can you tell me why it doesn't translate the date into numbers?
import datetime
from datetime import datetime
from datetime import timedelta
from django.db import models
def end_date():
return datetime.date.today() + timedelta(days=7)
def convert_date_str(add_date):
return datetime.strftime(add_date, "%m/%m/%Y")
class add(models.Model):
add_date = models.DateField('Дата додавання', auto_now_add=True)
end_date = models.DateField('Дата здачі', default=end_date)
as I do not output from the DB string month
First of all, strftime needs to be called on a datetime object so you need to enclose your addition statement in brackets.
from datetime import datetime, timedelta
def end_date():
return (datetime.today() + timedelta(days=7)).strftime("%d %B, %Y")
strptime is under datetime.datetime so you are going too deep. In any case, your second function doesn't make sense. You can only strptime from a string, not a datetime object. You also shouldn't use the same function and variable names. If you are trying to translate a datestring like 31/01/1999 to a datetime object, you can do
def convert_to_datetime(date_str):
return datetime.strptime(date_str, '%d/%m/%Y')
Your function names should also be more descriptive because at the moment I don't think any of them describes its actual functionality.
You're probably getting an Attribute error because timedelta has no strftime method. what you want is:
def end_date():
# add date object to timedelta first then get string
return (datetime.date.today() + timedelta(days=7)).strftime("%d %B, %Y")
second, strptime method takes a string and makes a datetime object. add_date is a date object not a string
def add_date():
add_date = datetime.date.today() # <-- this should be a string for strptime to work
# whats dt in your example?, also strptime argument must be a string
data_object = datetime.datetime.strptime(add_date, "%d %B, %Y").date()
return data_object
I want to add hours to a datetime and use:
date = date_object + datetime.timedelta(hours=6)
Now I want to add a time:
time='-7:00' (string) plus 4 hours.
I tried hours=time+4 but this doesn't work. I think I have to int the string like int(time) but this doesn't work either.
Better you parse your time like below and access datetime attributes for getting time components from the parsed datetime object
input_time = datetime.strptime(yourtimestring,'yourtimeformat')
input_seconds = input_time.second # for seconds
input_minutes = input_time.minute # for minutes
input_hours = input_time.hour # for hours
# Usage: input_time = datetime.strptime("07:00","%M:%S")
Rest you have datetime.timedelta method to compose the duration.
new_time = initial_datetime + datetime.timedelta(hours=input_hours,minutes=input_minutes,seconds=input_seconds)
See docs strptime
and datetime format
You need to convert to a datetime object in order to add timedelta to your current time, then return it back to just the time portion.
Using date.today() just uses the arbitrary current date and sets the time to the time you supply. This allows you to add over days and reset the clock to 00:00.
dt.time() prints out the result you were looking for.
from datetime import date, datetime, time, timedelta
dt = datetime.combine(date.today(), time(7, 00)) + timedelta(hours=4)
print dt.time()
Edit:
To get from a string time='7:00' to what you could split on the colon and then reference each.
this_time = this_time.split(':') # make it a list split at :
this_hour = this_time[0]
this_min = this_time[1]
Edit 2:
To put it all back together then:
from datetime import date, datetime, time, timedelta
this_time = '7:00'
this_time = this_time.split(':') # make it a list split at :
this_hour = int(this_time[0])
this_min = int(this_time[1])
dt = datetime.combine(date.today(), time(this_hour, this_min)) + timedelta(hours=4)
print dt.time()
If you already have a full date to use, as mentioned in the comments, you should convert it to a datetime using strptime. I think another answer walks through how to use it so I'm not going to put an example.
I have a date string formatted like this: "2017-05-31T06:44:13Z".
I need to check whether this date is within a one year span from today's date.
Which is the best method to do it: convert it into a timestamp and check, or convert into a date format?
Convert the timestamp to a datetime object so it can be compared with other datetime objects using <, >, =.
from datetime import datetime
from dateutil.relativedelta import relativedelta
# NOTE this format basically ignores the timezone. This may or may not be what you want
date_to_check = datetime.strptime('2017-05-31T06:44:13Z', '%Y-%m-%dT%H:%M:%SZ')
today = datetime.today()
one_year_from_now = today + relativedelta(years=1)
if today <= date_to_check <= one_year_from_now:
# do whatever
Use the datetime package together with timedelta:
import datetime
then = datetime.datetime.strptime("2017-05-31T06:44:13Z".replace('T',' ')[:-1],'%Y-%m-%d %H:%M:%S')
now = datetime.datetime.now()
d = datetime.timedelta(days = 365)
and simply check if now-d > then.
I am currently trying to convert a date in the following format YYYYmmddHHMMSS to a unix timestamp but I get an error (ValueError: year is out of range).
import datetime
def ts(date):
return datetime.datetime.fromtimestamp(date).strftime('%Y%m%d%H%M%S')
if __name__ == "__main__":
date = 20130814100000
print ts(date)
Your date should be a string. Here is how you do it. (If your date is an integer then just do date = str(date).
>>> import time
>>> from datetime import datetime
>>> date = '20130814100000'
>>> dt = datetime.strptime(date, '%Y%m%d%H%M%S')
>>> print dt
2013-08-14 10:00:00
>>> print time.mktime(dt.timetuple())
1376467200.0
time also has a strptime function but it returns a not so useful struct_time object. But if you only need a unix time, then you can use it too:
>>> time.mktime(time.strptime(date, '%Y%m%d%H%M%S'))
1376467200.0
I think the problem here is that .fromtimestamp() is expecting a Unix timestamp, not a date formatted as YYYYmmdd...
To parse the date information that you do have there, I'd recommend using .strptime() or the excellent python-dateutil package.
import datetime
def ts(date):
stamp = datetime.datetime.strptime(date, '%Y%m%d%H%M%S')
return stamp.strftime('%Y%m%d%H%M%S')
or
from dateutil.parser import parse
def ts(date):
stamp = parse(date)
return stamp.strftime('%Y%m%d%H%M%S')
http://labix.org/python-dateutil
The function that parses datetime is called strptime, not strftime (which formats time).
20130814100000 is not an UNIX timestamp
strptime takes string as argument
Overall, your code should look like:
import datetime
def ts(date):
return datetime.datetime.strptime(date, '%Y%m%d%H%M%S')
if __name__ == "__main__":
date = "20130814100000"
print ts(date)
You seem to have some confusion over the different ways that times are represented. The value you have assigned to date appears to already be a formatted timestring of "2013-08-14 10:00:00", but you're passing it into fromtimestamp. This function expects a Unix timestamp, which is simply the number of seconds that have elapsed since Midnight on Jan 1st 1970.
I believe something like this is what you're looking for:
import datetime
def ts(datestr):
return datetime.datetime.strptime(datestr, "%Y%m%d%H%M%S")
if __name__ == "__main__":
date = 20130814100000
print ts(date)
strftime like you had is for formatting times into strings. strptime is for parsing strings into times.
I have a string representing a unix timestamp (i.e. "1284101485") in Python, and I'd like to convert it to a readable date. When I use time.strftime, I get a TypeError:
>>>import time
>>>print time.strftime("%B %d %Y", "1284101485")
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: argument must be 9-item sequence, not str
Use datetime module:
from datetime import datetime
ts = int('1284101485')
# if you encounter a "year is out of range" error the timestamp
# may be in milliseconds, try `ts /= 1000` in that case
print(datetime.utcfromtimestamp(ts).strftime('%Y-%m-%d %H:%M:%S'))
>>> from datetime import datetime
>>> datetime.fromtimestamp(1172969203.1)
datetime.datetime(2007, 3, 4, 0, 46, 43, 100000)
Taken from http://seehuhn.de/pages/pdate
The most voted answer suggests using fromtimestamp which is error prone since it uses the local timezone. To avoid issues a better approach is to use UTC:
datetime.datetime.utcfromtimestamp(posix_time).strftime('%Y-%m-%dT%H:%M:%SZ')
Where posix_time is the Posix epoch time you want to convert
>>> import time
>>> time.ctime(int("1284101485"))
'Fri Sep 10 16:51:25 2010'
>>> time.strftime("%D %H:%M", time.localtime(int("1284101485")))
'09/10/10 16:51'
There are two parts:
Convert the unix timestamp ("seconds since epoch") to the local time
Display the local time in the desired format.
A portable way to get the local time that works even if the local time zone had a different utc offset in the past and python has no access to the tz database is to use a pytz timezone:
#!/usr/bin/env python
from datetime import datetime
import tzlocal # $ pip install tzlocal
unix_timestamp = float("1284101485")
local_timezone = tzlocal.get_localzone() # get pytz timezone
local_time = datetime.fromtimestamp(unix_timestamp, local_timezone)
To display it, you could use any time format that is supported by your system e.g.:
print(local_time.strftime("%Y-%m-%d %H:%M:%S.%f%z (%Z)"))
print(local_time.strftime("%B %d %Y")) # print date in your format
If you do not need a local time, to get a readable UTC time instead:
utc_time = datetime.utcfromtimestamp(unix_timestamp)
print(utc_time.strftime("%Y-%m-%d %H:%M:%S.%f+00:00 (UTC)"))
If you don't care about the timezone issues that might affect what date is returned or if python has access to the tz database on your system:
local_time = datetime.fromtimestamp(unix_timestamp)
print(local_time.strftime("%Y-%m-%d %H:%M:%S.%f"))
On Python 3, you could get a timezone-aware datetime using only stdlib (the UTC offset may be wrong if python has no access to the tz database on your system e.g., on Windows):
#!/usr/bin/env python3
from datetime import datetime, timezone
utc_time = datetime.fromtimestamp(unix_timestamp, timezone.utc)
local_time = utc_time.astimezone()
print(local_time.strftime("%Y-%m-%d %H:%M:%S.%f%z (%Z)"))
Functions from the time module are thin wrappers around the corresponding C API and therefore they may be less portable than the corresponding datetime methods otherwise you could use them too:
#!/usr/bin/env python
import time
unix_timestamp = int("1284101485")
utc_time = time.gmtime(unix_timestamp)
local_time = time.localtime(unix_timestamp)
print(time.strftime("%Y-%m-%d %H:%M:%S", local_time))
print(time.strftime("%Y-%m-%d %H:%M:%S+00:00 (UTC)", utc_time))
In Python 3.6+:
import datetime
timestamp = 1642445213
value = datetime.datetime.fromtimestamp(timestamp)
print(f"{value:%Y-%m-%d %H:%M:%S}")
Output (local time)
2022-01-17 20:46:53
Explanation
Line #1: Import datetime library.
Line #2: Unix time which is seconds since 1970-01-01.
Line #3: Converts this to a unix time object, check with: type(value)
Line #4: Prints in the same format as strp. Local time. To print in UTC see example below.
Bonus
To save the date to a string then print it, use this:
my_date = f"{value:%Y-%m-%d %H:%M:%S}"
print(my_date)
To output in UTC:
value = datetime.datetime.fromtimestamp(timestamp, tz=datetime.timezone.utc)
# 2022-01-17 18:50:52
Other than using time/datetime package, pandas can also be used to solve the same problem.Here is how we can use pandas to convert timestamp to readable date:
Timestamps can be in two formats:
13 digits(milliseconds) -
To convert milliseconds to date, use:
import pandas
result_ms=pandas.to_datetime('1493530261000',unit='ms')
str(result_ms)
Output: '2017-04-30 05:31:01'
10 digits(seconds) -
To convert seconds to date, use:
import pandas
result_s=pandas.to_datetime('1493530261',unit='s')
str(result_s)
Output: '2017-04-30 05:31:01'
For a human readable timestamp from a UNIX timestamp, I have used this in scripts before:
import os, datetime
datetime.datetime.fromtimestamp(float(os.path.getmtime("FILE"))).strftime("%B %d, %Y")
Output:
'December 26, 2012'
You can convert the current time like this
t=datetime.fromtimestamp(time.time())
t.strftime('%Y-%m-%d')
'2012-03-07'
To convert a date in string to different formats.
import datetime,time
def createDateObject(str_date,strFormat="%Y-%m-%d"):
timeStamp = time.mktime(time.strptime(str_date,strFormat))
return datetime.datetime.fromtimestamp(timeStamp)
def FormatDate(objectDate,strFormat="%Y-%m-%d"):
return objectDate.strftime(strFormat)
Usage
=====
o=createDateObject('2013-03-03')
print FormatDate(o,'%d-%m-%Y')
Output 03-03-2013
timestamp ="124542124"
value = datetime.datetime.fromtimestamp(timestamp)
exct_time = value.strftime('%d %B %Y %H:%M:%S')
Get the readable date from timestamp with time also, also you can change the format of the date.
Note that utcfromtimestamp can lead to unexpected results since it returns a naive datetime object. Python treats naive datetime as local time - while UNIX time refers to UTC.
This ambiguity can be avoided by setting the tz argument in fromtimestamp:
from datetime import datetime, timezone
dtobj = datetime.fromtimestamp(1284101485, timezone.utc)
>>> print(repr(dtobj))
datetime.datetime(2010, 9, 10, 6, 51, 25, tzinfo=datetime.timezone.utc)
Now you can format to string, e.g. an ISO8601 compliant format:
>>> print(dtobj.isoformat(timespec='milliseconds').replace('+00:00', 'Z'))
2010-09-10T06:51:25.000Z
Use the following codes, I hope it will solve your problem.
import datetime as dt
print(dt.datetime.fromtimestamp(int("1284101485")).strftime('%Y-%m-%d %H:%M:%S'))
Use datetime.strftime(format):
from datetime import datetime
unixtime = int('1284101485')
# Print with local time
print(datetime.fromtimestamp(unixtime).strftime('%Y-%m-%d %H:%M:%S'))
# Print with UTC time
print(datetime.utcfromtimestamp(unixtime).strftime('%Y-%m-%d %H:%M:%S'))
datetime.fromtimestamp(timestamp): Return the local date corresponding to the POSIX timestamp, such as is returned by time.time().
datetime.utcfromtimestamp(timestamp): Return the UTC datetime corresponding to the POSIX timestamp, with tzinfo None. (The resulting object is naive.)
import datetime
temp = datetime.datetime.fromtimestamp(1386181800).strftime('%Y-%m-%d %H:%M:%S')
print temp
Another way that this can be done using gmtime and format function;
from time import gmtime
print('{}-{}-{} {}:{}:{}'.format(*gmtime(1538654264.703337)))
Output: 2018-10-4 11:57:44
If you are working with a dataframe and do not want the series cannot be converted to class int error. Use the code below.
new_df= pd.to_datetime(df_new['time'], unit='s')
i just successfully used:
>>> type(tstamp)
pandas.tslib.Timestamp
>>> newDt = tstamp.date()
>>> type(newDt)
datetime.date
You can use easy_date to make it easy:
import date_converter
my_date_string = date_converter.timestamp_to_string(1284101485, "%B %d, %Y")
quick and dirty one liner:
'-'.join(str(x) for x in list(tuple(datetime.datetime.now().timetuple())[:6]))
'2013-5-5-1-9-43'