Question

在R中，我需要编写一个应用以下规则的函数：

fkt <- function(v1,v2){
    - find list name within listOfLists that contains v1       
    - go to column in df, where listname == colname  
    - return value from row where df$col2 == v2

}

例如：

df <- data.frame(Col1= c(1,2,3,4,5),
                   Col2= c("AAA","BBB","CCC","DDD","EEE"),
                   A = c("22","22","23","23","24"),
                   B = c("210","210","202","220","203"),
                   C = c("2000","2000","2010","2010","2200")
                   )



listOflists <- list(A <- c(1281, 1282, 1285, 1286, 1289),
                    B <- c(100,200,300,400,500,600,700,800,900,101,202,303,404,505,606),
                    C <-c(1000,1500,2000,2500,3000,3050,4000,4500,6000)
)

然后

fkt(800,"BBB") 
> 210

我尝试过

fkt<- function(v1,v2){
      r <- which(df$Col2== v1)
      s <- ifelse(v2 %in% A, df$A[r], 
                 ifelse( v2 %in% B ,df$A[r],df$C[r]))
  return(s)
}

A，结果是NA。而且编写许多ifelse（）语句效率不高-尤其是因为listOfLists可能包含50多个列表。

有人可以建议我如何如上所述以一种通用的，编程有效的方式编写此函数吗？

Answer 1

df <- data.frame(Col1= c(1,2,3,4,5),
               Col2= c("AAA","BBB","CCC","DDD","EEE"),
               A = c("22","22","23","23","24"),
               B = c("210","210","202","220","203"),
               C = c("2000","2000","2010","2010","2200"))
# Be cautious : = and <- are not equivalent, you were creating variables not named fields
listOflists <- list(A = c(1281, 1282, 1285, 1286, 1289),
                    B = c(100,200,300,400,500,600,700,800,900,101,202,303,404,505,606),
                    C = c(1000,1500,2000,2500,3000,3050,4000,4500,6000))

f <- function(v1) 
  for (i in 1:length(listOflists)) 
    if (v1 %in% listOflists[[i]]) return(names(listOflists)[i])

fkt <- function(v1,v2) df[df$Col2==v2,f(v1)]

fkt(800,"BBB")
#[1] 210
#Levels: 202 203 210 220

根据列表中的值从数据框中选择值的功能

1 个答案: