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The basic idea here is that I want to measure the number of seconds between two python datetime objects. However, I only want to count hours between 8:00 and 17:00, as well as skipping weekends (saturday and sunday). This works, but I wondered if anyone had clever ideas to make it cleaner.

START_HOUR = 8
STOP_HOUR = 17
KEEP = (STOP_HOUR - START_HOUR)/24.0

def seconds_between(a, b):

    weekend_seconds = 0

    current = a
    while current < b:
        current += timedelta(days = 1)
        if current.weekday() in (5,6):
            weekend_seconds += 24*60*60*KEEP



    a_stop_hour = datetime(a.year, a.month, a.day, STOP_HOUR)
    seconds = max(0, (a_stop_hour - a).total_seconds())
    b_stop_hour = datetime(b.year, b.month, b.day, STOP_HOUR)
    if b_stop_hour > b:
        b_stop_hour = datetime(b.year, b.month, b.day-1, STOP_HOUR)

    seconds += (b - b_stop_hour).total_seconds()

    return (b_stop_hour - a_stop_hour).total_seconds() * KEEP + seconds - weekend_seconds
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12
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1. Issues

Your code fails in the following corner cases:

  1. a and b on the same day, for example:

    >>> a = datetime(2012, 11, 22, 8)
    >>> a.weekday()
    3          # Thursday
    >>> seconds_between(a, a + timedelta(seconds = 100))
    54100.0    # Expected 100
    
  2. a or b at the weekend, for example:

    >>> a = datetime(2012, 11, 17, 8)
    >>> a.weekday()
    5          # Saturday
    >>> seconds_between(a, a + timedelta(seconds = 100))
    21700.0    # Expected 0
    
  3. a after STOP_HOUR or b before START_HOUR, for example:

    >>> a = datetime(2012, 11, 19, 23)
    >>> a.weekday()
    0          # Monday
    >>> seconds_between(a, a + timedelta(hours = 2))
    28800.0    # Expected 0
    

Also, you count the weekdays by looping over all the days between the start and end of the interval. That means that the computation time is proportional to the size of the interval:

>>> from timeit import timeit
>>> a = datetime(1, 1, 1)
>>> timeit(lambda:seconds_between(a, a + timedelta(days=999999)), number=1)
1.7254137992858887

For comparison, in this extreme case the revised code below is about 100,000 times faster:

>>> timeit(lambda:office_time_between(a, a + timedelta(days=999999)), number=100000)
1.6366889476776123

The break even point is about 4 days:

>>> timeit(lambda:seconds_between(a, a + timedelta(days=4)), number=100000)
1.5806620121002197
>>> timeit(lambda:office_time_between(a, a + timedelta(days=4)), number=100000)
1.5950188636779785

2. Improvements

barracel's answer has two very good ideas, which I adopted:

  1. compute the sum in seconds rather than days;

  2. add up whole days and subtract part days if necessary.

and I made the following additional improvements:

  1. handle corner cases correctly;

  2. run in constant time regardless of how far apart a and b are;

  3. compute the sum as a timedelta object rather than an integer;

  4. move common code out into functions for clarity;

  5. docstrings!

3. Revised code

from datetime import datetime, timedelta

def clamp(t, start, end):
    "Return `t` clamped to the range [`start`, `end`]."
    return max(start, min(end, t))

def day_part(t):
    "Return timedelta between midnight and `t`."
    return t - t.replace(hour = 0, minute = 0, second = 0)

def office_time_between(a, b, start = timedelta(hours = 8),
                        stop = timedelta(hours = 17)):
    """
    Return the total office time between `a` and `b` as a timedelta
    object. Office time consists of weekdays from `start` to `stop`
    (default: 08:00 to 17:00).
    """
    zero = timedelta(0)
    assert(zero <= start <= stop <= timedelta(1))
    office_day = stop - start
    days = (b - a).days + 1
    weeks = days // 7
    extra = (max(0, 5 - a.weekday()) + min(5, 1 + b.weekday())) % 5
    weekdays = weeks * 5 + extra
    total = office_day * weekdays
    if a.weekday() < 5:
        total -= clamp(day_part(a) - start, zero, office_day)
    if b.weekday() < 5:
        total -= clamp(stop - day_part(b), zero, office_day)
    return total
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  • 2
    \$\begingroup\$ This code is broken for a couple of cases, for instance if you give it a Monday and Friday in the same week it gives a negative timedelta. The issue is with the calculation of extra but I don't have a solution yet. \$\endgroup\$ – radman Mar 16 '17 at 6:22
7
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I think the initial calculation between the two dates looks cleaner using a generator expression + sum. The posterior correction is easier to understand if you do the intersection of hours by thinking in seconds of the day

from datetime import datetime
from datetime import timedelta

START_HOUR = 8 * 60 * 60
STOP_HOUR = 17 * 60 * 60
KEEP = STOP_HOUR - START_HOUR

def seconds_between(a, b):
    days = (a + timedelta(x + 1) for x in xrange((b - a).days))
    total = sum(KEEP for day in days if day.weekday() < 5)

    aseconds = (a - a.replace(hour=0, minute=0, second=0)).seconds
    bseconds = (b - b.replace(hour=0, minute=0, second=0)).seconds

    if aseconds > START_HOUR:
        total -= min(KEEP, aseconds - START_HOUR)

    if bseconds < STOP_HOUR:
        total -= min(KEEP, STOP_HOUR - bseconds)

    return total
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1
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The code below is a bit of a hybrid between the two approaches mentioned. I think it should work for all scenarios. No work done outside working hours is counted.

from datetime import datetime
from datetime import timedelta

def adjust_hour_delta(t, start, stop):

    start_hour = start.seconds//3600
    end_hour = stop.seconds//3600
    zero = timedelta(0)

    if t - t.replace(hour = start_hour, minute = 0, second = 0) < zero:
        t = t.replace(hour = start_hour, minute = 0, second = 0)
    elif t - t.replace(hour = end_hour, minute = 0, second = 0) > zero:
        t = t.replace(hour = end_hour, minute = 0, second = 0)
    # Now get the delta
    delta = timedelta(hours=t.hour, minutes=t.minute, seconds = t.second)

    return delta    

def full_in_between_working_days(a, b):
    working = 0
    b = b - timedelta(days=1)
    while b.date() > a.date():
        if b.weekday() < 5:
            working += 1
        b = b - timedelta(days=1)
    return working

def office_time_between(a, b, start = timedelta(hours = 8),
                        stop = timedelta(hours = 17)):
    """
    Return the total office time between `a` and `b` as a timedelta
    object. Office time consists of weekdays from `start` to `stop`
    (default: 08:00 to 17:00).
    """
    zero = timedelta(0)
    assert(zero <= start <= stop <= timedelta(1))
    office_day = stop - start
    working_days = full_in_between_working_days(a, b)

    total = office_day * working_days
    # Calculate the time adusted deltas for the the start and end days
    a_delta = adjust_hour_delta(a, start, stop)
    b_delta = adjust_hour_delta(b, start, stop)


    if a.date() == b.date():
        # If this was a weekend, ignore
        if a.weekday() < 5:
            total = total + b_delta - a_delta
    else:
        # We now consider if the start day was a weekend
        if a.weekday() > 4:
            a_worked = zero
        else:
            a_worked = stop - a_delta
        # And if the end day was a weekend
        if b.weekday() > 4:
            b_worked = zero
        else:
            b_worked = b_delta - start
        total = total + a_worked + b_worked

    return total
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  • \$\begingroup\$ Testing this for these two dates, I get 1 working day when it should be 4: 2018-02-26 18:07:17 and 2018-03-04 14:41:04 \$\endgroup\$ – oxtay Mar 6 '18 at 18:52

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