: 'QDateTime' object has no attribute 'toMSecsSinceEpoch'
I'm using PyQt 4.7.1 for Python 2.6
EDIT2 based on jonrsharpe's answer I've escaped string conversions:
date = self.__ui.dateTimeEdit.date().toPyDate()
seconds = time.mktime(date.timetuple())
result is the same.
EDIT3 even shorter solution based on falsetru's comment:
self.__ui.dateTimeEdit.dateTime().toTime_t()
Use QDateTime.toMSecsSinceEpoch:
>>> import PyQt4.QtCore
>>> d = PyQt4.QtCore.QDateTime(2014, 2, 20, 17, 10, 30)
>>> d.toMSecsSinceEpoch() / 1000
1392883830L
UPDATE
Alternative using QDateTime.toTime_t:
>>> d = PyQt4.QtCore.QDateTime(2014, 2, 20, 17, 10, 30)
>>> d.toTime_t()
1392883830L
The QDate you get from
self.__ui.dateTimeEdit.date()
has another method toPyDate that will save you the round trip through a string.