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Subtracting datetime in string format with datetime format

I have 2 variables.
One is datetime in string format and the other is datetime in datetime.datetime format.
For example -
2021-09-06T07:58:19.032Z # string
2021-09-05 14:58:10.209675 # datetime.datetime
I want to find out the difference between these 2 times in seconds.
I think we need to have both in datetime before we can do this subtraction.
I'm having a hard time converting the string to datetime.
Can someone please help.

You can convert the string into datetime object with strptime()
An example with your given dates:
from datetime import datetime
# Assuming this is already a datetime object in your code, you don't need this part
# I needed this part to be able to use it as a datetime object
date1 = datetime.strptime("2021-09-05 14:58:10.209675", "%Y-%m-%d %H:%M:%S.%f")
## The part where the string is converted to datetime object
# Since the string has "T" and "Z", we will have to remove them before we convert
formatted = "2021-09-06T07:58:19.032Z".replace("T", " ").replace("Z", "")
>>> 2021-09-06 07:58:19.032
# Finally, converting the string
date2 = datetime.strptime(formatted, "%Y-%m-%d %H:%M:%S.%f")
# Now date2 variable is a datetime object
# Performing a simple operation
print(date1 - date2)
>>> -1 day, 6:59:51.177675

Convert the str to datetime via strptime() and then get the difference of the 2 datetime objects in seconds via total_seconds().
from datetime import datetime, timezone
# Input
dt1_str = "2021-09-06T07:58:19.032Z" # String type
dt2 = datetime(year=2021, month=9, day=5, hour=14, minute=58, second=10, microsecond=209675, tzinfo=timezone.utc) # datetime type
# Convert the string to datetime
dt1 = datetime.strptime(dt1_str, "%Y-%m-%dT%H:%M:%S.%f%z")
# Subtract the datetime objects and get the seconds
diff_seconds = (dt1 - dt2).total_seconds()
print(diff_seconds)
Output
61208.822325

The first string time you mention could be rfc3339 format.
A module called python-dateutil could help
import dateutil.parser
dateutil.parser.parse('2021-09-06T07:58:19.032Z')
datetime module could parse this time format by
datetime.datetime.strptime("2021-09-06T07:58:19.032Z","%Y-%m-%dT%H:%M:%S.%fZ")
But this way may cause trouble when get a time in another timezone because it doesn't support timezone offset.

python json date object to python datetime

I have a JSON object with a date that returns
print row['ApplicationReceivedDateTime']
/Date(1454475600000)/
how do I process this using the pythons datetime module?
print type(row['ApplicationReceivedDateTime'])
returns <type 'unicode'>
print repr(row['ApplicationReceivedDateTime'])
returns u'/Date(1454475600000)/'

That looks like milliseconds. Try dividing by 1000.
import datetime as dt
>>> dt.datetime.fromtimestamp(1454475600000 / 1000)
datetime.datetime(2016, 2, 2, 21, 0)
If the date is in the string format per your question, extract the numeric portion using re.
date = '/Date(1454475600000)/'
>>> dt.datetime.fromtimestamp(int(re.findall(r"\d+", date)[0]) / 1000)
datetime.datetime(2016, 2, 2, 21, 0)

You probably want
datetime.datetime.strptime(string_date, "%Y-%m-%d %H:%M:%S.%f")
And the values of Year, Month, Day, Hour, Minute, Second and F, for that you can write a manual function for that like this
def generate_date_time_str(date_str):
"""Login to parse the date str"""
return date_str
the date_str will look link this
"%Y-%m-%d %H:%M:%S.%f"
There is no python module directly convert any random date str to DateTime object

You can use re to get the integer value and then use datetime.datetime.fromtimestamp to get the date value:
from datetime import datetime
import re
string_time = row['ApplicationReceivedDateTime']
parsed_time = int(re.search('\((\d+)\)', string_time)[1]) / 1e3 #1e3 == 1000
rcvd_date = datetime.fromtimestamp(parsed_time)
print(rcvd_date.strftime('%Y-%m-%d %H:%M:%S'))
Prints:
'2016-02-03 05:00:00'

Convert date to ordinal python?

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

Python datetime.strptime - Converting month in String format to Digit

I have a string that contains the date in this format:
full_date = "May.02.1982"
I want to use datetime.strptime() to display the date in all digits like: "1982-05-02"
Here's what I tried:
full_date1 = datetime.strptime(full_date, "%Y-%m-%d")
When I try to print this, I get garbage values like built-in-67732
Where am I going wrong? Does the strptime() method not accept string values?

Your format string is wrong, it should be this:
In [65]:
full_date = "May.02.1982"
import datetime as dt
dt.datetime.strptime(full_date, '%b.%d.%Y')
Out[65]:
datetime.datetime(1982, 5, 2, 0, 0)
You then need to call strftime on a datetime object to get the string format you desire:
In [67]:
dt.datetime.strptime(full_date, '%b.%d.%Y').strftime('%Y-%m-%d')
Out[67]:
'1982-05-02'
strptime is for creating a datetime format from a string, not to reformat a string to another datetime string.
So you need to create a datetime object using strptime, then call strftime to create a string from the datetime object.
The datetime format strings can be found in the docs as well as an explanation of strptime and strftime

Date Time Formats in Python

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"