- 原因:使用 dateutil 的 rrule 时,计算速度比较慢
| def axx(): |
| from dateutil import rrule |
| received_time = datetime.datetime.strptime('2019-04-21 23:00:00', '%Y-%m-%d %H:%M:%S') |
| complete_time = datetime.datetime.strptime('2019-04-22 01:00:00', '%Y-%m-%d %H:%M:%S') |
| workdays = [x for x in range(7) if x not in [5, 6]] |
| time_period = rrule.rrule(rrule.MINUTELY, dtstart=received_time, until=complete_time, byweekday=workdays).count() |
| print(time_period) |
- 尝试使用 pandas 的 bdate_range,但是发现只统计工作日天数,即便不足 1 天也是按 1 天算的,不符合需求,因为我要分钟
| def xxa(): |
| import pandas as pd |
| date = pd.bdate_range('2019-04-21 23:00:00', '2019-04-22 01:00:00', freq='min') |
| minutes = len(date) |
| print(minutes) |
| print(minutes/(60*60)) |
- 从 stackoverflow 找到一个方法
| def xax(): |
| from business_duration import businessDuration |
| import pandas as pd |
| received_time = pd.to_datetime('2019-04-21 23:00:00') |
| complete_time = pd.to_datetime('2019-04-22 01:15:00') |
| period = businessDuration(received_time, complete_time, unit='min') |
| print(period) |
- 自己使用 pandas 写的,还需测试
| def aaa(): |
| import pandas as pd |
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| received_time = '2019-04-19 23:00:00' |
| complete_time = '2019-04-22 01:00:00' |
| received_date = pd.to_datetime(received_time) |
| complete_date = pd.to_datetime(complete_time) |
| date_period = pd.bdate_range(received_time, complete_time) |
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| if date_period[0] == date_period[-1]: |
| if date_period[0] > received_date: |
| start = date_period[0] |
| end = complete_date |
| else: |
| start = received_date |
| end = date_period[0] + datetime.timedelta(days=1) |
| day_time = len(pd.date_range(start, end, freq='min')) - 1 |
| print('Workdays:' + str(day_time) + ' minutes') |
| else: |
| if (complete_date - date_period[-1]).days > 0: |
| end = date_period[-1] + datetime.timedelta(days=1) |
| else: |
| end = complete_date |
| if received_date < date_period[0]: |
| start = date_period[0] |
| else: |
| start = received_date |
| received_per = pd.date_range(start, date_period[0] + datetime.timedelta(days=1), freq='min') |
| complete_per = pd.date_range(date_period[-1], end, freq='min') |
| middle_time = (len(date_period) - 2) * 1440 |
| days_time = len(received_per) + middle_time + len(complete_per) - 2 |
| print('Workdays:' + str(days_time) + ' minutes') |
参考:https://stackoverflow.com/questions/46899627/business-hours-between-two-dates-in-pandas-dataframe-including-holidays?rq=1