R中的ridit变换序数变量

Question

Ridit 评分 (https://en.wikipedia.org/wiki/Ridit_scoring) 通常用于将有序分类变量转换为相对频率（低于给定值的个案比例加上该值比例的一半）。

在 R 中你会怎么做？

Answer 1

以下软件包可能会解决您的问题。特别是命令 Ridit::ridit 很有用，如下所述。

Kruskal-Wallis 检验的扩展，允许指定任意参考组。还提供每组的平均 Ridit。一组的平均 Ridit 是随机观察的概率估计来自该组的将大于或等于来自参考组的随机观察值。

https://cran.r-project.org/web/packages/Ridit/Ridit.pdf

另一种方法是使用二元选择模型，如 Probit、Logit 或 Exact Logit，并提取预测的自变量，即 0 或 1。

Answer 2

进一步更新 这些和其他几个功能现在在 CRAN 包 ridittools 中可用，由您真正维护。

更新删除涉及构建转换矩阵的相当愚蠢的代码，我忘记了 cumsum()

# Convert vector of counts to ridits

to.ridit <- function(v) {
  (cumsum(v) - .5 * v) / sum(v)
}

# Calculate mean ridit for vector of counts relative to reference group

mean.ridit <- function(v, ref) {
  sum(to.ridit(ref) * v ) / sum(v)
}

# Calculate mean ridits for several groups
# x is matrix of counts
# margin is 1 for groups in rows, 2 for groups in columns
# If ref is omitted, totals across groups are used as reference group
# If ref is a vector of counts, it's used as reference group
# Otherwise, ref is the number (or name if it exists) of the group to use as reference

ridits <- function(x, margin, ref=NULL) {
  if (length(ref) > 1) {
    refgroup <- ref
  } else if (length(ref) == 1) {
    if (margin==1) {
      refgroup <- x[ref,]
    } else {
      refgroup <- x[, ref]
    }
  } else {
    refgroup <- apply(x, 3-margin, sum)
  }
  apply(x, margin, mean.ridit, refgroup)
}

示例（Fleiss，1981 年：车祸的严重程度）：

to.ridit(c(17, 54, 60, 19, 9, 6, 14))

[1] 0.04748603 0.24581006 0.56424581 0.78491620 0.86312849 0.90502793 0.96089385

备注虽然我的代码不如另一个答案中提到的 Ridit::ridit 包灵活，但它似乎快了很多：

# Influenza subtypes by age as of week ending 2/24/18 (US CDC)

> flu.age
        BY  BV  BU   H3   H1
0-4    274  91  92 1808  500
5-24  1504 274 698 5090  951
25-64 1665 101 567 7538 1493
65+   1476  35 330 9541  515

# Using CRAN package

> system.time(ridit(flu.age,2))
   user  system elapsed 
  3.746   0.007   3.756 

# Using my code

> system.time(ridits(flu.age,2))
   user  system elapsed 
  0.001   0.000   0.000

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