数据框列表转换因子的自定义函数

Question

我正在尝试根据我之前的问题创建一个函数 converting columns of list of data frame to factor

但是在尝试创建一个函数时，我得到了错误，没有得到想要的输出

# Sample Data Frame

data<-data.frame( col1=c(1,1,NA,NA,NA,NA,NA,NA,1,NA,NA,NA,NA,NA,NA,NA,NA,1,NA,NA,NA,1,1,1,NA,1,1,NA,NA,NA,NA,1,NA,NA,NA,NA,1,NA,1),
                  col2=c(1,1,1,1,1,NA,NA,NA,NA,1,1,1,1,1,NA,NA,NA,1,1,1,NA,1,1,1,1,1,NA,NA,NA,1,1,1,1,1,1,1,NA,NA,NA),
                  col3=c(1,1,NA,NA,NA,NA,NA,1,NA,NA,NA,NA,NA,NA,NA,NA,1,NA,NA,NA,NA,NA,1,1,1,NA,NA,NA,1,NA,NA,1,1,1,1,1,NA,NA,1),
                  col4=c(1,NA,NA,NA,NA,NA,NA,NA,NA,NA,1,NA,NA,NA,NA,NA,NA,NA,1,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA),
                  col5=c(1,2,1,1,1,2,1,2,2,1,2,NA,1,1,2,2,2,1,1,1,2,NA,2,1,1,1,2,2,2,NA,1,2,2,1,1,1,2,2,2)
)  

data$col5<-factor(data$col5, levels=c(1,2), labels=c("Local","Overseas"))

# creating a list of dataframes
df<- data
df$cc1<-1
df2<- subset(df, col5 == 'Local')
df$cc2<-ifelse(df$col5 == 'Local',1,NA)
lst<-list(df$cc1, df$cc2)
ldat<-list("ALL" = df, "Local" =df2)

###############

# Created function in separate file and sourcing it in my rmd file
#"sub_function"

fac_conv <- function(x) {
  paste0(
    "factor(", col_names[[x]],
    ",labels=c('",
    paste0(labels[[x]],
           collapse = "','"
    ), "'))"
  )
}


fac_conv_lapply <- function(data,col_names,labels) {
  list_of_fac <- purrr::map_chr(seq_len(length(col_names)),
                                fac_conv)
  ldat <- map(data, ~  {
    .x[col_names] <- map2(.x %>% 
                            dplyr::select(col_names), 
                          labels, ~ factor(.x, labels= .y))
    .x} )
  
}

# NOw I am applying my function like below in my rmd file

col_name <- c("col1","col2","col3","col4")
label <- c("Sales","Ops","admin","HR") 

fac_conv_lapply(data = ldat,col_names = col_name, labels = label )

错误： paste0("factor(", col_names[[x]], ",labels=c('", paste0(labels[[x]], ) 中的错误： 找不到对象“col_names”

Answer 1

为了让您的代码正常工作，将fac_conv 移动到fac_conv_lapply：

# Sample Data Frame
library(purrr)
library(dplyr)

data$col5<-factor(data$col5, levels=c(1,2), labels=c("Local","Overseas"))

df<- data
df$cc1<-1
df2<- subset(df, col5 == 'Local')
df$cc2<-ifelse(df$col5 == 'Local',1,NA)
lst<-list(df$cc1, df$cc2)
ldat<-list("ALL" = df, "Local" =df2)

fac_conv_lapply <- function(data,col_names,labels) {
  
  fac_conv <- function(x) {
    paste0(
      "factor(", col_names[[x]],
      ",labels=c('",
      paste0(labels[[x]],
             collapse = "','"
      ), "'))"
    )
  }
  
  list_of_fac <- purrr::map_chr(seq_len(length(col_names)),
                                fac_conv)
  ldat <- map(data, ~  {
    .x[col_names] <- map2(.x %>% 
                            dplyr::select(col_names), 
                          labels, ~ factor(.x, labels= .y))
    .x} )

  ldat
}

col_name <- c("col1","col2","col3","col4")
label <- c("Sales","Ops","admin","HR") 

ldat1 <- fac_conv_lapply(data = ldat, col_names = col_name, labels = label )

lapply(ldat1, head)
#> $ALL
#>    col1 col2  col3 col4     col5 cc1 cc2
#> 1 Sales  Ops admin   HR    Local   1   1
#> 2 Sales  Ops admin <NA> Overseas   1  NA
#> 3  <NA>  Ops  <NA> <NA>    Local   1   1
#> 4  <NA>  Ops  <NA> <NA>    Local   1   1
#> 5  <NA>  Ops  <NA> <NA>    Local   1   1
#> 6  <NA> <NA>  <NA> <NA> Overseas   1  NA
#> 
#> $Local
#>     col1 col2  col3 col4  col5 cc1
#> 1  Sales  Ops admin   HR Local   1
#> 3   <NA>  Ops  <NA> <NA> Local   1
#> 4   <NA>  Ops  <NA> <NA> Local   1
#> 5   <NA>  Ops  <NA> <NA> Local   1
#> 7   <NA> <NA>  <NA> <NA> Local   1
#> 10  <NA>  Ops  <NA> <NA> Local   1

^{由reprex package (v0.3.0) 于 2020 年 10 月 10 日创建}

但是，我首选的解决方案是将fac_conv 重写为两个参数函数

fac_conv <- function(x, y) {
  paste0(
    "factor(", x,
    ",labels=c('",
    paste0(y,
      collapse = "','"
    ), "'))"
  )

然后这样称呼它：

list_of_fac <- purrr::map2_chr(col_names, labels, fac_conv)

编辑不知道你的意思'