r-使用离散化后分类变量级别的标签

Question

我尝试使用arules包使用离散化函数转换变量。但输出的标签非常尴尬。有人能建议如何将这些标签转换成类似"；"低"中等"高"；或者简单地为1、2、3。

library(arules)
#> Warning: package 'arules' was built under R version 3.6.3
#> Loading required package: Matrix
#> 
#> Attaching package: 'arules'
#> The following objects are masked from 'package:base':
#> 
#>     abbreviate, write
discretize(iris[,1], breaks = 3)
#>   [1] [4.3,5.4) [4.3,5.4) [4.3,5.4) [4.3,5.4) [4.3,5.4) [5.4,6.3) [4.3,5.4)
#>   [8] [4.3,5.4) [4.3,5.4) [4.3,5.4) [5.4,6.3) [4.3,5.4) [4.3,5.4) [4.3,5.4)
#>  [15] [5.4,6.3) [5.4,6.3) [5.4,6.3) [4.3,5.4) [5.4,6.3) [4.3,5.4) [5.4,6.3)
#>  [22] [4.3,5.4) [4.3,5.4) [4.3,5.4) [4.3,5.4) [4.3,5.4) [4.3,5.4) [4.3,5.4)
#>  [29] [4.3,5.4) [4.3,5.4) [4.3,5.4) [5.4,6.3) [4.3,5.4) [5.4,6.3) [4.3,5.4)
#>  [36] [4.3,5.4) [5.4,6.3) [4.3,5.4) [4.3,5.4) [4.3,5.4) [4.3,5.4) [4.3,5.4)
#>  [43] [4.3,5.4) [4.3,5.4) [4.3,5.4) [4.3,5.4) [4.3,5.4) [4.3,5.4) [4.3,5.4)
#>  [50] [4.3,5.4) [6.3,7.9] [6.3,7.9] [6.3,7.9] [5.4,6.3) [6.3,7.9] [5.4,6.3)
#>  [57] [6.3,7.9] [4.3,5.4) [6.3,7.9] [4.3,5.4) [4.3,5.4) [5.4,6.3) [5.4,6.3)
#>  [64] [5.4,6.3) [5.4,6.3) [6.3,7.9] [5.4,6.3) [5.4,6.3) [5.4,6.3) [5.4,6.3)
#>  [71] [5.4,6.3) [5.4,6.3) [6.3,7.9] [5.4,6.3) [6.3,7.9] [6.3,7.9] [6.3,7.9]
#>  [78] [6.3,7.9] [5.4,6.3) [5.4,6.3) [5.4,6.3) [5.4,6.3) [5.4,6.3) [5.4,6.3)
#>  [85] [5.4,6.3) [5.4,6.3) [6.3,7.9] [6.3,7.9] [5.4,6.3) [5.4,6.3) [5.4,6.3)
#>  [92] [5.4,6.3) [5.4,6.3) [4.3,5.4) [5.4,6.3) [5.4,6.3) [5.4,6.3) [5.4,6.3)
#>  [99] [4.3,5.4) [5.4,6.3) [6.3,7.9] [5.4,6.3) [6.3,7.9] [6.3,7.9] [6.3,7.9]
#> [106] [6.3,7.9] [4.3,5.4) [6.3,7.9] [6.3,7.9] [6.3,7.9] [6.3,7.9] [6.3,7.9]
#> [113] [6.3,7.9] [5.4,6.3) [5.4,6.3) [6.3,7.9] [6.3,7.9] [6.3,7.9] [6.3,7.9]
#> [120] [5.4,6.3) [6.3,7.9] [5.4,6.3) [6.3,7.9] [6.3,7.9] [6.3,7.9] [6.3,7.9]
#> [127] [5.4,6.3) [5.4,6.3) [6.3,7.9] [6.3,7.9] [6.3,7.9] [6.3,7.9] [6.3,7.9]
#> [134] [6.3,7.9] [5.4,6.3) [6.3,7.9] [6.3,7.9] [6.3,7.9] [5.4,6.3) [6.3,7.9]
#> [141] [6.3,7.9] [6.3,7.9] [5.4,6.3) [6.3,7.9] [6.3,7.9] [6.3,7.9] [6.3,7.9]
#> [148] [6.3,7.9] [5.4,6.3) [5.4,6.3)
#> attr(,"discretized:breaks")
#> [1] 4.3 5.4 6.3 7.9
#> attr(,"discretized:method")
#> [1] frequency
#> Levels: [4.3,5.4) [5.4,6.3) [6.3,7.9]
table(discretize(iris[,1], breaks = 3))
#> 
#> [4.3,5.4) [5.4,6.3) [6.3,7.9] 
#>        46        53        51

Answer 1

如果我正确阅读了您的目标，您可以使用基本的cut函数做同样的事情。例如，

cut(iris$Sepal.Length, breaks = c(4.3, 5.4, 6.3, 7.9), labels = c('lo', 'med', 'hi'))

如果您想用切割替换值：

cuts <- cut(iris$Sepal.Length, breaks = c(4.3, 5.4, 6.3, 7.9), labels = c('lo', 'med', 'hi'))
iris$Sepal.Length <- cuts

只需将标签替换为自己的标签即可。

Answer 2

对于一列，您可以执行以下操作：

discretize(iris[,1], breaks = 3,labels=c(letters[1:3]))

对于data.frame，您使用default=参数传递命令：

discretizeDF(iris, default = list(method = "interval", breaks = 3,labels=1:3))

这些可以在帮助页面中提供的示例中找到。