I'm given a timestamp (time since the epoch) and I need to convert it into this format:
yyyy/mm/dd hh:mm
I looked around and it seems like everyone else is doing this the other way around (date to timestamp).
If your answer involves dateutil that would be great.
Using datetime instead of dateutil:
import datetime as dt
dt.datetime.utcfromtimestamp(seconds_since_epoch).strftime("%Y/%m/%d %H:%M")
An example:
import time
import datetime as dt
epoch_now = time.time()
sys.stdout.write(str(epoch_now))
>>> 1470841955.88
frmt_date = dt.datetime.utcfromtimestamp(epoch_now).strftime("%Y/%m/%d %H:%M")
sys.stdout.write(frmt_date)
>>> 2016/08/10 15:09
EDIT: strftime() used, as the comments suggested.
Related
The following code converts a string into a timestamp. The timestamp comes out to: 1646810127.
However, if I use Excel to convert this date and time into a float I get: 44629,34.
I need the Excel's output from the Python script.
I have tried with a few different datetime strings to see if there is any pattern in between the two numbers, but cannot seem to find any.
Any thoughts on how I get the code to output 44629,34?
Much appreciated
import datetime
date_time_str = '2022-03-09 08:15:27'
date_time_obj = datetime.datetime.strptime(date_time_str, '%Y-%m-%d %H:%M:%S')
print('Date:', date_time_obj.date())
print('Time:', date_time_obj.time())
print('Date-time:', date_time_obj)
print(date_time_obj.timestamp())
>>output:
Date: 2022-03-09
Time: 08:15:27
Date-time: 2022-03-09 08:15:27
1646810127.0
calculate the timedelta of your datetime object versus Excel's "day zero", then divide the total_seconds of the timedelta by the seconds in a day to get Excel serial date:
import datetime
date_time_str = '2022-03-09 08:15:27'
UTC = datetime.timezone.utc
dt_obj = datetime.datetime.fromisoformat(date_time_str).replace(tzinfo=UTC)
day_zero = datetime.datetime(1899,12,30, tzinfo=UTC)
excel_serial_date = (dt_obj-day_zero).total_seconds()/86400
print(excel_serial_date)
# 44629.3440625
Note: I'm setting time zone to UTC here to avoid any ambiguities - adjust as needed.
Since the question is tagged pandas, you'd do the same thing here, only that you don't need to set UTC as pandas assumes UTC by default for naive datetime:
import pandas as pd
ts = pd.Timestamp('2022-03-09 08:15:27')
excel_serial_date = (ts-pd.Timestamp('1899-12-30')).total_seconds()/86400
print(excel_serial_date)
# 44629.3440625
See also:
background: What is story behind December 30, 1899 as base date?
inverse operation: Convert Excel style date with pandas
How can i get datetime in ISO 8601 date format (YYYY-MM-DDThh:mmTZD). Example 2019-02-26T09:30:46+03:00
I tried using
from datetime import datetime
d = datetime.now()
d.isoformat()
But the output is now correct
'2020-07-29T15:47:46.974744'
d = datetime.now()
d.strftime("%Y-%m-%dT%H:%M:%S%z")
Do note that %z will return empty if its a naive datetime object
This should give you the format you're looking for.
https://strftime.org/
This is a good reference on the available formats
Given I have a timestamp:
date_time_str = '2019-09-10T13:48:06+0200'
How can I calculate the time difference between the current time and this datetime?
I've got it so far with an impression of strong wrongdoing - this should be possible in a far simpler way:
from datetime import datetime, timezone
import time
date_time_str = '2019-09-10T13:48:06+0200'
format = '%Y-%m-%dT%H:%M:%S%z'
date_time_obj = datetime.strptime(date_time_str, format)
now = datetime.now()
now_time = now.strftime(format)
print(now_time)
now=datetime.strptime(datetime.fromtimestamp(int(time.time()), tz=timezone.utc).isoformat(), format)
print("now is: %s" % now)
print(now-time_obj)
The above program does not work because the current time comes out in a slightly different formatting:
'2019-09-10T15:56:11+00:00'
That is, if you run the above script for example Python 3.6.5, you get the error:
ValueError: time data '2019-09-10T18:18:09+00:00' does not match format '%Y-%m-%dT%H:%M:%S%z'
The mismatch is in the timezone format, "+00:00" vs. "+0200".
You can use datetime.now() to get the current datetime in utc:
# Same as your code
from datetime import datetime, timezone
date_time_str = '2019-09-10T13:48:06+0200'
format = '%Y-%m-%dT%H:%M:%S%z'
date_time_obj = datetime.strptime(date_time_str, format)
# Added:
print(datetime.now(tz=timezone.utc))
# 2019-09-10 18:35:48.066548+00:00
print(datetime.now(tz=timezone.utc) - date_time_obj)
# 6:47:42.066548
I'a m trying to extract only time (Hour:Minute) from datetime field
Example:
today_with_hour = fields.Datetime(
string=u'hora',
default=fields.Datetime.now,
)
I would like to know how get only hour from today_with_hour in format
17:10:20
This is one way to extract:
from datetime import datetime
now = datetime.now()
print(str(now.hour)+':'+str(now.minute)+':'+str(now.second))
This may be better way to do it
You can use strftime
Example:
from datetime import datetime
datetime.now().strftime("%H:%M:%S")
In your case you can follow like this:
from datetime import datetime
datetime.strptime('20/06/2019 17:28:52', "%d/%m/%Y %H:%M:%S").time()
Output will be:
17:28:52
Better way to do is by using strftime().
dt = datetime.strptime('20/06/2019 17:28:52', "%d/%m/%Y %H:%M:%S")
dt.strftime("%H:%M:%S")
output:
17:28:52
To get the current time from the datetime.now()
datetime.datetime.now().time().strftime("%H:%M:%S")
O/P:
'11:16:17'
if you want to get the time with milliseconds also, use isoformat()
datetime.datetime.now().time().isoformat()
O/P:
'11:20:34.978272'
I have an unusual datetime format in my dataset, which I need to convert to usable datetime object.
An example looks like: '1/3/2018 1:29:35 PM(UTC+0)'
I have tried to parse it with:
from dateutil.parser import parse
parse('1/3/2018 1:29:35 PM(UTC+0)')
but it doesn't recognize the format.
My current workaround is to parse the datetime column (the data is in pandas dataframe) using regex into two columns, like so:
and then depending on the value of the 'utc' column apply custom convert_to_eastern function.
I wonder if there is an easier way to accomplish it using datetime.datetime.strptime() ?
Following didn't work:
import datetime as dt
my_time='1/3/2018 1:29:35 PM(UTC+0)'
dt.datetime.strptime(my_time, '%m/%d/%Y %I:%M:%S %p(%z)')
Addition:
This is not a question: "How to convert UTC timezone into local timezone" My dataset has rows with UTC as well as Eastern time zone rows. The problem I have is that the format is not an ISO format, but some human-readable custom format.
Question: an easier way to accomplish it using datetime.datetime.strptime()
Split the datestring into parts: utc:[('1/3/2018 1:29:35 PM', '(UTC+0)', 'UTC', '+', '0')]
Rebuild the datestring, fixing the hour part padding with 0 to 2 digits.
I assume, there are no minutes in the UTC part, therefore defaults to 00.
If the datestring has more then 2 UTC digits, returns the unchanged datestring.
Note: The strptime format have to be %Z%z!
Documentation: strftime-and-strptime-behavior
from datetime import datetime
import re
def fix_UTC(s):
utc = re.findall(r'(.+?)(\((\w{3})(\+|\-)(\d{1,2})\))', s)
if utc:
utc = utc[0]
return '{}({}{}{})'.format(utc[0], utc[2], utc[3], '{:02}00'.format(int(utc[4])))
else:
return s
my_time = fix_UTC('1/3/2018 1:29:35 PM(UTC+0)')
date = datetime.strptime(my_time, '%m/%d/%Y %I:%M:%S %p(%Z%z)')
print("{} {}".format(date, date.tzinfo))
Output:
2018-01-03 13:29:35+01:00 UTC
Tested with Python: 3.4.2
The problem is with '+0' for your timezone 'UTC+0'. datetime only takes utc offset in the form of HHMM. Possible workaround:
import datetime as dt
my_time = '1/3/2018 1:29:35 PM(UTC+0)'
my_time=my_time.replace('+0','+0000')
dt.datetime.strptime(my_time, '%m/%d/%Y %I:%M:%S %p(%Z%z)')
It should be something like that:
import datetime as dt
my_time='1/3/2018 1:29:35 PM(UTC+0000)'
tmp = dt.datetime.strptime(my_time, '%m/%d/%Y %I:%M:%S %p(%Z%z)')
print(tmp)
Big "Z" for timezone (UTC, GMT etc), small "z" for delta. Also you should add more zeros to delta.