计算两个日期之间的星期几小时对 [英] Counting day-of-week-hour pairs between two dates

问题描述

考虑 24H 格式的星期几对以下列表：

Consider the following list of day-of-week-hour pairs in 24H format:

{
 'Mon': [9,23],
 'Thu': [12, 13, 14],
 'Tue': [11, 12, 14],
 'Wed': [11, 12, 13, 14]
 'Fri': [13],
 'Sat': [],
 'Sun': [],
}

和两个时间点，例如：

• 开始

datetime.datetime(2015, 7, 22, 17, 58, 54, 746784)

结束

End:

datetime.datetime(2015, 8, 30, 10, 22, 36, 363912)

说，我们需要知道上述两个数据时间（向上或向下舍入）中有多少个小时。

Say we need to know how many hours there are between these two datetimes (either rounding up or down) for each of the day-of-week-hour pairs specified above.

如何在Python中解决这个问题？我以一般的详细程度探索了 timedelta 和 relativedelta ，但我没有找到任何提供接近这个的东西。

How can I approach this problem in Python? I explored timedelta and relativedelta in a general level of detail but I didn't find anything that provides something close to this.

为简单起见，我们可以假设所有内容都是指相同的时区。

For simplicity, we can assume that everything refers to the same timezone.

也许一个更简单的问题是集中在一个单一的日间对，例如

Perhaps a simpler problem is to focus on a single day-hour pair, e.g. How many Wednesdays: 14 are there between two arbitrary datetimes?

推荐答案

几个星期三：14 可能是这样的：

from calendar import day_abbr
from datetime import datetime, timedelta


def solve(start, end, data):
    days = list(day_abbr)
    output = dict.fromkeys(days, 0)

    while start <= end:
        day = days[start.weekday()]
        if start.hour in data[day]:
            output[day] += 1
        start = start + timedelta(minutes=60)

    return output


data = {
 'Mon': [9, 23],
 'Thu': [12, 13, 14],
 'Tue': [11, 12, 14],
 'Wed': [11, 12, 13, 14],
 'Fri': [13],
 'Sat': [],
 'Sun': [],
}

start = datetime(2015, 7, 22, 17, 58, 54, 746784)
end = datetime(2015, 8, 30, 10, 22, 36, 363912)

print solve(start, end, data)
# {'Wed': 20, 'Sun': 0, 'Fri': 6, 'Tue': 15, 'Mon': 10, 'Thu': 18, 'Sat': 0} 

每小时计算一次：

from calendar import day_abbr
from collections import defaultdict
from datetime import datetime, timedelta
from pprint import pprint


def solve(start, end, data):
    days = list(day_abbr)
    output = defaultdict(lambda: defaultdict(int))

    while start <= end:
        day = days[start.weekday()]
        if start.hour in data[day]:
            output[day][start.hour] += 1
        start = start + timedelta(minutes=60)
    return {k: dict(v) for k, v in output.items()}


data = {
 'Mon': [9, 23],
 'Thu': [12, 13, 14],
 'Tue': [11, 12, 14],
 'Wed': [11, 12, 13, 14],
 'Fri': [13],
 'Sat': [],
 'Sun': [],
}

start = datetime(2015, 7, 22, 17, 58, 54, 746784)
end = datetime(2015, 8, 30, 10, 22, 36, 363912)

pprint(solve(start, end, data))
# output 
{'Fri': {13: 6},
 'Mon': {9: 5, 23: 5},
 'Thu': {12: 6, 13: 6, 14: 6},
 'Tue': {11: 5, 12: 5, 14: 5},
 'Wed': {11: 5, 12: 5, 13: 5, 14: 5}}

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