因此,经过挖掘并尝试不同的事情,这里是我想出了什么和对我有用。这个提示有点令人费解,而详细的巴士显然给出了确切的结果。在试图找到答案时,我学习了矢量化的想法(赦免我的德语口音),因为生成矢量化代码可以将计算结果的时间缩短到大约3分钟,而在大约96小时之后我停止计算而没有完成。
请注意,记录日期列表(并非每个医生都会完成他的班次记录)是一个简单日期的Excel表格。记录的时间工作时间间隔列表是日期和时间,有人开始在一列中看到病人,并且在另一列中停止看到病人。下一行将是类似的开始和停止时间和日期。
所有在文本中的变量都在德国或者德国词的缩写,但我希望我的意见就足以理解发生了什么事情。此外,很多代码都是针对特定于我的情况的问题。
特别感谢用户PhiSeu,谁与解决方案的不同方面帮助我user3507085。
#read dates
package(lubridate)
Daten<-read.csv2(„file.csv")
#convert start dates to POSIX
Daten$Beginn<-parse_date_time(Daten$Beginn,"dmy HM",tz="CET")
#prevent overlap by adding one second
Daten$Beginn<-Daten$Beginn+1
#convert end dates to POSIX
Daten$Ende<-parse_date_time(Daten$Ende,"dmy HM",tz="CET")
#remove empty rows
Daten<-na.omit(Daten)
#create intervals in which people worked
Daten$Intervall<-interval(Daten$Beginn,Daten$Ende)
#read dates on which people worked
doku<-read.csv2(„dates.csv“,header=FALSE)
doku<-parse_date_time(doku$V1,"%d.%m.%Y",tz="cet")
#create a start time of 09 A.M. for shifts
doku<-data.frame(cbind(doku,doku+32400))
#add column names
names(doku)<-c("Datum","Beginn")
#convert to POSIX
doku$Datum<-as.POSIXct(doku$Datum,origin="1970-01-01",tz="cet")
doku$Beginn<-as.POSIXct(doku$Beginn,origin="1970-01-01",tz="cet")
#Loop to create 15 min intervals for each documented shift spanning 24 hour against which actual working hours will be checked
begin <- as.POSIXct(doku$Beginn)
# copy begin time for loop
begin_new <- begin
# create duration object
aufl <- duration(15, "mins")
# count times for loop
times <- 24*60/15
# create dataframe with begin time
Intervall <- data.frame(begin,stringsAsFactors = FALSE)
for (i in 1:times){
cat("test",i,"\n")
# save old time for interval calculation
begin_start <- begin_new
# add 15 Minutes to original time
begin_new <- begin_new + aufl
cat(begin_new,"\n")
# create an interval object between
new_dur <- interval(begin_start,begin_new)
# bind to original dataframe
Intervall <- cbind(Intervall,new_dur)
}
# Add column names
vec_names <- paste0("v",c(1:(times+1)))
colnames(Intervall) <- vec_names
#create a matrix of the number of seconds worked in each of the above 15 intervals by checking the amount of intersection between 15 intervals and documented intervals of work
test<-vector()
Tabelle<-matrix(nrow=length(doku$Beginn),ncol=times)
Tabelle[is.na(Tabelle)]<-0
for (j in 1:length(doku$Beginn)){
for (k in 1:times){
test<-as.duration(intersect(Daten$Intervall,Intervall[j,k+1]))
test[is.na(test)]<-0
test<-sum(test)
Tabelle[j,k]<-test}}
#cadd start time to the above matrix
Ausw<-data.frame(cbind(Tabelle,begin))
#convert to POSIX
Ausw$begin<-as.POSIXct(Ausw$begin,origin="1970-01-01",tz="cet")
##analysis of data
#common to all days of the week
#create labels for 15 min intervals
Labels<-c("09","09:15","09:30","09:45","10","10:15","10:30","10:45","11","11:15","11:30","11:45","12","12:15","12:30","12:45","13","13:15","13:30","13:45","14","14:15","14:30","14:45","15","15:15","15:30","15:45","16","16:15","16:30","16:45","17","17:15","17:30","17:45","18","18:15","18:30","18:45","19","19:15","19:30","19:45","20","20:15","20:30","20:45","21","21:15","21:30","21:45","22","22:15","22:30","22:45","23","23:15","23:30","23:45","00","00:15","00:30","00:45","01","01:15","01:30","01:45","02","02:15","02:30","02:45","03","03:15","03:30","03:45","04","04:15","04:30","04:45","05","05:15","05:30","05:45","06","06:15","06:30","06:45","07","07:15","07:30","07:45","08","08:15","08:30","08:45")
##analysis for weekends
#how many percent people worked on average in any of the 15 min intervals on a saturday or sunday
Wochenende<-apply(Ausw[Ausw$wtag==c(1,7),1:times],MARGIN=2,FUN=sum)
Prozent<-Wochenende/length(Ausw$begin[Ausw$wtag==c(1,7)]) /as.numeric(aufl)*100
#add labels
names(Prozent)<-Labels
#plot as barplot and add axis labels
b=barplot(Prozent,axes = F,axisnames=F,main="Durchschnittliche Arbeitsbelastung am Wochenende",sub="über 100%: Übergabezeiten",xlab="Uhrzeit",ylab="Prozent")
axis(1,at=c(b[seq(1,length(Labels),4)],b[length(b)]+diff(b)[1]),labels = c(Labels[seq(1,length(Labels),4)],"09"))
axis(2,at=seq(0,160,25),las=2)
##analysos monday to friday
Woche<-apply(Ausw[Ausw$wtag==c(2,3,4,5,6),1:times],MARGIN=2,FUN=sum)
Prozent2<-Woche/length(Ausw$begin[Ausw$wtag==c(2,3,4,5,6)]) /as.numeric(aufl)*100
#add labels
names(Prozent2)<-Labels
#plot as barplot and add axis labels
b2=barplot(Prozent2,axes = F,axisnames=F,main="Durchschnittliche Arbeitsbelastung Montag - Freitag",,xlab="Uhrzeit",ylab="Prozent“,ylim=c(0,100))
axis(1,at=c(b2[seq(1,length(Labels),4)],b2[length(b2)]+diff(b2)[1]),labels = c(Labels[seq(1,length(Labels),4)],"09"))
axis(2,at=seq(0,160,25),las=2)
请发表你的最好的尝试代码。谢谢。 – lrnzcig
你能提供一些示例数据吗? 'dput(head(Daten))'的输出将非常有用,因为它允许我们重现您的一些工作数据。 – jdobres