R：具有子集的 T 统计量

我想要一个表作为输出，其中包含某些变量均值差异之间的 t 统计量并基于数据的两个特定子集。

我有以下数据：

structure(list(Name = c("A", "A", "A", "A", "B", "B", "B", "B", 
"C", "C", "C", "C", "D", "D", "D", "D"), Date = c("20.10.2018", 
"30.09.2018", "25.11.2019", "23.10.2020", "20.03.2018", "30.07.2018", 
"25.08.2019", "23.10.2020", "20.12.2018", "30.01.2018", "25.02.2019", 
"23.06.2020", "20.11.2018", "30.12.2018", "25.11.2019", "23.09.2020"
), Return = c(0.01, 0.05, 0.08, 0.07, 0.04, 0.03, 0.01, 0.03, 
0.03, 0.05, 0.06, 0.07, 0.07, 0.04, 0.06, 0.08), Age = c(5L, 
5L, 6L, 7L, 8L, 8L, 9L, 10L, 4L, 4L, 5L, 6L, 1L, 1L, 2L, 3L), 
    Size = c(53336L, 75768L, 86548L, 94567L, 40234L, 40240L, 
    50243L, 60352L, 5069L, 6069L, 7092L, 8024L, 2456L, 3046L, 
    4056L, 5600L), Rating = c(1L, 1L, 1L, 2L, 5L, 5L, 3L, NA, 
    4L, 5L, 4L, 5L, NA, 4L, 5L, 4L)), class = "data.frame", row.names = c(NA, 
-16L))

更具体地说，我想要一个表，其中对于评级为 1 和 5 的观测值的变量 Return、Age 和 Size 之间的均值差异有 t 统计量。t 统计量应该是 Rating 之间的列1 和评级 5，并且应包括表示 p 值的星号。

我尝试使用 t.test 函数，但仅将其用于子组并在评级 1 和评级 5 之间创建 t 统计列时遇到困难。

输出应具有如下布局：

structure(list(c("Return", "Age", "Size"), `Mean Rating 1` = c(NA, 
NA, NA), `t-statistics including p-value (indicated as stars)` = c(NA, 
NA, NA), `Mean Rating 5` = c(NA, NA, NA)), class = "data.frame", row.names = c(NA, 
-3L))

有人可以帮我这里的代码吗？

预先非常感谢您。

编辑 2022 年 4 月 22 日：

问题一： 如果我希望输出如下（现在没有值，只是为了说明我想要的布局），我需要如何调整答案中的代码：

structure(list(c("Return", "Age", "Size"), `Mean Rating 1` = c(NA, 
NA, NA), `Mean Rating2` = c(NA, NA, NA), `Mean Rating 3` = c(NA, 
NA, NA), `Mean Rating 4` = c(NA, NA, NA), `Mean Rating 5` = c(NA, 
NA, NA), `Mean Rating NA` = c(NA, NA, NA), `Difference in means Rating 5 and Rating 1` = c(NA, 
NA, NA), `p-value for differences in means Rating 5 and Rating 1` = c(NA, 
NA, NA), `stars for p-value for differences in means Rating 5 and Rating 1` = c(NA, 
NA, NA)), class = "data.frame", row.names = c(NA, -3L))

问题2： 当我想比较两组之间的均值差异时，使用 t 检验还是 F 检验更好？我选择了 t 检验，因为据我所知，如果我想比较两组之间的均值，t 检验是正确的检验。如果我想比较两组的两个标准差，则最好使用 F 检验。我的理解对吗？

您可以轻松地循环subset=.

t(with(mtcars, sapply(unique(cyl), \(i) t.test(am, subset=cyl == i))))
#      statistic parameter p.value      conf.int  estimate null.value stderr     alternative method              data.name
# [1,] 4.605489  31        6.632258e-05 numeric,2 0.40625  0          0.08820997 "two.sided" "One Sample t-test" "am"     
# [2,] 4.605489  31        6.632258e-05 numeric,2 0.40625  0          0.08820997 "two.sided" "One Sample t-test" "am"     
# [3,] 4.605489  31        6.632258e-05 numeric,2 0.40625  0          0.08820997 "two.sided" "One Sample t-test" "am"  

对于您的数据更具体，您可以这样做：

tcols <- c('Return', 'Age', 'Size')
r <- t(with(subset(dat, Rating %in% c(1, 5)), 
     sapply(setNames(tcols, tcols), \(i) unlist(
       t.test(reformulate('Rating', i))[
         c('estimate', 'statistic', 'p.value')]
       ))))
cbind(as.data.frame(r),
      ' '=c("   ", "*  ", "** ", "***")[
        rowSums(outer(r[, 'p.value'], c(Inf, 0.05, 0.01, 0.001), `<`))])
#        estimate.mean in group 1 estimate.mean in group 5 statistic.t   p.value    
# Return             4.666667e-02                     0.05  -0.1552301 0.8883096    
# Age                5.333333e+00                     5.60  -0.2198599 0.8353634    
# Size               7.188400e+04                 19724.60   4.0457818 0.0109848 *  

Note使用 R >= 4.1。

Edit

as.data.frame(t(with(subset(dat, Rating %in% c(1, 5)), 
       sapply(setNames(tcols, tcols), \(i) unlist(
         t.test(reformulate('Rating', i))[
           c('estimate', 'statistic', 'p.value')]
       ))))) |>
  {\(.) cbind(mean.diff.5.1=apply(.[1:2], 1, diff), .[3:4])}() |> 
  cbind(' '=c("   ", "*  ", "** ", "***")[
          rowSums(outer(r[, 'p.value'], c(Inf, 0.05, 0.01, 0.001), `<`))],
        `colnames<-`(t(aggregate(. ~ Rating, dat[3:6], mean)[-1]), 
                     paste0('mean.rating.', 1:5))) |>
  {\(.) .[c(5:9, 1:4)]}()
#        mean.rating.1 mean.rating.2 mean.rating.3 mean.rating.4 mean.rating.5 mean.diff.5.1 statistic.t   p.value    
# Return  4.666667e-02          0.07          0.01        0.0525          0.05  3.333333e-03  -0.1552301 0.8883096    
# Age     5.333333e+00          7.00          9.00        3.2500          5.60  2.666667e-01  -0.2198599 0.8353634    
# Size    7.188400e+04      94567.00      50243.00     5201.7500      19724.60 -5.215940e+04   4.0457818 0.0109848 *