R Cut()错误-';中断';它们不是独一无二的

R Cut()错误-';中断';它们不是独一无二的,r,R,我有以下数据帧: a ID a.1 b.1 a.2 b.2 1 1 40.00 100.00 NA 88.89 2 2 100.00 100.00 100 100.00 3 3 50.00 100.00 75 100.00 4 4 66.67 59.38 NA 59.38 5 5 37.50 100.00 NA 100.00 6 6 100.

我有以下数据帧:

 a         
    ID   a.1    b.1     a.2   b.2
1    1  40.00   100.00  NA    88.89
2    2  100.00  100.00  100   100.00
3    3  50.00   100.00  75    100.00
4    4  66.67   59.38   NA    59.38
5    5  37.50   100.00  NA    100.00
6    6  100.00  100.00  100   100.00
 temp <- do.call(rbind,strsplit(names(df)[-1],".",fixed=TRUE))
 dup.temp <- temp[duplicated(temp[,1]),]

 res <- lapply(dup.temp[,1],function(i) {
 breaks <- c(-Inf,quantile(a[,paste(i,1,sep=".")], na.rm=T),Inf)
 cut(a[,paste(i,2,sep=".")],breaks)
 })
当我将以下代码应用于此数据帧时:

 a         
    ID   a.1    b.1     a.2   b.2
1    1  40.00   100.00  NA    88.89
2    2  100.00  100.00  100   100.00
3    3  50.00   100.00  75    100.00
4    4  66.67   59.38   NA    59.38
5    5  37.50   100.00  NA    100.00
6    6  100.00  100.00  100   100.00
 temp <- do.call(rbind,strsplit(names(df)[-1],".",fixed=TRUE))
 dup.temp <- temp[duplicated(temp[,1]),]

 res <- lapply(dup.temp[,1],function(i) {
 breaks <- c(-Inf,quantile(a[,paste(i,1,sep=".")], na.rm=T),Inf)
 cut(a[,paste(i,2,sep=".")],breaks)
 })
但是,相同的代码在类似的数据帧上工作得非常好:

 varnames<-c("ID", "a.1", "b.1", "c.1", "a.2", "b.2", "c.2")

 a <-matrix (c(1,2,3,4, 5, 6, 7), 2,7)

 colnames (a)<-varnames

 df<-as.data.frame (a)


    ID  a.1  b.1  c.1  a.2  b.2  c.2
  1  1    3    5    7    2    4    6
  2  2    4    6    1    3    5    7

 res <- lapply(dup.temp[,1],function(i) {
 breaks <- c(-Inf,quantile(a[,paste(i,1,sep=".")], na.rm=T),Inf)
 cut(a[,paste(i,2,sep=".")],breaks)
 })

 res
[[1]]
[1] (-Inf,3] (-Inf,3]
Levels: (-Inf,3] (3,3.25] (3.25,3.5] (3.5,3.75] (3.75,4] (4, Inf]

[[2]]
[1] (-Inf,5] (-Inf,5]
Levels: (-Inf,5] (5,5.25] (5.25,5.5] (5.5,5.75] (5.75,6] (6, Inf]

[[3]]
[1] (5.5,7] (5.5,7]
Levels: (-Inf,1] (1,2.5] (2.5,4] (4,5.5] (5.5,7] (7, Inf]

varnames之所以会出现此错误,是因为数据中
b.1
a.2
b.2
列的分位数值对于某些级别是相同的,因此它们不能直接用作函数
cut()
中的断点值

解决此问题的一种方法是将
quantile()
放入
unique()
函数中,这样您将删除所有非唯一的分位数值。如果分位数不是唯一的,这当然会减少断点

res <- lapply(dup.temp[,1],function(i) {
  breaks <- c(-Inf,unique(quantile(a[,paste(i,1,sep=".")], na.rm=T)),Inf)
  cut(a[,paste(i,2,sep=".")],breaks)
})

[[1]]
[1] <NA>        (91.7,100]  (58.3,91.7] <NA>        <NA>        (91.7,100] 
Levels: (-Inf,37.5] (37.5,42.5] (42.5,58.3] (58.3,91.7] (91.7,100] (100, Inf]

[[2]]
[1] (59.4,100]  (59.4,100]  (59.4,100]  (-Inf,59.4] (59.4,100]  (59.4,100] 
Levels: (-Inf,59.4] (59.4,100] (100, Inf]

res如果你想保持分位数的数量,另一个选择是只添加一点抖动,例如

breaks = c(-Inf,quantile(a[,paste(i,1,sep=".")], na.rm=T),Inf)
breaks = breaks + seq_along(breaks) * .Machine$double.eps

您可以使用.bincode而不是cut,它接受非唯一的中断向量。

如果您说的是十分位数、四分位数等,实际上是指人口的10%或25%,而不是十分位数/四分位数桶的实际数值,您可以先对值进行排序,并在列组上应用
分位数
函数:

a <- c(1,1,1,2,3,4,5,6,7,7,7,7,99,0.5,100,54,3,100,100,100,11,11,12,11,0)
a_ranks <- rank(a, ties.method = "first")
decile <- cut(a_ranks, quantile(a_ranks, probs=0:10/10), include.lowest=TRUE, labels=FALSE)  
a