对于示例数据框:
df1 <- structure(list(X = 1:15, a = c(2L, 3L, 4L, 3L, 7L, 5L, NA, 2L,
9L, 7L, 0L, 1L, 20L, 15L, 14L)), .Names = c("X", "a"),
class = "data.frame", row.names = c(NA,
-15L))
我使用以下代码将列'a'分成四分位数:
cut.at.n.tile <- function(X , n = 4){
cut( X , breaks = quantile( X ,
probs = (0:n)/n , na.rm = TRUE ) , include.lowest = TRUE )}
df1$a.quartile <- cut.at.n.tile( df1$a , n = 4)
如何用1 - 4(1为最低)替换标签?我不希望简单地重新编码这些值,因为我将使用不同的连续变量运行很多次。
非常感谢任何帮助。
答案 0 :(得分:2)
将labels
参数用于cut
...
cut.at.n.tile <- function(X , n = 4){
cut( X , breaks = quantile( X ,
probs = (0:n)/n , na.rm = TRUE ) ,
labels = 1:n,
include.lowest = TRUE )}
cut.at.n.tile( df1$a , n = 4)
## [1] 1 2 2 2 3 3 <NA> 1 4 3
## 1 1 4 4 4
## Levels: 1 2 3 4
你可能也会对ggplot2::cut_number
感兴趣
基本上是一样的......
ggplot2::cut_number(df1$a, n = 4, labels = 1:4)
答案 1 :(得分:1)
您也可以在剪切的fcn内将标签设置为FALSE;即
cut(df1$a,
breaks = quantile( df1$a ,
probs = seq(0,1,.25) , na.rm = TRUE ),
labels = F, # setting labels to false just makes labels integer codes instead of factor levels
include.lowest = T)