用 "None" 替换具有 NA 因子水平的多个列

Question

我正在使用数据集 House Prices：Advanced Regression Techniques，其中包括多个因子变量，其水平之间具有 NA。考虑列 PoolQL、Alley 和 MiscFeatures。我想在一个函数中用 None 替换所有这些 NA，但我没有这样做。到目前为止试过这个：

MissingLevels <- function(x){
  for(i in names(x)){
  levels <- levels(x[i])
  levels[length(levels) + 1] <- 'None'
  x[i] <- factor(x[i], levels = levels)
  x[i][is.na(x[i])] <- 'None'
  return(x)
  }
}

MissingLevels(df[,c('Alley', 'Fence')])

apply(df[,c('Alley', 'Fence')], 2, MissingLevels)

https://www.kaggle.com/c/house-prices-advanced-regression-techniques/data

Answer 1

有几种方法例如：

x <- data.frame(another = 1:3, Alley = c("A", "B", NA), Fence = c("C", NA, NA))

选项 1：使用 forcats 包

x[,c("Alley", "Fence")] <- lapply(x[,c("Alley", "Fence")], fct_explicit_na, na_level = "None")

  another Alley Fence
1       1     A     C
2       2     B  None
3       3  None  None

选项 2：

x[,c("Alley", "Fence")] <- lapply(x[,c("Alley", "Fence")], function(x){`levels<-`(addNA(x), c(levels(x), "None"))})

PS：第二个答案灵感来自于@G。格洛腾迪克 post

用 "None" 替换具有 NA 因子水平的多个列

Replacing multiple columns with NA Factor Levels with "None"

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na

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