Python3获取纪元时间第一天和最后一天

Question

Given a month and a year, such as 03 2022 , I need to get the epoch time of the first day of the month and the last day of the month. I am not sure how to do that. Any help would be appreciated. Thank you.

Answer 1

• You can get the beginning of the month easily by setting the day to 1.
• To get the end of the month conveniently, you can calculate the first day of the next month, then go back one day.
• Then set the time zone (tzinfo) to UTC to prevent Python using local time.
• Finally a call to .timestamp()

import datetime

def date_to_endofmonth(
    dt: datetime.datetime, offset: datetime.timedelta = datetime.timedelta(days=1)
) -> datetime.datetime:
    """
    Roll a datetime to the end of the month.

    Parameters
    ----------
    dt : datetime.datetime
        datetime object to use.
    offset : datetime.timedelta, optional.
        Offset to the next month. The default is 1 day; datetime.timedelta(days=1).

    Returns
    -------
    datetime.datetime
        End of month datetime.
    """
    # reset day to first day of month, add one month and subtract offset duration
    return (
        datetime.datetime(
            dt.year + ((dt.month + 1) // 12), ((dt.month + 1) % 12) or 12, 1
        )
        - offset
    )


year, month = 2022, 11

# make datetime objects; make sure to set UTC
dt_month_begin = datetime.datetime(year, month, 1, tzinfo=datetime.timezone.utc)
dt_month_end = date_to_endofmonth(dt_month_begin).replace(tzinfo=datetime.timezone.utc)

ts_month_begin = dt_month_begin.timestamp()
ts_month_end = dt_month_end.timestamp()

print(ts_month_begin, ts_month_end)
# 1667260800.0 1701302400.0

Answer 2

Unable to comment due to reputation but @FObersteiner is excellent just I would recommend a small change.

For example running the current code it would produce this for Nov 2022

print(dt_month_begin.timestamp())
print(dt_month_begin)
print(dt_month_end.timestamp())
print(dt_month_end)

--->

1667260800.0
2022-11-01 00:00:00+00:00
1701302400.0
2023-11-30 00:00:00+00:00

Note the year field

I'd suggest the following

import datetime

def date_to_endofmonth(
    dt: datetime.datetime, offset: datetime.timedelta = datetime.timedelta(seconds=1)
) -> datetime.datetime:
    """
    Roll a datetime to the end of the month.

    Parameters
    ----------
    dt : datetime.datetime
        datetime object to use.
    offset : datetime.timedelta, optional.
        Offset to the next month. The default is 1 second; datetime.timedelta(seconds=1).

    Returns
    -------
    datetime.datetime
        End of month datetime.
    """
    # reset day to first day of month, add one month and subtract offset duration
    return (
        datetime.datetime(
            dt.year + ((dt.month + 1) // 13), ((dt.month + 1) % 12) or 12, 1
        )
        - offset
    )


year, month = 2022, 11

# make datetime objects; make sure to set UTC
dt_month_begin = datetime.datetime(year, month, 1, tzinfo=datetime.timezone.utc)
dt_month_end = date_to_endofmonth(dt_month_begin).replace(tzinfo=datetime.timezone.utc)

Differences being floor division by 13 instead of 12 to handle the month of November.

Changing the offset to seconds delta because I felt the user ( and myself who came looking for the answer) wanted the starting epoch time and the ending epoch time so

Nov 1st 00:00:00 --> Nov 30th 23:59:59 would be better than Nov 1st 00:00:00 --> Nov 30th 00:00:00 ( Losing a day worth of seconds)

Output of the above would be:

1667260800.0
2022-11-01 00:00:00+00:00
1669852799.0
2022-11-30 23:59:59+00:00