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base::levels帮助文件https://stat.ethz.ch/R-manual/R-devel/library/base/html/levels.html包含以下修改变量级别的示例：

z <- gl(3, 2, 12, labels = c("apple", "salad", "orange"))
z
levels(z) <- c("fruit", "veg", "fruit")
z

假设这些东西位于数据框内：

mydata <- data.frame(z=gl(3, 2, 12, labels = c("apple", "salad", "orange")), n=1:12)

我想编写一个函数来转换将数据框和变量名称作为输入的级别：

modify_levels <- function(df,varname,from,to) {
  ### MAGIC HAPPENS
}

这样就modify_levels(mydata,z,from=c("apple","orange"),to="fruit")完成了转换的一部分（并且modify_levels(mydata,z,from=c("salad","broccoli"),to="veg")完成了第二部分，即使该级别broccoli可能不存在于我的数据集中）。

使用一些非标准的评估巫毒，我可以缩小我需要修改的内容：

where_are_levels <- function(df,varname,from,to,verbose=FALSE) {
  # input checks
  if ( !is.data.frame(df) ) {
    stop("df is not a data frame")
  }
  if ( !is.factor(eval(substitute(varname),df)) ) {
    stop("df$varname is not a factor")
  }
  if (verbose==TRUE) {
    cat("df$varname is",
      paste0(substitute(df),"$",substitute(varname)))
    cat(" which evaluates to:\n")
    print( eval(substitute(varname),df) )
  }
  if (length(to)!=1) {
    stop("Substitution is ambiguous")
  }
  # figure out what the cases are with the supplied source values
  for (val in from) {
    r <- (eval(substitute(varname),df) == val)
    if (verbose==TRUE) {
      print(r)
      cat( paste0(substitute(df),"$",substitute(varname)),"==",val)
      cat(": ",sum(r), "case(s)\n")
    }
  }
}

到目前为止，一切都很好（to尽管该选项没有任何作用）：

> where_are_levels(mydata,z,from=c("apple","orange"),to="",verbose=TRUE)

## df$varname is mydata$z which evaluates to:
## [1] apple  apple  salad  salad  orange orange apple  apple  salad  salad  orange orange
## Levels: apple salad orange
## [1]  TRUE  TRUE FALSE FALSE FALSE FALSE  TRUE  TRUE FALSE FALSE FALSE FALSE
## mydata$z == apple:  4 case(s)
## [1] FALSE FALSE FALSE FALSE  TRUE  TRUE FALSE FALSE FALSE FALSE  TRUE  TRUE
## mydata$z == orange:  4 case(s)

现在，对于下一步，我认为我需要做的是将目标变量的级别附加一个附加级别，并更改该变量的值。在互动工作中，我会

# to <- "fruit" # passed as a function argument
l1 <- levels(mydata$z)
levels(mydata$z) <- union(l1,to)
mydata[r,"z"] <- to

其中我只能在val循环中以编程方式获得第一行：

l1 <- levels(eval(substitute(varname),df))

这将在val循环内发生。

请注意，我想保留现有的苹果和橙子级别，而不是仅仅改变整个事情（如帮助文件中的大修示例中所做的那样）。

如果解决方案更容易通过dplyr从头开始编程来实现，那对我来说很好（尽管我的理解是带有它的 NSE 甚至dplyr比基础 R 中更核心）。

4

3 回答 3

3

不需要所有的替换，一个就足够了。我会保留你所有的信息

where_are_levels <- function(df,varname,from,to,verbose=FALSE) {
  # input checks
  varname <- substitute(varname)

  if (!is.data.frame(df)) {
    stop("df is not a data frame")
  }
  if (!is.factor(df[[varname]])) {
    stop("df$varname is not a factor")
  }
  if (verbose) {
    cat("df$varname is", paste0(substitute(df),"$",varname))
    cat(" which evaluates to:\n")
    print(df[[varname]])
  }
  if (length(to) != 1) {
    stop("Substitution is ambiguous")
  }
  # figure out what the cases are with the supplied source values
  r <- df[[varname]] %in% from
  new_levels <- union(levels(df[[varname]]), to)
  df[[varname]] <- factor(df[[varname]], new_levels)
  df[[varname]] <- replace(df[[varname]], r, to)
  if (verbose) {
    print(r)
    cat( paste0(df[[varname]]),"==",from)
    cat(": ",sum(r), "case(s)\n")
  }
  return(df)
}

where_are_levels(mydata,z,from=c("apple","orange"),to="fruit")
       z  n
1  fruit  1
2  fruit  2
3  salad  3
4  salad  4
5  fruit  5
6  fruit  6
7  fruit  7
8  fruit  8
9  salad  9
10 salad 10
11 fruit 11
12 fruit 12
于 2018-01-02T23:33:37.817 回答
2

我认为不需要非标准评估或任何 tidyverse 魔法。只需使用普通的“[[”和levels<-

modify_levels <- function(dfrm, cname, from=NA,to=NA) { 
                          pos <- which( from %in% levels(dfrm[[cname]]) )
                         levels(dfrm[[cname]])[pos] <- to
                          dfrm[[cname]]}  # be sure to assign the result back

利用：

> modify_levels(mydata,'z',from=c("salad","broccoli"),to="veg")
 [1] fruit fruit veg   veg   fruit fruit fruit fruit veg   veg   fruit fruit
Levels: fruit veg

但确实需要分配结果：

> mydata$z <- modify_levels(mydata,'z',from=c("salad","broccoli"),to="veg")
> mydata
       z  n
1  fruit  1
2  fruit  2
3    veg  3
4    veg  4
5  fruit  5
6  fruit  6
7  fruit  7
8  fruit  8
9    veg  9
10   veg 10
11 fruit 11
12 fruit 12
于 2018-01-02T23:44:05.323 回答
1

您可以将您的功能更改为：

where_are_levels<-function(mydata,varname,from, to, additional){
  mydata[[varname]]<-plyr::mapvalues(mydata[[varname]], from = from, to = to)
  mydata[[varname]]<-factor(mydata[[varname]],levels=c(levels(mydata[[varname]]),additional))
  return(mydata)
}

例子：

varname="z"
from = c("apple", "salad","orange")
to = c("fruit", "veg", "fruit")
additional="Milk"    
a<-where_are_levels(mydata,varname,from, to, additional)
于 2018-01-02T23:31:28.620 回答