Question

在R中，我有3个数据帧，类似于我在下面提供的示例版本。第一个Data是主要数据集，TW和UW数据框具有与Data（MN-mapping_for_N）类似的变量，然后是1000个不同的值每个变量（N48）等我在这里为我的目的提供了3个。

Data<-matrix(c(4720,44.29,"Work or Private Clinic",N48,2659,55.05,"Hospital",N1,1612,59.99,"No Care",N48),ncol = 4,byrow=TRUE)
colnames(Data)<-c("studyid", "Pred_ex", "wherecare", "MN-mapping_for_N")
Data<-data.frame(Data)


TW<-matrix(c("N48",0.07,0.08,0.09,"N1",0.10,0.11,0.12,"N2",0.02,0.03,0.04,"N3",0.04,0.05,0.06),ncol = 4, byrow = TRUE)
colnames(TW)<-c("MN-mapping_for_N","draw1","draw2","draw3")`
TW<-data.frame(TW)

    UW<-matrix(c("N48",0.71,0.81,0.91,"N1",0.11,0.111,0.131,"N2",0.021,0.031,0.041,"N3",0.041,0.051,0.061),ncol = 4, byrow = TRUE)
colnames(UW)<-c("MN-mapping_for_N","draw1","draw2","draw3")`
UW<-data.frame(UW)

我的目标是使用UT和TW数据中随机选择的列创建一个新列，正确绘制的值取决于Data$wherecare中的值

我一直在使用dplyr和匹配功能的组合以及我自己创建的几个功能。目前看起来像

drawselect<-function(x) {
samplepick<-sample(2:1001,1)
select(x,1,num_range("draw",samplepick))
 }

DALY_FX_LT_NR<-function(x){
 draw_T_DW<-drawselect(TW)
  draw_UT_DW<-drawselect(UW)
  drawnames.TW<-colnames((draw_T_DW))
  drawnames.UT<-colnames(draw_UT_DW)
  UT.draw<-drawnames.UT[2]
  T.draw<-drawnames.T[2]
  print(UT.draw)       
  print(T.draw)
  newdf<-x %>% mutate(DW=NA)
  for(i in 1:nrow(newdf)){
if(newdf$wherecare[i]!= "No Care"){
  newdf$DW=draw_T_DW[,2][match(newdf$`MN-mapping_for_N`,draw_T_DW$`MN-mapping_for_N`)]
  next
}else if(newdf$wherecare[i]=="No Care"){
  newdf$DW=draw_UT_DW[,2][match(newdf$`MN-mapping_for_N`,draw_UT_DW$`MN-mapping_for_N`[i])]
}
 }
newdf
}

代码运行，但我似乎无法让它实际逐行迭代，以便从正确的数据框中拉出绘制值（即UT或TW。 drawselect功能）。

所以我看起来像：

-------------------------------------------------------------


studyid   Pred_ex        wherecare         MN-mapping_for_N     DW
--------- --------- ---------------------- ------------------ ------
  4720      44.29   Work or Private Clinic        N48          0.08

  2659      55.05          Hospital                N1          0.11

  1612      59.99          No Care                N48          0.08
--------------------------------------------------------------------

当我应该得到：

     studyid   Pred_ex        wherecare         MN-mapping_for_N    DW
    --------- --------- ---------------------- ------------------ ------
      4720      44.29   Work or Private Clinic        N48          0.08

      2659      55.05          Hospital                N1          0.11

      1612      59.99          No Care                N48          0.81
    --------------------------------------------------------------------

关键区别是右下角的0.81，样本数据不是很大，但实际数据是几百行长，所以我想让函数“正确决定”拉出哪个数据集从。此值可能为0.71,0.81或0.91，UT的任何N48值均可用。

最终目标是在计算中使用该值乘以Pred_ex列，我可以这样做，然后多次重新运行此函数来引导数据，但直到我能得到这些{{ 1}}语句正常工作我不能这样做。我也尝试使用if来匹配这些并且在条件语句不起作用时遇到了类似的问题。我认为dplyr::left_join函数的编写效果会更好，但我肯定会对任何事情持开放态度。

非常感谢任何帮助。

另外，感谢大家一般堆栈溢出，阅读其他问题的答案是我得到这个目标的主要原因。

Answer 1

因此，您不需要新功能（我保留drawselect，您可以执行以下操作：

for (i in 1:nrow(Data)){
    if (Data$wherecare[i] != "No Care"){
        Data$DW[i]<- drawselect(TW)[which(drawselect(TW)$MN.mapping_for_N == as.character(Data$MN.mapping_for_N[i])), 2]
    } else {
        Data$DW[i]<- drawselect(UW)[which(drawselect(UW)$MN.mapping_for_N == as.character(Data$MN.mapping_for_N[i])), 2]
    }
}

> Data
  studyid Pred_ex              wherecare MN.mapping_for_N   DW
1    4720   44.29 Work or Private Clinic              N48 0.08
2    2659   55.05               Hospital               N1 0.11
3    1612   59.99                No Care              N48 0.81

如果您想将所有内容都包装在一个函数中（包括drawselect），请尝试以下几行：

    DALY_FX_LT_NR<-function(x, y, z){ #x would be Data, y would be TW, z would be UW
 samplepick<-sample(2:(ncol(y)-1),1) 
 for (i in 1:nrow(x)){
    if (x$wherecare[i] != "No Care"){
        x$DW[i]<- y[which(y$MN.mapping_for_N==as.character(x$MN.mapping_for_N[i])), paste0("draw", samplepick)]
    } else {
        x$DW[i]<- z[which(z$MN.mapping_for_N==as.character(x$MN.mapping_for_N[i])), paste0("draw", samplepick)]
    }
  }
  return(x)
}

> DALY_FX_LT_NR(x = Data, y = TW, z = UW)
  studyid Pred_ex              wherecare MN.mapping_for_N   DW
1    4720   44.29 Work or Private Clinic              N48 0.09
2    2659   55.05               Hospital               N1 0.12
3    1612   59.99                No Care              N48 0.91

R-如何跨多个数据帧应用for循环？

1 个答案: