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Provides basic datetime related tasks used by the Horae resource planning system

Project Description

Introduction

horae.timeaware provides basic datetime related tasks used by the Horae resource planning system.

Dependencies

Horae

Third party

Guided user story

Let’s first make an object aware of time (holding multiple time entries)

>>> from horae.timeaware.timeaware import TimeAware, TimeEntry
>>> from datetime import datetime, timedelta
>>> entries = TimeAware()
>>> entry = TimeEntry()
>>> entry.date_start = datetime(2011, 1, 1, 8)
>>> entry.date_end = datetime(2011, 1, 1, 8)+timedelta(hours=5)
>>> entry.hours()
5.0
>>> entries.append(entry)
>>> entries.hours()
5.0
>>> entry = TimeEntry()
>>> entry.date_start = datetime(2011, 1, 1, 14)
>>> entry.date_end = datetime(2011, 1, 1, 14)+timedelta(hours=4, minutes=30)
>>> entry.hours()
4.5
>>> entries.append(entry)
>>> entries.hours()
9.5

Overlapping entries

Entries are always flattened before doing any calculations. Overlapping entries get combined.

>>> overlapping = TimeAware()
>>> entry = TimeEntry()
>>> entry.date_start = datetime(2011, 1, 1, 9)
>>> entry.date_end = datetime(2011, 1, 1, 9)+timedelta(hours=5)
>>> entry.hours()
5.0
>>> overlapping.append(entry)
>>> overlapping = entries + overlapping
>>> len(overlapping.entries())
1
>>> overlapping.hours()
10.5

Repeating time entries

Now let’s add some repeating time entries

Yearly

>>> from horae.timeaware import interfaces
>>> entry = TimeEntry()
>>> entry.date_start = datetime(2011, 2, 1, 8)
>>> entry.date_end = datetime(2011, 2, 1, 8)+timedelta(hours=5)
>>> entry.repeat = interfaces.REPEAT_YEARLY
>>> entry.entries()
Traceback (most recent call last):
...
ValueError: ('daterange', 'daterange required for infinite repeating events')
>>> entry.hours()
Traceback (most recent call last):
...
ValueError: ('daterange', 'daterange required for infinite repeating events')
>>> nonrepeating = entry.entries((datetime(2011, 2, 1), datetime(2012, 3, 1)))
>>> len(nonrepeating)
2
>>> nonrepeating[0].date_start == entry.date_start
True
>>> nonrepeating[0].date_end == entry.date_end
True
>>> nonrepeating[1].date_start == entry.date_start+timedelta(days=365)
True
>>> nonrepeating[1].date_end == entry.date_end+timedelta(days=365)
True

What about leap years?

>>> nonrepeating = entry.entries((datetime(2012, 3, 1), datetime(2013, 3, 1)))
>>> len(nonrepeating)
1
>>> nonrepeating[0].date_start == entry.date_start+timedelta(days=365)+timedelta(366)
True
>>> nonrepeating[0].date_end == entry.date_end+timedelta(days=365)+timedelta(366)
True

The entries returned are intersected with the provided daterange

>>> nonrepeating = entry.entries((datetime(2011, 2, 1, 9), datetime(2012, 2, 1, 11)))
>>> len(nonrepeating)
2
>>> nonrepeating[0].date_start == datetime(2011, 2, 1, 9)
True
>>> nonrepeating[0].hours()
4.0
>>> nonrepeating[1].date_end == datetime(2012, 2, 1, 11)
True
>>> nonrepeating[1].hours()
3.0

Now let’s have a look at the other repeating possibilities

Monthly

>>> entry.repeat = interfaces.REPEAT_MONTHLY
>>> nonrepeating = entry.entries((datetime(2011, 2, 1), datetime(2012, 3, 1)))
>>> len(nonrepeating)
13
>>> [(e.date_start.year, e.date_start.month) for e in nonrepeating]
[(2011, 2), (2011, 3), (2011, 4), (2011, 5), (2011, 6), (2011, 7), (2011, 8), (2011, 9), (2011, 10), (2011, 11), (2011, 12), (2012, 1), (2012, 2)]

Weekly

>>> entry.repeat = interfaces.REPEAT_WEEKLY
>>> nonrepeating = entry.entries((datetime(2011, 2, 1), datetime(2011, 3, 1)))
>>> len(nonrepeating)
4
>>> [(e.date_start.month, e.date_start.day) for e in nonrepeating]
[(2, 1), (2, 8), (2, 15), (2, 22)]

On work days

>>> entry.repeat = interfaces.REPEAT_WORKDAYS
>>> nonrepeating = entry.entries((datetime(2011, 2, 1), datetime(2011, 2, 13)))
>>> len(nonrepeating)
9
>>> [(e.date_start.month, e.date_start.day) for e in nonrepeating]
[(2, 1), (2, 2), (2, 3), (2, 4), (2, 7), (2, 8), (2, 9), (2, 10), (2, 11)]

Daily

>>> entry.repeat = interfaces.REPEAT_DAILY
>>> nonrepeating = entry.entries((datetime(2011, 2, 1), datetime(2011, 2, 10)))
>>> len(nonrepeating)
9
>>> [(e.date_start.month, e.date_start.day) for e in nonrepeating]
[(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (2, 7), (2, 8), (2, 9)]

Adding and subtracting

Combining time aware objects is possible by adding them

>>> entries2 = TimeAware()
>>> entry = TimeEntry()
>>> entry.date_start = datetime(2011, 1, 2, 7)
>>> entry.date_end = datetime(2011, 1, 2, 12, 30)
>>> entries2.append(entry)
>>> entry = TimeEntry()
>>> entry.date_start = datetime(2011, 1, 2, 14)
>>> entry.date_end = datetime(2011, 1, 2, 17, 30)
>>> entries2.append(entry)
>>> len(entries2.entries())
2
>>> entries2.hours()
9.0
>>> len(entries.entries())
2
>>> entries.hours()
9.5
>>> combined = entries + entries2
>>> len(combined.entries())
4
>>> combined.hours()
18.5

Subtracting is done similarly but is only possible for time aware objects containing no infinitely repeating entries otherwise the subtract method has to be used and passed a daterange and returns a list of time entries

>>> entry = TimeEntry()
>>> entry.date_start = datetime(2011, 1, 4, 8)
>>> entry.date_end = datetime(2011, 1, 4, 18, 30)
>>> entry2 = TimeEntry()
>>> entry2.date_start = datetime(2011, 1, 4, 12)
>>> entry2.date_end = datetime(2011, 1, 4, 18, 30)
>>> subtracted = entry - entry2
>>> len(subtracted)
1
>>> subtracted[0].date_start == entry.date_start
True
>>> subtracted[0].date_end == entry2.date_start
True
>>> entry2 = TimeEntry()
>>> entry2.date_start = datetime(2011, 1, 4, 8)
>>> entry2.date_end = datetime(2011, 1, 4, 12, 30)
>>> subtracted = entry - entry2
>>> len(subtracted)
1
>>> subtracted[0].date_start == entry2.date_end
True
>>> subtracted[0].date_end == entry.date_end
True
>>> subtracted = entry2 - entry
>>> len(subtracted)
0
>>> entry3 = TimeEntry()
>>> entry3.date_start = datetime(2011, 1, 4, 10, 15)
>>> entry3.date_end = datetime(2011, 1, 4, 14)
>>> subtracted = entry - entry3
>>> len(subtracted)
2
>>> subtracted[0].date_start == entry.date_start
True
>>> subtracted[0].date_end == entry3.date_start
True
>>> subtracted[1].date_start == entry3.date_end
True
>>> subtracted[1].date_end == entry.date_end
True

Now let’s try to subtract from a repeating entry

>>> entry.repeat = interfaces.REPEAT_DAILY
>>> entry - entry3
Traceback (most recent call last):
...
ValueError: ('daterange', 'daterange required for infinite repeating events')
>>> subtracted = entry.subtract(entry3, (datetime(2011, 1, 4), datetime(2011, 1, 6)))
>>> len(subtracted)
3
>>> subtracted[0].date_start == entry.date_start
True
>>> subtracted[0].date_end == entry3.date_start
True
>>> subtracted[1].date_start == entry3.date_end
True
>>> subtracted[1].date_end == entry.date_end
True
>>> subtracted[2].date_start == entry.date_start+timedelta(days=1)
True
>>> subtracted[2].date_end == entry.date_end+timedelta(days=1)
True

What about subtracting repeating entries from repeating entries?

>>> entry3.repeat = interfaces.REPEAT_DAILY
>>> subtracted = entry.subtract(entry3, (datetime(2011, 1, 4), datetime(2011, 1, 6)))
>>> len(subtracted)
4
>>> subtracted[0].date_start == entry.date_start
True
>>> subtracted[0].date_end == entry3.date_start
True
>>> subtracted[1].date_start == entry3.date_end
True
>>> subtracted[1].date_end == entry.date_end
True
>>> subtracted[2].date_start == entry.date_start+timedelta(days=1)
True
>>> subtracted[2].date_end == entry3.date_start+timedelta(days=1)
True
>>> subtracted[3].date_start == entry3.date_end+timedelta(days=1)
True
>>> subtracted[3].date_end == entry.date_end+timedelta(days=1)
True

Subtracting from time awares containing multiple time entries is also possible

>>> entries = TimeAware()
>>> entry = TimeEntry()
>>> entry.date_start = datetime(2011, 1, 5, 8, 15)
>>> entry.date_end = datetime(2011, 1, 5, 12, 30)
>>> entries.append(entry)
>>> entry = TimeEntry()
>>> entry.date_start = datetime(2011, 1, 5, 14)
>>> entry.date_end = datetime(2011, 1, 5, 18, 30)
>>> entries.append(entry)
>>> entries.hours()
8.75
>>> entries2 = TimeAware()
>>> entry = TimeEntry()
>>> entry.date_start = datetime(2011, 1, 5, 12)
>>> entry.date_end = datetime(2011, 1, 5, 14, 30)
>>> entries2.append(entry)
>>> entry = TimeEntry()
>>> entry.date_start = datetime(2011, 1, 5, 16)
>>> entry.date_end = datetime(2011, 1, 5, 17)
>>> entries2.append(entry)
>>> entries2.hours()
3.5
>>> subtracted = entries - entries2
>>> subtracted.hours()
6.75
>>> entries = subtracted.entries()
>>> len(entries)
3
>>> entries[0].date_start == datetime(2011, 1, 5, 8, 15)
True
>>> entries[0].date_end == datetime(2011, 1, 5, 12)
True
>>> entries[1].date_start == datetime(2011, 1, 5, 14, 30)
True
>>> entries[1].date_end == datetime(2011, 1, 5, 16)
True
>>> entries[2].date_start == datetime(2011, 1, 5, 17)
True
>>> entries[2].date_end == datetime(2011, 1, 5, 18, 30)
True

Now for a slightly more realistic example. Let’s say John is working from 8.00 to 12.00 and from 13.30 to 18.00 on all work days.

>>> worktime = TimeAware()
>>> morning = TimeEntry()
>>> morning.date_start = datetime(2011, 1, 3, 8)
>>> morning.date_end = datetime(2011, 1, 3, 12)
>>> morning.repeat = interfaces.REPEAT_WORKDAYS
>>> worktime.append(morning)
>>> afternoon = TimeEntry()
>>> afternoon.date_start = datetime(2011, 1, 3, 13, 30)
>>> afternoon.date_end = datetime(2011, 1, 3, 18)
>>> afternoon.repeat = interfaces.REPEAT_WORKDAYS
>>> worktime.append(afternoon)

Let’s see how many hours John is working during one day and through the whole 2011.

>>> worktime.hours((datetime(2011, 1, 3), datetime(2011, 1, 4)))
8.5
>>> worktime.hours((datetime(2011, 1, 1), datetime(2012, 1, 1)))
2210.0

Now let’s say every 4th Friday he takes one day off. Additionally he is on vacation from February 21st 2011 till 13th of March.

>>> absence = TimeAware()
>>> friday = TimeEntry()
>>> friday.date_start = datetime(2011, 1, 7)
>>> friday.date_end = datetime(2011, 1, 8)
>>> friday.repeat = interfaces.REPEAT_4WEEKS
>>> absence.append(friday)
>>> vacation = TimeEntry()
>>> vacation.date_start = datetime(2011, 2, 21)
>>> vacation.date_end = datetime(2011, 3, 13)
>>> absence.append(vacation)

Let’s see what happens to the hours John is working in 2011 which should result in a reduction of the 13 Fridays and 15 days of the vacation where one of them is one of the Fridays already counted. Which leads us to 27 days multiplied by 8.5 hours resulting in 229.5 hours and thus 1980.5 total work hours in 2011.

>>> effective_worktime = worktime.subtract(absence, (datetime(2011, 1, 1), datetime(2012, 1, 1)))
>>> effective_worktime.hours()
1980.5

Changelog

1.0a1 (2012-01-16)

  • Initial release
Release History

Release History

This version
History Node

1.0a1

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