I want to change date format using Python. I don't know How to do that.
I have date format as shown below
2020-10-22 12:14:41.293000+00:00
I want to change above date format into below format:
date(2020, 10, 22)
I want this format because I want to get different between two dates.
with above format I can get difference by using following code,
d0 = date(2017, 8, 18)
d1 = date(2017, 10, 26)
delta = d1 - d0
print(delta.days)
So how to Change date Format as I discussed above.and also let me know if anyone know other way to get difference between these two dates 2020-10-22 12:14:41.293000+00:00 and 2020-10-25 12:14:41.293000+00:00 without changing its formet.I will be thankful if anyone can help me with this issue.
use fromisoformat and date() to get only the date, without the time:
from datetime import datetime
d = datetime.fromisoformat('2020-10-22 12:14:41.293000+00:00').date()
# d
# datetime.date(2020, 10, 22)
To convert a string to a datetime object, we use the datetime module.
from datetime import datetime
def str_to_date(s):
s = s.split()[0] # Separate the str by spaces and get the first item (the date)
s = datetime.strptime(s, "%Y-%m-%d")
return s # You now have a datetime object to do operations on.
If the format is always in the form YYYY-MM-DD ..... then you can strip the first 10 characters of the date and use datetime to convert into datetime object
from datetime import datetime
unformatted_date = "2020-10-22 12:14:41.293000+00:00"
date_obj = datetime.strptime(unformatted_date[0:11], "%Y-%m-%d").date()
print(date_obj)
Change string to datetime format:
from datetime import datetime
date1 = datetime.strptime('2020-10-22 12:14:41.293000+00:00', '%Y-%m-%d %H:%M:%S.%f%z').date()
date2 = datetime.strptime('2020-10-23 12:14:41.293000+00:00', '%Y-%m-%d %H:%M:%S.%f%z').date()
print(date2-date1)
#1 day, 0:00:00
Related
I have this column where the string has date, month, year and also time information. I need to take the date, month and year only.
There is no space in the string.
The string is on this format:
date
Tuesday,August22022-03:30PMWIB
Monday,July252022-09:33PMWIB
Friday,January82022-09:33PMWIB
and I expect to get:
date
2022-08-02
2022-07-25
2022-01-08
How can I get the date, month and year only and change the format into yyyy-mm-dd in python?
thanks in advance
Use strptime from datetime library
var = "Tuesday,August22022-03:30PMWIB"
date = var.split('-')[0]
formatted_date = datetime.strptime(date, "%A,%B%d%Y")
print(formatted_date.date()) #this will get your output
Output:
2022-08-02
You can use the standard datetime library
from datetime import datetime
dates = [
"Tuesday,August22022-03:30PMWIB",
"Monday,July252022-09:33PMWIB",
"Friday,January82022-09:33PMWIB"
]
for text in dates:
text = text.split(",")[1].split("-")[0]
dt = datetime.strptime(text, '%B%d%Y')
print(dt.strftime("%Y-%m-%d"))
An alternative/shorter way would be like this (if you want the other date parts):
for text in dates:
dt = datetime.strptime(text[:-3], '%A,%B%d%Y-%I:%M%p')
print(dt.strftime("%Y-%m-%d"))
The timezone part is tricky and works only for UTC, GMT and local.
You can read more about the format codes here.
strptime() only accepts certain values for %Z:
any value in time.tzname for your machine’s locale
the hard-coded values UTC and GMT
You can convert to datetime object then get string back.
from datetime import datetime
datetime_object = datetime.strptime('Tuesday,August22022-03:30PM', '%A,%B%d%Y-%I:%M%p')
s = datetime_object.strftime("%Y-%m-%d")
print(s)
You can use the datetime library to parse the date and print it in your format. In your examples the day might not be zero padded so I added that and then parsed the date.
import datetime
date = 'Tuesday,August22022-03:30PMWIB'
date = date.split('-')[0]
if not date[-6].isnumeric():
date = date[:-5] + "0" + date[-5:]
newdate = datetime.datetime.strptime(date, '%A,%B%d%Y').strftime('%Y-%m-%d')
print(newdate)
# prints 2022-08-02
I have an variable integer that is in the format YYYYMMDD.
How do i convert that variable to datetime while maintaining the format of YYYYMMDD?
date = 20200930
nextday = date + 1
How do i fix this, so the nextday variable displays as 20201001
If I'm understanding what you need to do correctly, you can easily do it using the datetime package.
First, convert your variable to a date:
import datetime
date = 20200930
dt = datetime.datetime.strptime(str(date), '%Y%m%d')
Now, when you print this, you will see:
datetime.datetime(2020, 9, 30, 0, 0).
You can then add a day to it:
dt_new = dt + datetime.timedelta(1)
And specify the format you want to see the new date variable in:
print ('{date: %Y%m%d}'.format(date=dt_new))
which will give:
20201001.
I using:
s = "20200113"
final = datetime.datetime.strptime(s, '%Y%m%d')
I need convert a number in date format (2020-01-13)
but when I print final:
2020-01-13 00:00:00
Tried datetime.date(s, '%Y%m%d') but It's returns a error:
an integer is required (got type str)
Is there any command to get only date without hour?
Once you have a datetime object just use strftime
import datetime
d = datetime.datetime.now() # Some datetime object.
print(d.strftime('%Y-%m-%d'))
which gives
2020-02-20
You can use strftime to convert back in the format you need :
import datetime
s = "20200113"
temp = datetime.datetime.strptime(s, '%Y%m%d')
# 2020-01-13 00:00:00
final = temp.strftime('%Y-%m-%d')
print(final)
# 2020-01-13
Use datetime.date(year, month, day). Slice your string and convert to integers to get the year, month and day. Now it is a datetime.date object, you can use it for other things. Here, however, we use .strftime to convert it back to text in your desired format.
s = "20200113"
year = int(s[:4]) # 2020
month = int(s[4:6]) # 1
day = int(s[6:8]) # 13
>>> datetime.date(year, month, day).strftime('%Y-%m-%d')
'2020-01-13'
You can also convert directly via strings.
>>> f'{s[:4]}-{s[4:6]}-{s[6:8]}'
'2020-01-13'
You can use .date() on datetime objects to 'remove' the time.
my_time_str = str(final.date())
will give you the wanted result
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"
I have a web form which has 2 input fields, "StartDate" and "StartTime". I convert the StartDate field's string into a Python datetime object, no problem. The StartTime field is passed in as a string in the form "0130" for 1:30am. What is the best way to convert the StartTime string and combine it with the StartDate datetime object so that both are stored as a single datetime?
Use datetime.combine:
import datetime as dt
mytime = dt.datetime.strptime('0130','%H%M').time()
mydatetime = dt.datetime.combine(dt.date.today(), mytime)
If you can load the time into a datetime.time, you can use the following code
import datetime
dt = datetime.datetime(2012, 2, 12)
tm = datetime.time(1, 30)
combined = dt.combine(dt, tm)
print(combined)
Output
2012-02-12 01:30:00
Just a short version:
from datetime import datetime
print datetime.combine(datetime.strptime("5 Mar 12", "%d %b %y"), datetime.strptime("0130","%H%M").time())
Output
2012-03-05 01:30:00
import datetime
def time_tango(date, time):
return datetime.datetime.combine(date, time)