考虑分位数不起作用的关系，找到将变量分成相等组的最佳截止点

Question

我有一个变量

x

，我想将其分为具有相同观察值的三组。然而，由于平局，使用分位数并不会产生最平等的组，因为分位数截止点可能会导致平局被分配到多个组中。我正在寻找一种函数或算法，可以找到最佳分界点，同时确保关系不会分散在多个组中。

x = c(26, 34, 27, 26, 38, 40, 34, 28, 27, 36, 29, 30, 29, 44, 30, 
      34, 32, 30, 26, 29, 34, 32, 38, 27, 35, 29, 28, 34, 26, 27, 27, 
      30, 27, 28, 27, 28, 28, 27, 29, 29, 28, 29, 29, 28, 29, 29, 28, 
      27, 29, 27, 36, 34, 34, 39, 34, 31, 31, 33, 35, 31, 31, 32, 37, 
      38, 32, 31, 28, 33, 33, 28, 27, 27, 30, 31, 32, 28, 27, 31, 36, 
      27, 33, 31, 34, 31, 35, 38, 37, 36, 39, 33, 33, 28, 41, 34, 35, 
      37, 37, 41, 32, 37, 30, 34, 38, 30, 40, 35, 31, 30, 30, 29, 29, 
      30, 29, 35, 28, 27, 27, 27, 29, 27, 28, 27, 27, 27, 26, 28, 28, 
      27, 29, 29, 27, 27, 27, 27, 29, 27, 28, 27, 28, 34, 29, 28, 28, 
      28, 29, 38, 33, 39, 28, 27, 28, 27, 29, 34, 29, 32, 70, 26, 29, 
      43, 48, 30, 30, 27, 26, 29, 27, 27, 27, 27, 28, 28, 27, 28, 28, 
      27, 28, 28, 38, 52, 26, 31, 56, 29, 29, 36, 28, 35, 32, 34, 35, 
      28, 27, 37, 26, 26, 32, 26, 27, 30, 28, 28, 30, 29, 30, 29, 29, 
      28, 26, 33, 39, 26, 31, 27, 28, 30, 30, 28, 28, 29, 26, 27, 26, 
      29, 28, 28, 27, 27, 27, 28, 27, 28, 28, 28, 28, 28, 27, 27, 29, 
      27, 26, 28, 28, 27, 27, 28, 27, 28, 28, 30, 27, 30, 28, 32, 34, 
      28, 27, 28, 28, 27, 28, 27, 27, 27, 28, 27, 28, 27, 27, 28, 27, 
      27, 27, 27, 27, 28, 27, 27, 27, 26, 27, 27, 30, 28, 27, 30, 30, 
      42, 26, 27, 40, 33, 29, 29, 29, 52, 58, 44, 32, 43, 30, 27, 38, 
      30, 27, 30, 27, 31, 39, 35, 32, 32, 34, 45, 31, 44, 42, 29, 29, 
      30, 30, 50, 30, 33, 31, 35, 27, 28, 27, 28, 55, 28, 28, 28, 27, 
      27, 28, 29, 27, 28, 27, 28, 28, 28, 28, 27, 28, 29, 34, 45, 27, 
      29, 61, 38, 62, 29, 36, 36, 30, 31, 45, 27, 30, 28, 29, 44, 45, 
      42, 52, 50, 52, 42, 38, 42, 32, 27, 37, 40, 52, 27, 36, 38, 39, 
      34, 30, 29, 34, 29, 26, 35, 43, 33, 40, 35, 33, 41, 61, 45, 35, 
      52, 50, 38, 43, 29, 35, 38, 39, 31, 28, 28, 29, 34, 27, 30, 32, 
      28, 26, 28, 27, 26, 29, 27, 26, 29, 29, 27, 29, 27, 27, 29, 27, 
      30, 29, 25, 30, 27, 29, 29, 30, 30, 27, 30, 28, 28, 27, 29, 29, 
      30, 29, 27, 28, 28, 28, 29, 28, 28, 27, 28, 29, 28, 29, 27, 28, 
      28, 28, 30, 27, 27, 28, 26, 28, 27, 27, 28, 28, 28, 28, 27, 27, 
      28, 27, 28, 27, 35, 27, 27, 28, 29, 27, 27, 28, 26, 27, 28, 28, 
      28, 27, 27, 27, 28, 32, 27, 28, 28, 29, 28, 28, 27, 28, 28, 30, 
      29, 28, 25, 27, 28, 30, 28, 30, 30, 28, 30, 30, 28, 29, 30, 28, 
      28, 26, 27, 28, 45, 36, 40, 28, 50, 45, 30, 45, 40, 30, 45, 45, 
      29, 45, 35, 40, 40, 30, 30, 30, 45, 40, 40, 40, 40, 40, 40, 35, 
      34, 49, 40, 30, 61, 35, 40, 30, 36, 35, 29, 27, 48, 28, 27, 27, 
      26, 27, 29, 27, 26, 27, 31, 27, 27, 28, 29, 28, 27, 28, 29, 38, 
      30, 26, 36, 40, 58, 57, 30, 33, 56, 35, 39, 37, 38, 46, 37, 39, 
      39, 45, 35, 46, 58, 65, 60, 45, 32, 36, 43, 32, 68, 39, 28, 31, 
      27, 28, 27, 37, 38, 30, 30, 28, 36, 45, 28, 26, 28, 28, 28, 27, 
      26, 28, 27, 26, 26, 27, 28, 31, 32, 37, 35, 29, 33, 35, 29, 41, 
      32, 36, 29, 28, 28, 28, 37, 36, 37, 35, 31, 32, 30, 27, 31, 32, 
      31, 33, 28, 33, 29, 27, 28, 31, 28, 31, 28, 34, 27, 27, 28, 27, 
      27, 27, 27, 26, 26, 26, 27, 27, 28, 26, 31, 26, 29, 31, 29, 29, 
      30, 29, 30, 31, 32, 29, 30, 27, 32, 27, 26, 31, 31, 31, 27, 27, 
      33, 27, 28, 28, 28, 26, 27, 27, 28, 30, 27, 27, 30, 29, 26, 27, 
      28, 27, 26, 26, 28, 27, 26, 28, 28, 26, 28, 27, 29, 27, 28, 28, 
      26, 26, 29, 28, 27, 27, 27, 28, 26, 25, 27, 29, 30, 36, 40, 28, 
      38, 26, 27, 27, 50, 27, 45, 27, 28, 26, 25, 35, 35, 44, 30, 27, 
      31, 27, 28, 27, 27, 28, 28, 28, 35, 33, 30, 28, 28, 29, 29, 36, 
      32, 36, 34, 32, 28, 28, 29, 28, 28, 32, 30, 35, 33, 36, 32, 30, 
      32, 36, 34)

quantile(x, probs = c(0.333, 0.666))
#> 33.3% 66.6% 
#>    28    31

l = cut(x, breaks = c(-Inf, 28, 31, Inf))
table(l)
#> l
#> (-Inf,28]   (28,31] (31, Inf] 
#>       387       185       246

#using different cut-off points yielded more equal groups
l = cut(x, breaks = c(-Inf, 28, 32, Inf))
table(l)
#> l
#> (-Inf,28]   (28,32] (32, Inf] 
#>       387       214       217

#again using different cut-off points which yielded more equal groups
l = cut(x, breaks = c(-Inf, 27, 32, Inf))
table(l)
#> l
#> (-Inf,27]   (27,32] (32, Inf] 
#>       222       379       217

^{创建于 2024-10-07，使用 reprex v2.1.1}

编辑： 我认为“等于”这个词不清楚，所以我想我可以说，我寻求最合适的分配，使最高和最低组观察数之间的差异最小，仅对连续数字进行分组，且不存在多个关系组

Answer 1

这不是一个明智的方法，但希望可能会有所帮助

n <- 2
lst <- c()
repeat {
    q <- quantile(x, probs = seq.int(n - 1) / n)
    if (!anyDuplicated(q)) {
        lst <- c(lst,list(table(cut(x, breaks = c(-Inf, q, Inf)))))
    } else {
        break
    }
    n <- n + 1
}
lst[[which.min(sapply(lst, var))]]

这给出了

(-Inf,27]   (27,28]   (28,30]   (30,35] (35, Inf]
      222       165       153       132       146