当营业时间根据R中的日期而变化时，如何计算两个日期之间的营业时间?

本文介绍了当营业时间根据R中的日期而变化时，如何计算两个日期之间的营业时间?的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我正在尝试计算两个日期之间的营业时间.营业时间因当天而异.

I'm trying to calculate business hours between two dates. Business hours vary depending on the day.

工作日有15个工作时间( 8:00-23:00 )，星期六和星期日有12个工作时间( 9:00-21:00 ).

Weekdays have 15 business hours (8:00-23:00), saturdays and sundays have 12 business hours (9:00-21:00).

例如:开始日期 07/24/2020 22:20 (星期五)和结束日期 07/25/2020 21:20 (星期六)，因为我我只对营业时间感兴趣，结果应该是 12.67 hours.

For example: start date 07/24/2020 22:20 (friday) and end date 07/25/2020 21:20 (saturday), since I'm only interested in the business hours the result should be 12.67hours.

以下是数据框和所需输出的示例:

Here an example of the dataframe and desired output:

start_date end_date business_hours 07/24/2020 22:20 07/25/2020 21:20 12.67 07/14/2020 21:00 07/16/2020 09:30 18.50 07/18/2020 08:26 07/19/2020 10:00 13.00 07/10/2020 08:00 07/13/2020 11:00 42.00

推荐答案

以下是您可以尝试使用 lubridate 的方法.我编辑了我认为可能会有所帮助的另一个功能.

Here is something you can try with lubridate. I edited another function I had I thought might be helpful.

首先在两个感兴趣的日期之间创建一个日期序列.然后根据营业时间创建时间间隔，检查每个日期是否在周末.

First create a sequence of dates between the two dates of interest. Then create intervals based on business hours, checking each date if on the weekend or not.

然后，钳位"使用 pmin 和 pmax 的允许的营业时间时间间隔开始和结束时间.

Then, "clamp" the start and end times to the allowed business hours time intervals using pmin and pmax.

您可以使用 time_length 来获取时间间隔的时间量度；总结一下，将为您提供总的使用时间.

You can use time_length to get the time measurement of the intervals; summing them up will give you total time elapsed.

library(lubridate) library(dplyr) calc_bus_hours <- function(start, end) { my_dates <- seq.Date(as.Date(start), as.Date(end), by = "day") my_intervals <- if_else(weekdays(my_dates) %in% c("Saturday", "Sunday"), interval(ymd_hm(paste(my_dates, "09:00"), tz = "UTC"), ymd_hm(paste(my_dates, "21:00"), tz = "UTC")), interval(ymd_hm(paste(my_dates, "08:00"), tz = "UTC"), ymd_hm(paste(my_dates, "23:00"), tz = "UTC"))) int_start(my_intervals[1]) <- pmax(pmin(start, int_end(my_intervals[1])), int_start(my_intervals[1])) int_end(my_intervals[length(my_intervals)]) <- pmax(pmin(end, int_end(my_intervals[length(my_intervals)])), int_start(my_intervals[length(my_intervals)])) sum(time_length(my_intervals, "hour")) } calc_bus_hours(as.POSIXct("07/24/2020 22:20", format = "%m/%d/%Y %H:%M", tz = "UTC"), as.POSIXct("07/25/2020 21:20", format = "%m/%d/%Y %H:%M", tz = "UTC")) [1] 12.66667

编辑:对于西班牙语，请使用 c(sábado"，"domingo")代替 c("Saturday"，星期天")).

Edit: For Spanish language, use c("sábado", "domingo") instead of c("Saturday", "Sunday")

对于数据框示例，您可以使用 mapply 将选定的两个列作为参数来调用该函数.试试:

For the data frame example, you can use mapply to call the function using the two selected columns as arguments. Try:

df$business_hours <- mapply(calc_bus_hours, df$start_date, df$end_date) start end business_hours 1 2020-07-24 22:20:00 2020-07-25 21:20:00 12.66667 2 2020-07-14 21:00:00 2020-07-16 09:30:00 18.50000 3 2020-07-18 08:26:00 2020-07-19 10:00:00 13.00000 4 2020-07-10 08:00:00 2020-07-13 11:00:00 42.00000