I'm trying to get current date so I can pass it to the DATE field in SQL. I'm using datetime.datetime, below is my code:
from datetime import datetime
dt = datetime.strptime(datetime.today().date(), "%m/%d/%Y").date()
However, i'm getting this error:
TypeError: strptime() argument 1 must be str, not datetime.datetime
How can I fix the issue above? I'm still confused about datetime and datetime.datetime, and i want to keep using from datetime import datetime not import datetime.
How can I fix the issue above? thank you
If you see closely, the result of following statement,
>>> datetime.today().date()
datetime.date(2019, 9, 30)
>>> str(datetime.today().date())
'2019-09-30'
You'll notice that the datetime returned is - seperated and you'll have to convert it explicitly to a string value. Hence, for the above statement to work, change it to :
dt = datetime.strptime(str(datetime.today().date()), "%Y-%M-%d").date()
Then change it to whatever format you desire for using strftime (in your case >>> "%d/%m/%Y")
>>> dt.strftime("%d/%m/%Y")
'30/01/2019'
Just use datetime.strftime:
from datetime import datetime
dt = datetime.today().strftime("%m/%d/%Y")
print(dt)
Prints:
'09/30/2019'
strptime takes a string and makes a datetime object. Whereas strftime does exactly the opposite, taking a datetime object and returning a string.
Related
Consider this:
from datetime import datetime
now = datetime.now()
now.strftime("%p") # returns 'PM'
'{0.day}'.format(now) # returns 22
'{0.strftime("%p")}'.format(now)
# gives
# AttributeError: 'datetime.datetime' object has no attribute 'strftime("%p")'
This seems to imply that I can't call a class method inside the format (I guess that's what strftime is).
What's a workaround for this (assuming I need to call the method inside the string, and keep using a format) ?
You could do this.
>>> from datetime import datetime
>>> now = datetime.now()
>>> '{0:%p}'.format(now)
'PM'
This will also work with f-strings.
>>> f"{now:%p}"
'PM'
You can use the f-string:
from datetime import datetime
now = datetime.now()
now.strftime("%p") # returns 'PM'
'{0.day}'.format(now) # returns 22
print(f'{now.strftime("%p")}')
Else:
from datetime import datetime
now = datetime.now()
now.strftime("%p") # returns 'PM'
'{0.day}'.format(now) # returns 22
print('{0:%p}'.format(now))
Documentation
strftime() and strptime() Behavior
You could use f-strings like this:
from datetime import datetime
now = datetime.now()
now.strftime("%p")
print(f'{now.day}')
print(f'{now.strftime("%p")}')
How can one make 2020/09/06 15:59:04 out of 06-09-202015u59m04s.
This is my code:
my_time = '06-09-202014u59m04s'
date_object = datetime.datetime.strptime(my_time, '%d-%m-%YT%H:%M:%S')
print(date_object)
This is the error I receive:
ValueError: time data '06-09-202014u59m04s' does not match format '%d-%m-%YT%H:%M:%S'
>>> from datetime import datetime
>>> my_time = '06-09-202014u59m04s'
>>> dt_obj = datetime.strptime(my_time,'%d-%m-%Y%Hu%Mm%Ss')
Now you need to do some format changes to get the answer as the datetime object always prints itself with : so you can do any one of the following:
Either get a new format using strftime:
>>> dt_obj.strftime('%Y/%m/%d %H:%M:%S')
'2020/09/06 14:59:04'
Or you can simply use .replace() by converting datetime object to str:
>>> str(dt_obj).replace('-','/')
'2020/09/06 14:59:04'
As your error says what you give does not match format - %d-%m-%YT%H:%M:%S - means you are expecting after year: letter T hour:minutes:seconds when in example show it is houruminutesmsecondss without T, so you should do:
import datetime
my_time = '06-09-202014u59m04s'
date_object = datetime.datetime.strptime(my_time, '%d-%m-%Y%Hu%Mm%Ss')
print(date_object)
Output:
2020-09-06 14:59:04
You need to always make sure that your desired date format should match up with your required format.
from datetime import datetime
date_object = datetime.strptime("06-09-202015u59m04s", '%d-%m-%Y%Hu%Mm%Ss')
print(date_object.strftime('%Y/%m/%d %H:%M:%S'))
Output
2020/09/06 15:59:04
I need to assign to a variable the current datetime string in isoformat like the following:
2018-09-27T16:19:16+02:00
What I'm doing is:
import datetime
....
print(datetime.datetime.utcnow().replace(tzinfo=datetime.timezone.utc, microsecond=0).isoformat())
But this is going to print the string with utc tz:
2018-09-28T07:05:35+00:00
Not clear yet to me what's the clean way I should change tzinfo param to set wanted tz to UTC+02:00 ?
Thanks
utcnow() already gives you the the time at +00:00, if you'd like to recieve the time at a specific timezone, you should provide the timezone as an argument to now([tz]).
https://docs.python.org/3/library/datetime.html
>>> import datetime as dt
>>> dt.datetime.now(tz = dt.timezone(offset = dt.timedelta(hours = 2))).replace(microsecond = 0).isoformat()
'2018-09-28T09:20:19+02:00'
I have a time series that I have pulled 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 +10' (+10 being GMT: +10)
time = nc_data.variables['time'][:]
time_idx = 0 # first timestamp
print time[time_idx]
9465.0
My desired output is a datetime object like so (also GMT +10):
"2015-12-01 00:00:00"
I have tried converting this using the time module without much success although I believe I may be using wrong (I'm still a novice in python and programming).
import time
time_datetime = time.strftime('%Y-%m-%d %H:%M:%S', time.gmtime(time[time_idx]*24*60*60))
Any advice appreciated,
Cheers!
The datetime module's timedelta is probably what you're looking for.
For example:
from datetime import date, timedelta
days = 9465 # This may work for floats in general, but using integers
# is more precise (e.g. days = int(9465.0))
start = date(1990,1,1) # This is the "days since" part
delta = timedelta(days) # Create a time delta object from the number of days
offset = start + delta # Add the specified number of days to 1990
print(offset) # >>> 2015-12-01
print(type(offset)) # >>> <class 'datetime.date'>
You can then use and/or manipulate the offset object, or convert it to a string representation however you see fit.
You can use the same format as for this date object as you do for your time_datetime:
print(offset.strftime('%Y-%m-%d %H:%M:%S'))
Output:
2015-12-01 00:00:00
Instead of using a date object, you could use a datetime object instead if, for example, you were later going to add hours/minutes/seconds/timezone offsets to it.
The code would stay the same as above with the exception of two lines:
# Here, you're importing datetime instead of date
from datetime import datetime, timedelta
# Here, you're creating a datetime object instead of a date object
start = datetime(1990,1,1) # This is the "days since" part
Note: Although you don't state it, but the other answer suggests you might be looking for timezone aware datetimes. If that's the case, dateutil is the way to go in Python 2 as the other answer suggests. In Python 3, you'd want to use the datetime module's tzinfo.
netCDF num2date is the correct function to use here:
import netCDF4
ncfile = netCDF4.Dataset('./foo.nc', 'r')
time = ncfile.variables['time'] # do not cast to numpy array yet
time_convert = netCDF4.num2date(time[:], time.units, time.calendar)
This will convert number of days since 1900-01-01 (i.e. the units of time) to python datetime objects. If time does not have a calendar attribute, you'll need to specify the calendar, or use the default of standard.
We can do this in a couple steps. First, we are going to use the dateutil library to handle our work. It will make some of this easier.
The first step is to get a datetime object from your string (1990-01-01 00:00:00 +10). We'll do that with the following code:
from datetime import datetime
from dateutil.relativedelta import relativedelta
import dateutil.parser
days_since = '1990-01-01 00:00:00 +10'
days_since_dt = dateutil.parser.parse(days_since)
Now, our days_since_dt will look like this:
datetime.datetime(1990, 1, 1, 0, 0, tzinfo=tzoffset(None, 36000))
We'll use that in our next step, of determining the new date. We'll use relativedelta in dateutils to handle this math.
new_date = days_since_dt + relativedelta(days=9465.0)
This will result in your value in new_date having a value of:
datetime.datetime(2015, 12, 1, 0, 0, tzinfo=tzoffset(None, 36000))
This method ensures that the answer you receive continues to be in GMT+10.
How can I calculate the next day from a string like 20110531 in the same YYYYMMDD format? In this particular case, I like to have 20110601 as the result. Calculating "tomorrow" or next day in static way is not that tough, like this:
>>> from datetime import date, timedelta
>>> (date.today() + timedelta(1)).strftime('%Y%m%d')
'20110512'
>>>
>>> (date(2011,05,31) + timedelta(1)).strftime('%Y%m%d')
'20110601'
But how can I use a string like dt = "20110531" to get the same result as above?
Here is an example of how to do it:
import time
from datetime import date, timedelta
t=time.strptime('20110531','%Y%m%d')
newdate=date(t.tm_year,t.tm_mon,t.tm_mday)+timedelta(1)
print newdate.strftime('%Y%m%d')
>>> from datetime import datetime
>>> print datetime.strptime('20110531', '%Y%m%d')
2011-05-31 00:00:00
And then do math on that date object as you show in your question.
The datetime library docs.
You are most of the way there! along with the strftime function which converts a date to a formatted string, there is also a strptime function which converts back the other way.
To solve your problem you can just replace date.today() with strptime(yourDateString, '%Y%m%d').
ED: and of course you will also have to add strptime to the end of your from datetime import line.