我想應用這個測驗，不僅x1像我在這個例子中所做的那樣應用于 column ，而且應用于df. 在這種情況下x1和x2。
我試圖把這段代碼放在一個函式中并使用purrr::map，但我做錯了。
library(tidyverse)
df <- tibble(skul = c(rep('a',60), rep('b', 64)),
x1 = sample(1:10, 124, replace = TRUE),
x2 = sample(1:10, 124, replace = TRUE),
i_f = c(rep(0, 30), rep(1, 30), rep(0, 32), rep(1, 32)))
lapply(split(df, factor(df$skul)),
function(x)wilcox.test(data=x, x1 ~ i_f,
paired=FALSE))
#> Warning in wilcox.test.default(x = c(10L, 5L, 8L, 4L, 6L, 3L, 10L, 2L, 10L, :
#> cannot compute exact p-value with ties
#> Warning in wilcox.test.default(x = c(3L, 3L, 4L, 9L, 8L, 10L, 5L, 5L, 4L, :
#> cannot compute exact p-value with ties
#> $a
#>
#> Wilcoxon rank sum test with continuity correction
#>
#> data: x1 by i_f
#> W = 546, p-value = 0.1554
#> alternative hypothesis: true location shift is not equal to 0
#>
#>
#> $b
#>
#> Wilcoxon rank sum test with continuity correction
#>
#> data: x1 by i_f
#> W = 565, p-value = 0.4781
#> alternative hypothesis: true location shift is not equal to 0
Created on 2022-04-13 by the reprex package (v2.0.1)
uj5u.com熱心網友回復：
一種方法是將感興趣的列作為嵌套的內部回圈在 之后回圈split，使用創建公式reformulate并應用wilcox.test
out <- lapply(split(df, df$skul), function(x)
lapply(setNames(c("x1", "x2"), c("x1", "x2")), function(y)
wilcox.test(reformulate("i_f", response = y), data = x)))
-輸出
> out$a
$x1
Wilcoxon rank sum test with continuity correction
data: x1 by i_f
W = 452, p-value = 0.9822
alternative hypothesis: true location shift is not equal to 0
$x2
Wilcoxon rank sum test with continuity correction
data: x2 by i_f
W = 404.5, p-value = 0.5027
alternative hypothesis: true location shift is not equal to 0
如果我們想使用tidyverse
library(dplyr)
df %>%
group_by(skul) %>%
summarise(across(c(x1, x2),
~list(broom::tidy(wilcox.test(reformulate("i_f", cur_column()))))))
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