我正在尝试使用R函数quantcut()将数值变量重新编码为具有与分位数对应的级别的因子。例如:
> X
[1] 6 4 9 6 1 2 5 3 5 7 10 7 2 7 7 5 6 6 3 4 6 4 2 7 6 7
[27] 4 3 5 3 7 6 8 12 4 4 0 1 7 6 7 4 7 1 1 1 2 3 3 1 1 6
[53] 5 3 1 1 1 3 3 3 1 1 3 1 1 1 3 3 0 1 3 1 8 5 3 0 0 2
[79] 1 3 8 0 1 4 1 1 1 1 1 1 3 2 1 4 1 5 5 12 7 2 6 6 2 6
[105] 0 1 4 1 4 0 7 3 2 1 1 8 5 5 3 0 5 6 2 4 2 2 2 6 4 2
[131] 2 2 2 6 8 5 1 2 8 3 2 7 4 6 6 6 7 5 1 5 5 6 1 4 4 5
[157] 6 2 4 7 2 4 10 6 3 5 2 2 6 6 2 4 5 7 4 5 11 6 6 8 2 4
[183] 4 6 12 16 9 7 14 13 11 5 5 2 2 7 7 6 4 3 4 3 5 4 5 7 9 4
[209] 3 12 4 4 4 8 7 6 1 3 6 7 5 5 6 9 6 4 7 8 5 6 3 6 4 7
[235] 3 3 4 7 5 7 5 9 5 8 3 4 3 2 5 2 4 3 8 4 2 2 1 5 3 5
[261] 8 5 6 4 5 1 1 2 6 2 7 2 4 4 3 3 4 10 5 6 10 2 5 5 0 1
[287] 6 2 5 4 6 6 9 5 5 6 3 8 1 5 1 8 5 2 5 2 4 2 4 4
bins=10
labels = 1:bins
library(gtools)
x2 = quantcut(X, q = seq(0, 1, by=1/bins), labels=labels)
我收到错误:“cut.default中的错误(x [!flag],breaks = newquant,include.lowest = TRUE,: “break”不是唯一的。“我认为这是因为在分位数中存在联系,但是quantcut的文档特别显示了函数如何通过使用更少的间隔来处理关系的示例。无论我是否指定了错误,都会发生错误。标签参数。
非常感谢任何建议。
编辑:这是输入变量X的代码:
X = c(6L, 4L, 9L, 6L, 1L, 2L, 5L, 3L, 5L, 7L, 10L, 7L, 2L, 7L, 7L,
5L, 6L, 6L, 3L, 4L, 6L, 4L, 2L, 7L, 6L, 7L, 4L, 3L, 5L, 3L, 7L,
6L, 8L, 12L, 4L, 4L, 0L, 1L, 7L, 6L, 7L, 4L, 7L, 1L, 1L, 1L,
2L, 3L, 3L, 1L, 1L, 6L, 5L, 3L, 1L, 1L, 1L, 3L, 3L, 3L, 1L, 1L,
3L, 1L, 1L, 1L, 3L, 3L, 0L, 1L, 3L, 1L, 8L, 5L, 3L, 0L, 0L, 2L,
1L, 3L, 8L, 0L, 1L, 4L, 1L, 1L, 1L, 1L, 1L, 1L, 3L, 2L, 1L, 4L,
1L, 5L, 5L, 12L, 7L, 2L, 6L, 6L, 2L, 6L, 0L, 1L, 4L, 1L, 4L,
0L, 7L, 3L, 2L, 1L, 1L, 8L, 5L, 5L, 3L, 0L, 5L, 6L, 2L, 4L, 2L,
2L, 2L, 6L, 4L, 2L, 2L, 2L, 2L, 6L, 8L, 5L, 1L, 2L, 8L, 3L, 2L,
7L, 4L, 6L, 6L, 6L, 7L, 5L, 1L, 5L, 5L, 6L, 1L, 4L, 4L, 5L, 6L,
2L, 4L, 7L, 2L, 4L, 10L, 6L, 3L, 5L, 2L, 2L, 6L, 6L, 2L, 4L,
5L, 7L, 4L, 5L, 11L, 6L, 6L, 8L, 2L, 4L, 4L, 6L, 12L, 16L, 9L,
7L, 14L, 13L, 11L, 5L, 5L, 2L, 2L, 7L, 7L, 6L, 4L, 3L, 4L, 3L,
5L, 4L, 5L, 7L, 9L, 4L, 3L, 12L, 4L, 4L, 4L, 8L, 7L, 6L, 1L,
3L, 6L, 7L, 5L, 5L, 6L, 9L, 6L, 4L, 7L, 8L, 5L, 6L, 3L, 6L, 4L,
7L, 3L, 3L, 4L, 7L, 5L, 7L, 5L, 9L, 5L, 8L, 3L, 4L, 3L, 2L, 5L,
2L, 4L, 3L, 8L, 4L, 2L, 2L, 1L, 5L, 3L, 5L, 8L, 5L, 6L, 4L, 5L,
1L, 1L, 2L, 6L, 2L, 7L, 2L, 4L, 4L, 3L, 3L, 4L, 10L, 5L, 6L,
10L, 2L, 5L, 5L, 0L, 1L, 6L, 2L, 5L, 4L, 6L, 6L, 9L, 5L, 5L,
6L, 3L, 8L, 1L, 5L, 1L, 8L, 5L, 2L, 5L, 2L, 4L, 2L, 4L, 4L)
答案 0 :(得分:2)
好的,问题可以追溯到这里,正如你所说,70%和80%的分位数是相同的。 quantile
quantcut
quantile(X,probs=seq(0,1,0.1))
0% 10% 20% 30% 40% 50% 60% 70% 80% 90% 100%
0.0 1.0 2.0 3.0 3.6 4.0 5.0 6.0 6.0 8.0 16.0
我无法看到如何使用quantcut
本身来解决此问题,但您可以始终只使用cut
和quantile
以及unique
来对其进行排序。从我所知道的情况来看,无论如何,这都是quantcut
在内部所做的事情。
result <- cut(X,unique(quantile(X,probs=seq(0,1,0.1))),include.lowest=TRUE)
> result[2:10]
[1] (3.6,4] (8,16] (5,6] [0,1] (1,2] (4,5] (2,3] (4,5] (6,8]
#Levels: [0,1] (1,2] (2,3] (3,3.6] (3.6,4] (4,5] (5,6] (6,8] (8,16]
> X[2:10]
[1] 4 9 6 1 2 5 3 5 7