为了评估服务台处理的故障单，我想知道故障单有多少工作时间。我可以轻松减去时间并获得总小时数。但是，应该计算的唯一时间是08:30至18:00。
例如：如果在11/23/2015 10:20创建了一个故障单，并在11/24/2015 17:20上完成了故障单，那么31＆＃39;正常＆＃39;几个小时过去了。我只对已经过的营业时间（8:30到18:00）感兴趣;在这种情况下16 hours and 30 minutes
答案 0 :(得分：3)
library(lubridate)
tickets <- data.frame(open = as.POSIXct(strptime(df$open, "%m/%d/%Y %H:%M")),
closed = as.POSIXct(strptime(df$closed, "%m/%d/%Y %H:%M"))
excludeDayCount <- Vectorize(function(open, close) {
# Check if the ticket was open and closed on the same day
if (identical(as.Date(open), as.Date(close))) return (0)
# All the holidays to be excluded need to be put here
holidays <- as.POSIXct(strptime(c("12/24/2015", "12/25/2015"),
"%m/%d/%Y"))
# Dates between open and close
day_seq <- floor_date(seq(open + days(1), close, by = "days"), "day")
# Count holidays / weekend days
return(sum(day_seq %in% holidays | wday(day_seq) %in% c(1,7)))
})
bizHrDiff <- function(open, close) {
# Hours from the end of one work day until the start of another
hours_between_days <- dhours(6) + dhours(8.5)
# Number of days to exclude
excl_days <- excludeDayCount(open, close)
# Number of days in include
reg_days <- as.integer(as.Date(close) - as.Date(open)) - excl_days
# Total duration between dates
span <- as.duration(interval(open, close))
# Remove the number of holidays and weekends
span <- span - ddays(excl_days)
# Remove out of office hours
span <- span - (reg_days * hours_between_days)
# Return in hours
return(time_length(span, unit = "hour"))
}
bizHrDiff(tickets$open, tickets$closed)
答案 1 :(得分：2)
I ran into similar needs recently and created a package based on @mvan 's script above.
https://github.com/janzzon/difftimeOffice
Input needs to be as POSIX time.
Output is in seconds.
Specify helpdesk open hours by argument working_hours = c(8.5, 18).
Additional holliday days is not implemented (yet).
答案 2 :(得分：2)
我已经尝试过@mvan和@Steffan Jansson的解决方案。不幸的是，他们都不能满足我的需求。第一个返回不正确的信息。后者是为了减缓我的要求，并没有考虑夏令时。
我创建了一个更快的功能，并将夏令时考虑在内。您可以指定营业时间和假期。
<强>用法
该函数接受5个参数。其中3个是可选的。
示例：
start <- as.POSIXct('2014-09-27 10:12:37', tz = 'Europe/Amsterdam')
end <- as.POSIXct('2016-12-10 20:04:18', tz = 'Europe/Amsterdam')
biz_hrs(start, end, '10:00', '17:00')
您也可以在dataframe列上运行它。确保正确格式化每个值。如果结束值早于起始值，则该函数返回NA。要更改该编辑行6
<强>代码
library(lubridate)
biz_hrs <- Vectorize(function(start, end, starting_time = '9:00', ending_time = '17:00', holidays = NULL){
if(end < start){
return(NA)
} else {
start_datetime <- as.POSIXct(paste0(substr(start,1,11), starting_time, ':00'))
end_datetime <- as.POSIXct(paste0(substr(end,1,11), ending_time, ':00'))
if(as.Date(start) == as.Date(end) & !as.Date(start) %in% holidays & !format(as.Date(start), "%u") %in% c(6,7)){ #if starting time stamp is on same day as ending time stamp and if day is not a holiday or weekend
if(start > start_datetime & end < end_datetime){ #if starting time stamp is later than start business hour and ending time stamp is earlier then ending business hour.
return(as.numeric(difftime(end, start), units = 'hours'))
} else if(start > start_datetime & end > end_datetime & start < end_datetime){ #if starting time stamp is later than end business hour and ending time stamp is earlier then ending business hour.
return(as.numeric(difftime(as.POSIXct(paste0(substr(start,1,11), ending_time, ':00')), start), units = 'hours'))
} else if(start < start_datetime & end < end_datetime & end > start_datetime){ #if starting time stamp is earlier than end business hour and ending time stamp is later than starting business hour.
return(as.numeric(difftime(end, start_datetime), units = 'hours'))
} else if(start > end_datetime & end > end_datetime){ #if starting time stamp is later than end business hour and ending time stamp is later than ending business hour.
return(0)
} else if(start < start_datetime & end < start_datetime){ #if starting time stamp is earlier than start business hour and ending time stamp is earlier than starting business hour.
return(0)
} else {
return(as.numeric(difftime(end_datetime, start_datetime), units = 'hours'))
}
} else { #if starting time stamp and ending time stamp occured on a different day.
business_hrs <- as.numeric(difftime(as.POSIXct(paste0('2017-01-01', ending_time, ':00')),
as.POSIXct(paste0('2017-01-01', starting_time, ':00')) #calculate business hours range by specified parameters
), units = 'hours')
start_day_hrs <- ifelse(start < as.POSIXct(paste0(substr(start,1,11), ending_time, ':00')) & !as.Date(start) %in% holidays & !format(as.Date(start), "%u") %in% c(6,7), #if start time stamp is earlier than specified ending time
as.numeric(difftime(as.POSIXct(paste0(substr(start,1,11), ending_time, ':00')), start), units = 'hours'), #calculate time between time stamp and specified ending time
0 #else set zero
) #calculate amount of time on starting day
start_day_hrs <- pmin(start_day_hrs, business_hrs) #cap the maximum amount of hours dertermined by the specified business hours
start_day_hrs
end_day_hrs <- ifelse(end > as.POSIXct(paste0(substr(end,1,11), starting_time, ':00')) & !as.Date(end) %in% holidays & !format(as.Date(end), "%u") %in% c(6,7), #if end time stamp is later than specified starting time
as.numeric(difftime(end, as.POSIXct(paste0(substr(end,1,11), starting_time, ':00'))), units = 'hours'), #calculate time between time stamp and specified starting time
0) #calculate amount of time on ending day
end_day_hrs <- pmin(end_day_hrs, business_hrs) #cap the maximum amount of hours dertermined by the specified business hours
days_between <- seq(as.Date(start), as.Date(end), by = 1) #create a vector of dates (from and up to including) the starting time stamp and ending time stamp
business_days <- days_between[!days_between %in% c(as.Date(start), as.Date(end)) & !days_between %in% holidays & !format(as.Date(days_between), "%u") %in% c(6,7)] #remove weekends and holidays from vector of dates
return(as.numeric(((length(business_days) * business_hrs) + start_day_hrs + end_day_hrs))) #multiply the remaining number of days in the vector (business days) by the amount of business hours and add hours from the starting and end day. Return the result
}
}
})