R:匹配数据帧中的值

R:匹配数据帧中的值,r,list,dataframe,match,R,List,Dataframe,Match,我有以下数据帧列表: a<-data.frame( Data0=c("Y","Y","Y","Y","Y","Y","N","N","N","N","N","N"), Data1=c(16,18,19,20,21,50,16,18,19,20,21,50), Data2=c(2.2291,2.0743,1.9369,1.8148,1.7064,1.6102,2.2291,2.0743,1.9369,1.8148,1.7064,1.6102) ) b<-data.frame

我有以下数据帧列表:

a<-data.frame(
  Data0=c("Y","Y","Y","Y","Y","Y","N","N","N","N","N","N"),
  Data1=c(16,18,19,20,21,50,16,18,19,20,21,50),
  Data2=c(2.2291,2.0743,1.9369,1.8148,1.7064,1.6102,2.2291,2.0743,1.9369,1.8148,1.7064,1.6102)
)
b<-data.frame(
  Data0=c(-2 , 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 ,10 ,11) ,
  Data1=c(0.8891 ,0.8891,0.9051,1,0.8891,0.8891,0.7907,0.8891,0.9929,0.8891,0.8891,0.8891,0.8891)
)
dfl<-list(a,b)

好的,我终于想出了一个办法,但它不是那么优雅,所以我们将不胜感激

get_value<-function(vector){
  if (length(vector)==1) {
    from<-append(head(unlist(dfl[[2]][1],use.names=FALSE),-1),-9999,0)
    to  <-unlist(dfl[[2]][1],use.names=FALSE)

    match_from<-vector > from
    match_to<-vector <=  to

    pos<-which(match_from==match_to)
    return(unlist(dfl[[2]][pos,2]))
  }
  else{

    pos_1<-which(unlist(dfl[[1]][1],use.names=FALSE)==vector[1])

    from<-append(head(as.numeric(unlist(dfl[[1]][2],use.names=FALSE))[pos_1],-1),-9999,0)
    to  <-as.numeric(unlist(dfl[[1]][2],use.names=FALSE))[pos_1]

    match_from<-as.numeric(vector[2]) > from
    match_to  <-as.numeric(vector[2]) <=to
    pos<-pos_1[match_from == match_to]
    return(unlist(dfl[[1]][pos,3]))
  }
}

get_value使用
tidyverse
函数的替代方法如下:

library(tidyverse)
library(magrittr)

get_value<-function(vector){
  if (length(vector)==1) 
  { df <- dfl[[2]]
    k <- df %>%
      arrange(Data0)%>%
      filter(Data0 >= vector) %>%
      select(Data1) %>%
      head(1)
  }
  else
  {df <- dfl[[1]]
  k <- df %>%
    arrange(Data0)%>%
    filter(Data0 == vector[1]) %>%
    filter(Data1 >= vector[2]) %>%
    select(Data2) %>%
    head(1)
  }
  return(k)
}

我没有得到这个列表,因为你的数据帧是如此的不同。是否在函数中指定列表中要查找的数据框?或者函数是否需要全部尝试。在表中输入和输出一些示例将非常有用。如果您将“Y”18发送给a或“N”18会怎么样?你给b发什么,像6这样的数字?您是否也发送“a”或“b”?@Gregor:您可以事先指定需要哪个数据帧,但如果这有意义的话,我希望两个数据帧都使用相同的函数。“can”听起来是可选的,这增加了复杂性。请给出一个输入/输出表。另外,如果您已经解决了
b
的问题,请显示该代码,以便我们更好地了解您所做的工作@格雷戈:好的,两秒钟这似乎很合理。好吧,谢谢你!也许不漂亮,但我会采取合理的:)非常感谢!它看起来很干净!我会坐下来了解你今天晚些时候做了什么:)
get_value<-function(vector){
  if (length(vector)==1) {
    from<-append(head(unlist(dfl[[2]][1],use.names=FALSE),-1),-9999,0)
    to  <-unlist(dfl[[2]][1],use.names=FALSE)

    match_from<-vector > from
    match_to<-vector <=  to

    pos<-which(match_from==match_to)
    return(unlist(dfl[[2]][pos,2]))
  }
  else{

    pos_1<-which(unlist(dfl[[1]][1],use.names=FALSE)==vector[1])

    from<-append(head(as.numeric(unlist(dfl[[1]][2],use.names=FALSE))[pos_1],-1),-9999,0)
    to  <-as.numeric(unlist(dfl[[1]][2],use.names=FALSE))[pos_1]

    match_from<-as.numeric(vector[2]) > from
    match_to  <-as.numeric(vector[2]) <=to
    pos<-pos_1[match_from == match_to]
    return(unlist(dfl[[1]][pos,3]))
  }
}
library(tidyverse)
library(magrittr)

get_value<-function(vector){
  if (length(vector)==1) 
  { df <- dfl[[2]]
    k <- df %>%
      arrange(Data0)%>%
      filter(Data0 >= vector) %>%
      select(Data1) %>%
      head(1)
  }
  else
  {df <- dfl[[1]]
  k <- df %>%
    arrange(Data0)%>%
    filter(Data0 == vector[1]) %>%
    filter(Data1 >= vector[2]) %>%
    select(Data2) %>%
    head(1)
  }
  return(k)
}
> get_value(c(-44))
   Data1
1 0.8891
> get_value(c('N',16))
   Data2
1 2.2291
> get_value(c('Y',16))
   Data2
1 2.2291
> get_value(c('N',12))
   Data2
1 2.2291
> get_value(c('Y',11))
   Data2
1 2.2291
> get_value(c('Y',18))
   Data2
1 2.0743