如何按 R 中的组对所有连续的列对进行测试

Question

我正在尝试将 wilcoxon test 应用于 df 的所有顺序列对。我使用 col<-seq(2,ncol(df),by=2).

没问题

但我不知道如何分组，在这种情况下 skul。

library(tidyverse)

df <- tibble(skul = c(rep('a',15), rep('b', 16)),
             x1i = sample(1:10, 31, replace = TRUE),
             x1f = sample(1:10, 31, replace = TRUE),
             x2i = sample(1:10, 31, replace = TRUE),
             x2f = sample(1:10, 31, replace = TRUE))

col<-seq(2,ncol(df),by=2)

# I have to select out the column "skul"

map2(df[,col], df[,c(-col, -1)], wilcox.test)
#> Warning in wilcox.test.default(.x[[i]], .y[[i]], ...): cannot compute exact p-
#> value with ties

#> Warning in wilcox.test.default(.x[[i]], .y[[i]], ...): cannot compute exact p-
#> value with ties
#> $x1i
#> 
#>  Wilcoxon rank sum test with continuity correction
#> 
#> data:  .x[[i]] and .y[[i]]
#> W = 486, p-value = 0.9435
#> alternative hypothesis: true location shift is not equal to 0
#> 
#> 
#> $x2i
#> 
#>  Wilcoxon rank sum test with continuity correction
#> 
#> data:  .x[[i]] and .y[[i]]
#> W = 401, p-value = 0.2613
#> alternative hypothesis: true location shift is not equal to 0

由 reprex 包 (v2.0.1) 创建于 2022-04-13

Answer 1

这是您要找的吗：

df <- dplyr::tibble(skul = c(rep('a',15), rep('b', 16)),
                    x1i = sample(1:10, 31, replace = TRUE),
                    x1f = sample(1:10, 31, replace = TRUE),
                    x2i = sample(1:10, 31, replace = TRUE),
                    x2f = sample(1:10, 31, replace = TRUE))


stems <- unique(gsub("[if]", "", names(df)[-1]))
out <- by(df, list(df$skul), function(data){
  lapply(stems, function(x){
    inds <- grep(x, names(df))
    n <- names(df)[inds]
    cmd <- paste0("wilcox.test(data$", n[1], ", data$", n[2], ")")
    eval(parse(text=cmd))
  })})
#> Warning in wilcox.test.default(data$x1i, data$x1f): cannot compute exact p-value
#> with ties
#> Warning in wilcox.test.default(data$x2i, data$x2f): cannot compute exact p-value
#> with ties
#> Warning in wilcox.test.default(data$x1i, data$x1f): cannot compute exact p-value
#> with ties
#> Warning in wilcox.test.default(data$x2i, data$x2f): cannot compute exact p-value
#> with ties
out
#> : a
#> [[1]]
#> 
#>  Wilcoxon rank sum test with continuity correction
#> 
#> data:  data$x1i and data$x1f
#> W = 110, p-value = 0.9334
#> alternative hypothesis: true location shift is not equal to 0
#> 
#> 
#> [[2]]
#> 
#>  Wilcoxon rank sum test with continuity correction
#> 
#> data:  data$x2i and data$x2f
#> W = 75.5, p-value = 0.1266
#> alternative hypothesis: true location shift is not equal to 0
#> 
#> 
#> ------------------------------------------------------------ 
#> : b
#> [[1]]
#> 
#>  Wilcoxon rank sum test with continuity correction
#> 
#> data:  data$x1i and data$x1f
#> W = 113.5, p-value = 0.5952
#> alternative hypothesis: true location shift is not equal to 0
#> 
#> 
#> [[2]]
#> 
#>  Wilcoxon rank sum test with continuity correction
#> 
#> data:  data$x2i and data$x2f
#> W = 152, p-value = 0.3719
#> alternative hypothesis: true location shift is not equal to 0

^{由 reprex package (v2.0.1)}

于 2022-04-13 创建

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