Question

我有一个数据集，其中一些记录被分成两行（Bolus Type = Dual），我试图根据每个记录之间最多10分钟合并它们。这里有一个数据示例：

<!-- https://mvnrepository.com/artifact/xalan/serializer -->
    <dependency>
        <groupId>xalan</groupId>
        <artifactId>serializer</artifactId>
        <version>2.7.2</version>
    </dependency>

我目前使用的代码有效但运行速度太慢，无法在我的某些文件上运行（> 30MB）。我考虑对数据进行子集化，以便它只显示Time Bolus Type Bolus Volume 1 0.0 Dual (normal part only) 1 2 0.2 Dual (square part only) 2 3 0.4 Normal 3 4 0.6 Dual (normal part only) 2 5 0.8 Dual (square part only) 1 6 1.0 Normal 3包含“Dual *”的记录，但后来我不知道如何返回原始记录以使推注类型合并。

Bolus Type

编辑：期望的输出（根据当前代码输出）

#GENERATING SAMPLE DATA
a<-seq(0,1,length.out=6)
b<-c("Dual (normal part only)","Dual (square part only)","Normal","Dual (normal part only)","Dual (square part only)","Normal")
c<-c(1,2,3,2,1,3)
pt<-data.frame(a,b,c, stringsAsFactors = FALSE)
colnames(pt)<-c('Time', 'Bolus Type','Bolus Volume')

countDual=0
countSquare=0
countNormal=0
min10<-0.3

#FIND EACH "DUAL NORMAL" PART
for(i in 1:nrow(pt)) {
  if(!is.na(pt$`Bolus Type`[i])&&pt$`Bolus Type`[i]=="Dual (normal part only)"){
    j<-i
    time_lim<-pt$`Time`[i]+min10

    found_square<-0

    #LOOK AT ROWS AHEAD OF I TO FIND NEAREST SQUARE WITHIN TIME PERIOD (0.2)
    while(pt$`Time`[j]<time_lim&&found_square!=1){
      if(!is.na(pt$`Bolus Type`[j])&&pt$`Bolus Type`[j]=="Dual (square part only)"){
        date<-pt[i,"Date"]
        time<-pt[i,"Time"]
        total_DB<-pt[i,"Bolus Volume"]+pt[j,"Bolus Volume"]

        #ADD DUAL TOTAL ROW AT BOTTOM
        row_num<-nrow(pt) + 1
        pt[row_num,"Time"] = time
        pt[row_num,"Bolus Type"] = "Dual (total)"
        pt[row_num,"Bolus Volume"] = total_DB


        found_square<-1 #Exit loop when finds first square within 10 minutes
        countDual<-countDual+1

        #MARK THE "LINKED" RECORDS
        pt[j,"Bolus Type"]<-"Dual (square part)"
        pt[i,"Bolus Type"]<-"Dual (normal part)"

      }
      j<-j+1
    }
  }
}

#MESSAGE OUT RESULTS
countNormalOnly<-sum(pt$`Bolus Type`[!is.na(pt$`Bolus Type`)]=="Dual (normal part only)")
countSquareOnly<-sum(pt$`Bolus Type`[!is.na(pt$`Bolus Type`)]=="Dual (square part only)")
message(paste(c("Dual:",countDual," Square only:",countSquareOnly," Normal only:",countNormalOnly)))

Answer 1

library(dplyr)
library(stringr)
library(data.table)

pt_joined <-
  pt %>%
  filter(`Bolus Type` %like% 'Dual') %>%
  mutate(Index = cumsum(`Bolus Type` %like% 'normal')) %>%
  group_by(Index) %>%
  mutate(time_diff = max(Time - lag(Time), na.rm = TRUE)) %>%
  filter(time_diff <= .2) %>%
  mutate(`Bolus Type` = str_remove_all(`Bolus Type`, " only")) %>%
  ungroup %>% select(-time_diff)

pt_dual <- pt_joined %>%
  group_by(`Index`) %>%
  summarize(
    Time = min(Time),
    `Bolus Type` = 'Dual',
    `Bolus Volume` = sum(`Bolus Volume`)
  )

pt_new <-
  pt %>%
  left_join(pt_joined, by = c("Time", "Bolus Volume")) %>%
  mutate(`Bolus Type` =
           ifelse(is.na(`Bolus Type.y`), `Bolus Type.x`, `Bolus Type.y`)) %>%
  select(-contains(".")) %>%
  bind_rows(pt_dual) %>% select(-Index)

在R中组合链接数据行的最佳方法是什么？

1 个答案: