I have the following code and am getting the above error. Since I'm new to python I'm having trouble understanding the syntax here and how I can fix the error:
if not start or date < start: start = date
There is a datetime.date() method for converting from a datetime to a date.
To do the opposite conversion, you could use this function datetime.datetime(d.year, d.month, d.day)
You can use the datetime.datetime.combine method to compare the date object to datetime object, then compare the converted object with the other datetime object.
import datetime
dt1 = datetime.datetime(2011, 03, 03, 11, 12)
day = datetime.date(2011, 03, 02)
dt2 = datetime.datetime.combine(day, datetime.time(0, 0))
print dt1 > dt2
Assuming start is a datetime, Use it like this:
if not start or date < start.date(): start = date
I don't think there is a need to convert date to datetime in python, as you can just do the opposite and compare.
Or else you have other methods to create a new datetime by using the date to convert and time at 00:00.
Wow, question and answers are too old, it needs update. Converting datetime.datetime object to datetime.date object is just easy:
somestringtext = '7.04.2021'
datetime_datetime_object = datetime.strptime(somestringtext, '%d.%m.%Y')
### returns datetime.datetime(2021, 4, 7, 0, 0)
datetime_date_object = datetime.date(datetime_datetime_object)
And datetime object is not same as date object, you cant compare
datetime_datetime_object == datetime_date_object
### returns False
unless you convert them to same format:
datetime.date(datetime_datetime_object) == datetime_date_object
### returns True
I was receiving the above error while using pandas, however, because the date_column was the string I wasted a lot of time without realizing I was formatting the wrong thing:
# didnt work
df[(df.date_column > parse_datestr('2018-01-01'))]
# works
df['date_column'] = pd.to_datetime(df['date_column'])
df[(df.date_column > '2018-01-01') & (df.date_column < '2018-02-28')]
This problem arises when you are trying to compare a date field (DateField) and a datetime field (DateTimeField).
The solution would be check where you defined the fields in your models and ensure that the types are uniform.
I would suggest you replace all DateField with DateTimeField.
Your variables start and date are of different type I guess. One is a datetime and one is a date. You may have to show more code in order to get decent help.
But look at this: http://docs.python.org/library/datetime.html#available-types
It tells you that datetime.datetime has attributes like day, month and year, just like datetime.date.
I solved it using .date() function available with datetime object. Here is how:
date_object = datetime_object.date()
and then you can compare it with any datetime.date object. Hope this helps.
Related
I have the following two date/time which are date_time1 and date_time2 respectively:
2017-04-15 00:00:00
2017-04-17 15:35:19+00:00
parsed1 = dateutil.parser.parse(date_time1)
parsed2 = dateutil.parser.parse(date_time2)
and would if I were to receive another date/time called input_date_time (e.g. 2017-04-16 12:11:42+00:00), would like to do the following:
# Would like to check if `input_date_time` is within the range
if parsed1 <= input_date_time <= parsed2:
…
And got an error: TypeError: can't compare offset-naive and offset-aware datetimes
Thought up of breaking it down to just year, month, day, hour, minute, and second, and compare every single one.
What would be the proper way to do so?
here is my edited (again) example
I think we should provide timezone data to every datetime object
assume that date_time1 is a local time.
I think we should add timezone data to date_time1 instead of clear other tzinfo (my first example)
import dateutil.parser
import datetime
from pytz import utc
date_time1 ='2017-04-15 00:00:00'
date_time2 ='2017-04-17 15:35:19+00:00'
input_date_time = '2017-04-16 12:11:42+00:00'
parsed1 = dateutil.parser.parse(date_time1).astimezone(utc)
parsed2 = dateutil.parser.parse(date_time2)
input_parsed = dateutil.parser.parse(input_date_time)
if parsed1 <= input_parsed <= parsed2:
print('input is between')
this can check if input is between parsed1 and parsed2
Assuming you have python datetime obejcts,
two objects in python can be compared with the "<", "==", and ">" signs.
You don't need to parse them to compare them.
if date_time1 <= input_date_time <= datetime_2:
#do work
If you don't have datetime objects, there is also a method called datetime in the datetime class, which will allow you to create datetime objects, if you'll find that useful.
You need to apply a timezone to the 'naive ' datetime object (2017-04-15 00:00:00 in your example) (to make it TZ aware) OR convert the 'aware' datetime object (2017-04-17 15:35:19+00:00 in your example) to a 'naive' object and the date you are trying to compare.
Then your TypeError will disappear.
Since your second date has a timezone offset of +00:00 and your input_datetime is also +00:00, let's apply UTC to the naive first date (assuming that it's the correct timezone) and then convert it to whatever timezone you need (you can skip the conversion if UTC is correct - the comparison will now work.)
parsed1 = dateutil.parser.parse(date_time1)
parsed2 = dateutil.parser.parse(date_time2)
# make parsed1 timezone aware (UTC)
parsed1 = parsed1.replace(tzinfo=pytz.utc)
Now your comparison should work.
If you want to apply another timezone to any of the dates, you can use the astimezone function. Lets change the timezone to that applicable to Sydney, Australia. Here is a list of timezones https://gist.github.com/heyalexej/8bf688fd67d7199be4a1682b3eec7568
syd_tz = pytz.timezone('Australia/Sydney')
syd_parsed1 = parsed1.astimezone(syd_tz)
You can now check what timezone is applied to each of your datetime objects using the %zand %Z parameters for strftime. Using %c will print it in the local time format as will %x and %X.
Using Python3+:
print("Local time: %s" % syd_parsed1.strftime('%c'))
print("Offset-Timezone-Date-Time: %s" % syd_parsed1.strftime("%z-%Z-%x-%X))
Hope that helps, the timezone functions did my head in when I used them the first time when I didn't know about %c.
I want to convert
2010-03-01 to 733832
I just found this toordinal code
d=datetime.date(year=2010, month=3, day=1)
d.toordinal()
from this
But i want something more like
d=datetime.date('2010-03-01')
d.toordinal()
Thanks in advance
You'll need to use strptime on the date string, specifying the format, then you can call the toordinal method of the date object:
>>> from datetime import datetime as dt
>>> d = dt.strptime('2010-03-01', '%Y-%m-%d').date()
>>> d
datetime.date(2010, 3, 1)
>>> d.toordinal()
733832
The call to the date method in this case is redundant, and is only kept for making the object consistent as a date object instead of a datetime object.
If you're looking to handle more date string formats, Python's strftime directives is one good reference you want to check out.
like this:
datetime.strptime("2016-01-01", "%Y-%m-%d").toordinal()
You need to firstly convert the time string to datetime object using strptime(). Then call .toordinal() on the datetime object
>>> from datetime import datetime
>>> date = datetime.strptime('2010-03-01', '%Y-%M-%d')
>>> date.toordinal()
733773
It is even better to create a function to achieve this as:
def convert_date_to_ordinal(date):
return datetime.strptime(date, '%Y-%M-%d').toordinal()
convert_date_to_ordinal('2010-03-01')
#returns: 733773
I am trying to set a variable to equal today's date.
I looked this up and found a related article:
Set today date as default value in the model
However, this didn't particularly answer my question.
I used the suggested:
dt.date.today
But after
import datetime as dt
date = dt.date.today
print date
<built-in method today of type object at 0x000000001E2658B0>
Df['Date'] = date
I didn't get what I actually wanted which as a clean date format of today's date...in Month/Day/Year.
How can I create a variable of today's day in order for me to input that variable in a DataFrame?
You mention you are using Pandas (in your title). If so, there is no need to use an external library, you can just use to_datetime
>>> pandas.to_datetime('today').normalize()
Timestamp('2015-10-14 00:00:00')
This will always return today's date at midnight, irrespective of the actual time, and can be directly used in pandas to do comparisons etc. Pandas always includes 00:00:00 in its datetimes.
Replacing today with now would give you the date in UTC instead of local time; note that in neither case is the tzinfo (timezone) added.
In pandas versions prior to 0.23.x, normalize may not have been necessary to remove the non-midnight timestamp.
If you want a string mm/dd/yyyy instead of the datetime object, you can use strftime (string format time):
>>> dt.datetime.today().strftime("%m/%d/%Y")
# ^ note parentheses
'02/12/2014'
Using pandas: pd.Timestamp("today").strftime("%m/%d/%Y")
pd.datetime.now().strftime("%d/%m/%Y")
this will give output as '11/02/2019'
you can use add time if you want
pd.datetime.now().strftime("%d/%m/%Y %I:%M:%S")
this will give output as '11/02/2019 11:08:26'
strftime formats
You can also look into pandas.Timestamp, which includes methods like .now and .today.
Unlike pandas.to_datetime('now'), pandas.Timestamp.now() won't default to UTC:
import pandas as pd
pd.Timestamp.now() # will return California time
# Timestamp('2018-12-19 09:17:07.693648')
pd.to_datetime('now') # will return UTC time
# Timestamp('2018-12-19 17:17:08')
i got the same problem so tried so many things
but finally this is the solution.
import time
print (time.strftime("%d/%m/%Y"))
simply just use pd.Timestamp.now()
for example:
input: pd.Timestamp.now()
output: Timestamp('2022-01-12 14:43:05.521896')
I know all you want is Timestamp('2022-01-12') you don't anything after
thus we could use replace to remove hour, minutes , second and microsecond
here:
input: pd.Timestamp.now().replace(hour=0, minute=0, second=0, microsecond=0)
output: Timestamp('2022-01-12 00:00:00')
but looks too complicated right, here is a simple way use normalize
input: pd.Timestamp.now().normalize()
output: Timestamp('2022-01-12 00:00:00')
Easy solution in Python3+:
import time
todaysdate = time.strftime("%d/%m/%Y")
#with '.' isntead of '/'
todaysdate = time.strftime("%d.%m.%Y")
import datetime
def today_date():
'''
utils:
get the datetime of today
'''
date=datetime.datetime.now().date()
date=pd.to_datetime(date)
return date
Df['Date'] = today_date()
this could be safely used in pandas dataframes.
There are already quite a few good answers, but to answer the more general question about "any" period:
Use the function for time periods in pandas. For Day, use 'D', for month 'M' etc.:
>pd.Timestamp.now().to_period('D')
Period('2021-03-26', 'D')
>p = pd.Timestamp.now().to_period('D')
>p.to_timestamp().strftime("%Y-%m-%d")
'2021-03-26'
note: If you need to consider UTC, you can use: pd.Timestamp.utcnow().tz_localize(None).to_period('D')...
From your solution that you have you can use:
import pandas as pd
pd.to_datetime(date)
using the date variable that you use
I'm working on an image upload utility, and part of the functionality is to parse the IPTC and EXIF data of the images.
IPTCInfo gets the information I need, but the date fields are in the format 20130925.
Now, I can break that integer up into 2013 09 25 and create a date object. Before I do so, is there already existing functionality to solve this issue?
The date class doesn't have a string-parsing function, but the datetime class does, strptime.
So, first make a datetime, then extract the date part of it:
>>> s = '20130925'
>>> dt = datetime.datetime.strptime(s, '%Y%m%d')
>>> d = dt.date()
>>> d
datetime.date(2013, 9, 25)
If you don't understand where the '%Y%m%d' comes from, see strftime() and strptime() Behavior.
You can use datetime.strptime:
>>> import datetime
>>> datetime.datetime.strptime("20130925","%Y%m%d").date()
datetime.date(2013, 9, 25)
What are these date-time formats? I need to convert them to the same format, to check if they are the same. These are just two coming from a separate data source, so I need to find a way to make them the same format. Any ideas?
2013-07-12T07:00:00Z
2013-07-10T11:00:00.000Z
Thanks in advance
That extra .000 is micro seconds.
This will convert a date string of a format to datetime object.
import datetime
d1 = datetime.datetime.strptime("2013-07-12T07:00:00Z","%Y-%m-%dT%H:%M:%SZ")
d2 = datetime.datetime.strptime("2013-07-10T11:00:00.000Z","%Y-%m-%dT%H:%M:%S.%fZ")
Then convert them into any format depending on your requirement, by using:
new_format = "%Y-%m-%d"
d1.strftime(new_format)
perhaps use .isoformat()
string in ISO 8601 format, YYYY-MM-DDTHH:MM:SS[.mmmmmm][+HH:MM]
>>> import datetime
>>> datetime.datetime.utcnow().isoformat() + "Z"
'2013-07-11T22:26:51.564000Z'
>>>
Z specifies "zulu" time or UTC.
You can also add the timezone component by making your datetime object timezone aware by applying the appropriate tzinfo object. With the tzinfo applied the .isoformat() method will include the appropriate utc offset in the output:
>>> d = datetime.datetime.utcnow().replace(tzinfo=datetime.timezone.utc)
>>> d.isoformat()
'2019-11-11T00:52:43.349356+00:00'
You can remove the microseconds by change the microseconds value to 0:
>>> no_ms = d.replace(microsecond=0)
>>> no_ms.isoformat()
'2019-11-11T00:52:43+00:00'
Also, as of python 3.7 the .fromisoformat() method is available to load an iso formatted datetime string into a python datetime object:
>>> datetime.datetime.fromisoformat('2019-11-11T00:52:43+00:00')
datetime.datetime(2019, 11, 11, 0, 52, 43, tzinfo=datetime.timezone.utc)
http://www.ietf.org/rfc/rfc3339.txt
you can try to trim the string
data = "2019-10-22T00:00:00.000-05:00"
result1 = datetime.datetime.strptime(data[0:19],"%Y-%m-%dT%H:%M:%S")
result2 = datetime.datetime.strptime(data[0:23],"%Y-%m-%dT%H:%M:%S.%f")
result3 = datetime.datetime.strptime(data[0:9], "%Y-%m-%d")
use datetime module.
For a variable
import datetime
def convertDate(d):
new_date = datetime.datetime.strptime(d,"%Y-%m-%dT%H:%M:%S.%fZ")
return new_date.date()
convertDate("2019-12-23T00:00:00.000Z")
you can change the ".date()" to ".year", ".month", ".day" etc...
Output: # is now a datetime object
datetime.date(2019, 12, 23)
For a DataFrame column, use apply()
df['new_column'] = df['date_column'].apply(convertDate)
* Short and best way:
str(datetime.datetime.now()).replace(' ','T')
or
str(datetime.datetime.now()).replace(' ','T') + "Z"