如何在间隔中打破数字

Question

我正在尝试将信用评分降低到范围内，以便为其分配评级。

我试过这个：

x <- sample(510:780, 50, replace=TRUE)
cut(x, breaks=c(300, 579, 669, 739, 799), include.lowest=TRUE)

结果：

[300,579] (579,669] (669,739] (739,799]

但我想要这样的东西：

(300, 579] (580, 669] (670, 739] (740,799] (739,799]

类似于这里的范围： https : //www.experian.com/blogs/ask-experian/credit-education/score-basics/what-is-a-good-credit-score/

Answer 1

也许你可以试试下面的代码：

bks <- c(300, 579, 669, 739, 799)
bks <- sort(c(bks, bks[-c(1,length(bks))]+1))
r <- cut(x, breaks= bks, include.lowest=F)
f <- levels(r)[-seq(2,length(levels(r)),by = 2)]
r <- factor(r[r %in% f])

这样你得到：

> levels(r)
[1] "(300,579]" "(580,669]" "(670,739]" "(740,799]"

Answer 2

我认为你不需要在这里展示的东西。 根据间隔的数据应该属于哪个组，您可以使用findInterval或cut如下：

findInterval(579, c(300, 579, 669, 739, 799), left.open = TRUE) 
#[1] 1

findInterval(579, c(300, 579, 669, 739, 799))
#[1] 2

带cut

cut(579, c(300, 579, 669, 739, 799))
#[1] (300,579]
#Levels: (300,579] (579,669] (669,739] (739,799]

cut(579, c(300, 579, 669, 739, 799), right = FALSE)
#[1] [579,669)
#Levels: [300,579) [579,669) [669,739) [739,799)