将 class "perm.htest" 个对象从（多重排列 t 检验）转换为数据框

Question

我有一些来自 perm.t.test 的结果，我想将其转换为数据框。我已经尝试了很多并进行了很多搜索，但我无法解决这个问题，我还是 R 的新手。

这是我的数据集的简化版本：

treat = c("C","C","C","C","C","C","C","C","C","C","C","C","C",
         "C","C","C","C","C","C","C","T","T","T","T","T","T",
         "T","T","T","T","T","T","T","T","T","T","T","T","T","T")
subj = c("B16","B17","B18","B19","B20","B16","B17","B18","B19",
        "B20","B16","B17","B18","B19","B20","B16","B17","B18",
        "B19","B20","B1","B2","B3","B4","B5","B1","B2","B3","B4"
        ,"B5","B1","B2","B3","B4","B5","B1","B2","B3","B4","B5")
t = c("T0","T0","T0","T0","T0","T1","T1","T1","T1","T1","T2",
      "T2","T2","T2","T2","T3","T3","T3","T3","T3","T0","T0",
      "T0","T0","T0","T1","T1","T1","T1","T1","T2","T2","T2",
      "T2","T2","T3","T3","T3","T3","T3")
exparat = c(0.11,0.27,0.04,0.47,-0.11,-0.05,-0.05,0.33,-0.11,
            0.47,-0.01,0.43,0.47,0.33,-0.11,-0.09,0.20,-0.11,
                0.47,0.33,0.19,0.02,0.33,0.47,-0.11,0.42,0.13,0.47,
                -0.11,0.33,0.42,0.19,-0.11,0.33,0.47,0.42,0.17,
                0.33,0.47,-0.11)

data = data.frame(treat, subj, t, exparat)

data$treat <- factor(data$treat)
data$t <- factor(data$t,levels=unique(data$t))

head(data)

  treat subj  t exparat
1     C  B16 T0    0.11
2     C  B17 T0    0.27
3     C  B18 T0    0.04
4     C  B19 T0    0.47
5     C  B20 T0   -0.11
6     C  B16 T1   -0.05

I 运行多个 MKinfer::perm.t.test 时间组合之间 (t) 独立于每个治疗 (treat)，使用此功能（感谢 Ronak Shah）：

library(MKinfer)

    combn(levels(data$t), 2, function(x) {
      perm.t.test(exparat~t,data = subset(data, t %in% x), R=999, paired = T)
    }, simplify = FALSE) -> result

但是现在我有两个问题：

1- 我的结果是 class "perm.htest" 对象的列表，但我需要将其转换为数据框，其中每一行都是一个测试，每一列都是一个输出测试的（即统计，参数，p.value）。所以我可以轻松地检查我的结果，更正多重比较的 p 值并导出它们。

有人建议我使用：

map_df(result, broom::tidy) %>%
  mutate(combination = combn(levels(data$t), 2, paste0, collapse = ' vs '), .before = 1)

问题是 MKinfer::perm.t.test 同时执行排列 t 检验和参数 t 检验。所以在结果中，既有p.value又有perm.p.value，有conf.int也有perm.conf.int。因此，如果我使用 broom::tidy，它会将其识别为 htest，我将只获得参数 t 检验的结果。

也许类似的东西会起作用：

dtlist <- combn(levels(data$t), 2, function(x) {
  dt <- MKinfer::perm.t.test(respvar~t,data = subset(data, t %in% x), R=999, paired = T)
  df <- data.frame(
    perm.p = dt$perm.p.value,
    stat = dt$statistic)
  return(df)
})

但我需要一个数据框作为输出，然后我可以将其导出为 csv。

有人有解决办法吗？

提前致谢。

Answer 1

library(tidyverse)

tidy_result <- function(result) {
  tibble(
    estimate = result$estimate,
    statistic = result$statistic,
    conf.low = result$conf.int[1],
    conf.high = result$conf.int[2],
    perm.p.value = result$perm.p.value
  )
}

combn(levels(data$t), 2, function(x) {
  perm.t.test(exparat~t,data = subset(data, t %in% x), R=999, paired = T)
}, simplify = FALSE) -> result

map_dfr(result, tidy_result)


# # A tibble: 6 x 5
#   estimate statistic conf.low conf.high perm.p.value
#      <dbl>     <dbl>    <dbl>     <dbl>        <dbl>
# 1   -0.015    -0.116   -0.307    0.277         0.918
# 2   -0.073    -0.764   -0.289    0.143         0.470
# 3   -0.04     -0.670   -0.175    0.0950        0.528
# 4   -0.058    -0.482   -0.330    0.214         0.593
# 5   -0.025    -0.220   -0.282    0.232         0.824
# 6    0.033     0.292   -0.222    0.288         0.748

Answer 2

你可以使用-

do.call(rbind, combn(levels(data$t), 2, function(x) {
  dt <- MKinfer::perm.t.test(exparat~t,
            data = subset(data, t %in% x), R=999, paired = T)
  data.frame(perm.p = dt$perm.p.value,stat = dt$statistic)
}, simplify = FALSE)) -> result

result$combination <- combn(levels(data$t), 2, paste0, collapse = ' vs ')

将 class "perm.htest" 个对象从（多重排列 t 检验）转换为数据框

Transform a class "perm.htest" objects, from (multiple permutation t test) into a data frame

r

permutation

dataframe

tidyverse