找到每个患者最接近的匹配时间

时间:2017-02-01 14:54:41

标签: r dataframe time logic

我有两组数据:

第一集:

 patient<-c("A","A","B","B","C","C","C","C")
 arrival<-c("11:00","11:00","13:00","13:00","14:00","14:00","14:00","14:00")
 lastRow<-c("","Yes","","Yes","","","","Yes")

 data1<-data.frame(patient,arrival,lastRow)

另一组数据:

 patient<-c("A","A","A","A","B","B","B","C","C","C")
 availableSlot<-c("11:15","11:35","11:45","11:55","12:55","13:55","14:00","14:00","14:10","17:00")

 data2<-data.frame(patient, availableSlot)

我想创建为第一个数据集添加一列,以便对于每个患者的每个最后一行,它显示可用的插槽 最接近抵达时间:

结果将是:

  patient arrival lastRow availableSlot
       A   11:00        
       A   11:00     Yes     11:15
       B   13:00        
       B   13:00     Yes     12:55
       C   14:00        
       C   14:00        
       C   14:00        
       C   14:00     Yes     14:00

如果有人能告诉我如何在R中实现这一点,我将不胜感激。

2 个答案:

答案 0 :(得分:8)

我使用data.table,首先通过转换为ITime进行清理并忽略冗余行:

library(data.table)
setDT(data1)[, arrival := as.ITime(as.character(arrival))]
setDT(data2)[, availableSlot := as.ITime(as.character(availableSlot))]
DT1 = unique(data1, by="patient", fromLast=TRUE)

然后你可以做一个&#34;滚动加入&#34;:

res = data2[DT1, on=.(patient, availableSlot = arrival), roll="nearest", 
  .(patient, availableSlot = x.availableSlot)]

#    patient availableSlot
# 1:       A      11:15:00
# 2:       B      12:55:00
# 3:       C      14:00:00

工作原理

语法为x[i, on=, roll=, j]

  • on=是合并列。
  • 这是一个加入:对于i的每一行,我们正在寻找x中的匹配。
  • 使用roll="nearest"on=中的最后一列是&#34;滚动&#34;到最近的一场比赛。
  • 原始表格中的on=列可以使用x.*i.*前缀引用。
  • j参数应该提供列的列表,.()list()的别名。

http://r-datatable.com/Getting-started查看软件包的介绍材料,并输入?data.table以获取与滚动连接相关的文档。

我会停在res,但如果你真的想把它放回原来的桌子......

# a very nonstandard step:
data1[lastRow == "Yes", availableSlot := res$availableSlot ]

#    patient  arrival lastRow availableSlot
# 1:       A 11:00:00                  <NA>
# 2:       A 11:00:00     Yes      11:15:00
# 3:       B 13:00:00                  <NA>
# 4:       B 13:00:00     Yes      12:55:00
# 5:       C 14:00:00                  <NA>
# 6:       C 14:00:00                  <NA>
# 7:       C 14:00:00                  <NA>
# 8:       C 14:00:00     Yes      14:00:00

现在,data1在新列中有availableSlot,类似于data1$col <- val时的情况。

答案 1 :(得分:1)

以下是基于R基础的解决方案(基于joel.wilson's answer我的问题)

#Convert dates to POSIXct format
data1$arrival = as.POSIXct(data1$arrival, format = "%H:%M")
data2$availableSlot = as.POSIXct(data2$availableSlot, format = "%H:%M")

#Lookup times from data2$availableSlot closest to data1$arrival
data1$availableSlot = sapply(data1$arrival, function(x)
                    data2$availableSlot[which.min(abs(x - data2$availableSlot))])

#Keep just hour and minutes
data1$availableSlot = strftime(as.POSIXct(data1$availableSlot, 
                                origin = "1970-01-01"), format = "%H:%M")
data1$arrival = strftime(as.POSIXct(data1$arrival), format = "%H:%M")

#Remove times when lastrow is empty
data1$availableSlot[which(data1$lastRow != "Yes")] = ""