R 如何对2个列表元素的所有组合进行操作

您可以使用combn()生成组合，然后使用lapply()迭代组合:

animals <- list(
  dog = data.frame(
    col_1 = sample(c(100:200), size = 10, replace = TRUE),
    col_2 = sample(c(100:200), size = 10, replace = TRUE),
    col_3 = sample(c(100:200), size = 10, replace = TRUE),
    col_4 = sample(c(100:200), size = 10, replace = TRUE)
  ),
  cat = data.frame(
    col_1 = sample(c(100:200), size = 10, replace = TRUE),
    col_2 = sample(c(100:200), size = 10, replace = TRUE),
    col_3 = sample(c(100:200), size = 10, replace = TRUE),
    col_4 = sample(c(100:200), size = 10, replace = TRUE)
  ),
  mouse = data.frame(
    col_1 = sample(c(100:200), size = 10, replace = TRUE),
    col_2 = sample(c(100:200), size = 10, replace = TRUE),
    col_3 = sample(c(100:200), size = 10, replace = TRUE),
    col_4 = sample(c(100:200), size = 10, replace = TRUE)
  )
)


combinations <- asplit(combn(names(animals), 2), 2)

print(combinations)
#> [[1]]
#> [1] "dog" "cat"
#> 
#> [[2]]
#> [1] "dog"   "mouse"
#> 
#> [[3]]
#> [1] "cat"   "mouse"

lapply(combinations, function(x) {
  generic_function(
    animals[[x[1]]][, c("col_1", "col_2")],
    animals[[x[2]]][, c("col_1", "col_2")]
  )
})

^{创建于2024-03-14年第reprex v2.1.0页}