应用于分组数据中组的两个日期之间的工作日数

Question

我试图在分组df上使用gapply来获得项目时间输入的时间线。

下面我想得到一个专栏，将有一个人的可用工作时间，根据工作时间，从最早的日期，他们预订的时间，他们的时间。

library("dplyr")
library("stringr")
library("bizdays")
library("nlme")

time_df %>% dput()
structure(list(uID = c(2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L), hours = c(39, 
39, 39, 39, 19.5, 39, 31.2, 39, 39, 39, 39, 39, 39, 39, 39, 31.2, 
39, 39, 39, 39, 31.2, 39, 39, 39, 31.2, 39, 39, 39, 39, 39, 39, 
39, 39, 15.6, 39, 39, 39, 23.4, 39, 39, 23.4, 3.9, 3.9, 31.2, 
7.8, 3.9, 3.9, 3.9, 3.9, 3.9), onset = structure(c(16090, 16111, 
16097, 16083, 16125, 16104, 16076, 16118, 16139, 16216, 16209, 
16181, 16167, 16160, 16174, 16188, 16146, 16153, 16265, 16251, 
16223, 16244, 16258, 16230, 16237, 16307, 16363, 16328, 16349, 
16314, 16335, 16321, 16356, 16391, 16384, 16370, 16398, 16412, 
16377, 16405, 16433, 16139, 16160, 16153, 16209, 16251, 16272, 
16230, 16342, 16314), class = "Date"), terminus = c("2014-01-24", 
"2014-02-14", "2014-01-31", "2014-01-17", "2014-02-28", "2014-02-07", 
"2014-01-10", "2014-02-21", "2014-03-14", "2014-05-30", "2014-05-23", 
"2014-04-25", "2014-04-11", "2014-04-04", "2014-04-18", "2014-05-02", 
"2014-03-21", "2014-03-28", "2014-07-18", "2014-07-04", "2014-06-06", 
"2014-06-27", "2014-07-11", "2014-06-13", "2014-06-20", "2014-08-29", 
"2014-10-24", "2014-09-19", "2014-10-10", "2014-09-05", "2014-09-26", 
"2014-09-12", "2014-10-17", "2014-11-21", "2014-11-14", "2014-10-31", 
"2014-11-28", "2014-12-12", "2014-11-07", "2014-12-05", "2014-12-31", 
"2014-03-14", "2014-04-04", "2014-03-28", "2014-05-23", "2014-07-04", 
"2014-07-25", "2014-06-13", "2014-10-03", "2014-09-05")), row.names = c(NA, 
-50L), class = c("tbl_df", "tbl", "data.frame"))

#creating demo calendar to exclude weekends
create.calendar(name = "demo", weekdays = c("saturday","sunday"))

#function should calculate working hours between first time entry and last time entry
#ideally this will be applied to each group
timeentry = function(x){
  #creates an end_date variable from further end date in the group
  end_date = max(x$terminus)

  #creates a start_date from earliest start date in the group
  start_date = min(x$onset) 

  #returns weekdays between star and end 
  #then multiplies by 8 to get work hours
  start_date %>% bizdays(end_date, cal = "demo") * 8 
}

#group by uID and summarize
time_group = time_df %>% group_by(uID)
time_util = time_group %>% gapply(.,timeentry, which = c(onset,terminus))

Error in getGroups.data.frame(object, form, level) : 
  invalid formula for groups

我测试了我的功能，以确保它按预期工作。

> time_group %>% timeentry()
[1] 2056

> time_group$terminus %>% max()
[1] "2014-12-31"
> time_group$onset %>% min()
[1] "2014-01-06"

> bizdays("2014-01-06","2014-12-31",cal = "demo") * 8
[1] 1952

我不明白他们怎么能给出不同的结果。

我知道关于gapply和我所写的函数有一些基本的东西，但我并不真正理解。gapply说我应该得到一个数据输出。我想用我的原始数据加入这个输出，这样我就可以计算出人们的利用率。

任何想法都将不胜感激。

Answer 1

library(dplyr)
library(bizdays)

#create calendar
create.calendar(name = "demo", weekdays = c("saturday","sunday"))

#calculate working hour
time_df %>% 
  group_by(uID) %>% 
  summarise(onset_earliest  = min(onset), 
            terminus_latest = max(terminus), 
            working_hour    = bizdays(onset_earliest, terminus_latest, cal="demo") * 8)

输出是

# A tibble: 2 x 4
    uID onset_earliest terminus_latest working_hour
  <int> <date>         <chr>                  <dbl>
1     2 2014-01-06     2014-12-31              2056
2     5 2014-03-10     2014-10-03              1192

输入数据预览：

> head(time_df)
# A tibble: 6 x 4
    uID hours onset      terminus  
  <int> <dbl> <date>     <chr>     
1     2  39   2014-01-20 2014-01-24
2     2  39   2014-02-10 2014-02-14
3     2  39   2014-01-27 2014-01-31
4     2  39   2014-01-13 2014-01-17
5     2  19.5 2014-02-24 2014-02-28
6     2  39   2014-02-03 2014-02-07