Python-算法查找时隙 [英] Python - Algorithm find time slots

问题描述

比方说，今天我的工作时间是09:00至18:00，我有3个约会：

Let's say today I work from 09:00 to 18:00, I have 3 appointments:

• 10:00-10： 30


• 12:00-13:00


• 15:30-17:10

我需要找到一天中1小时的可用时隙列表。

这是我应该获得的列表

I need to find a list of available time slots of 1 hour in the day.
Here is the list that I should get

• 09:00-10:00


• 10:30-11:30


• 13:00-14:00


• 14:00-15:00

• 09:00 - 10:00
• 10:30 - 11:30
• 13:00 - 14:00
• 14:00 - 15:00

我有已经在php中实现了该功能，我只是尝试将其放入python中。

这是我的尝试：

I have already implemented that in php and I just tried to put it in python.
Here is my try:

def get_slots(areas, duration):
    slots = []
    for area in areas:
        if area['start'] == area['end']:
            continue
        i = area['start']
        while (i + duration) <= area['end']:
            i += duration
            slots.append({
                'start': (i - duration),
                'end': i,
            })
    return slots

def get_areas(day_start, day_end, appts):
    areas = []
    old_end = day_start
    for appt in appts:
        if appt['start'] > old_end:
            areas.append({
                'start': old_end,
                'end': appt['start'],
            })
        old_end = appt['end']
        if old_end > day_end:
            return areas
    areas.append({
        'start': old_end,
        'end': day_end,
    })
    return areas

测试：

>>> day_start = datetime.datetime(2012, 5, 22, 9)
>>> day_end = datetime.datetime(2012, 5, 22, 18)
>>> appts = [{
    'start': datetime.datetime(2012, 5, 22, 10),
    'end': datetime.datetime(2012, 5, 22, 10, 30),
  },{
    'start': datetime.datetime(2012, 5, 22, 12),
    'end': datetime.datetime(2012, 5, 22, 13),
  },{
    'start': datetime.datetime(2012, 5, 22, 15, 30),
    'end': datetime.datetime(2012, 5, 22, 17, 10),
  },]
>>> duration = datetime.timedelta(hours=1)
>>> pprint.pprint(get_slots(get_areas(day_start, day_end, appts), duration))

但是我只是从php移植了代码。

所以我不确定这是一种Python方式。

It works, but I simply ported the code from php.
So I'm not sure it's a pythonist way to do it.

您能告诉我哪里吗？我可以改善吗？

Can you show me where I can improve ?

推荐答案

#time_slots.py
from datetime import datetime, timedelta

appointments = [(datetime(2012, 5, 22, 10), datetime(2012, 5, 22, 10, 30)),
                (datetime(2012, 5, 22, 12), datetime(2012, 5, 22, 13)),
                (datetime(2012, 5, 22, 15, 30), datetime(2012, 5, 22, 17, 10))]

hours = (datetime(2012, 5, 22, 9), datetime(2012, 5, 22, 18))

def get_slots(hours, appointments, duration=timedelta(hours=1)):
    slots = sorted([(hours[0], hours[0])] + appointments + [(hours[1], hours[1])])
    for start, end in ((slots[i][1], slots[i+1][0]) for i in range(len(slots)-1)):
        assert start <= end, "Cannot attend all appointments"
        while start + duration <= end:
            print "{:%H:%M} - {:%H:%M}".format(start, start + duration)
            start += duration

if __name__ == "__main__":
    get_slots(hours, appointments)


% python time_slots.py 
09:00 - 10:00
10:30 - 11:30
13:00 - 14:00
14:00 - 15:00

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