使用rapply：

rapply(mylist, \(x) all(x == "TRUE"), how = "list")

输出

[[1]]
[[1]][[1]]
[1] FALSE

[[1]][[2]]
[1] FALSE

[[1]][[3]]
[1] FALSE

[[2]]
[[2]][[1]]
[1] TRUE

[[2]][[2]]
[1] TRUE

[[2]][[3]]
[1] FALSE

[[3]]
[[3]][[1]]
[1] TRUE

[[3]][[2]]
[1] FALSE

[[3]][[3]]
[1] FALSE

或者，如果要获得更紧凑的结果：

rapply(mylist, \(x) all(x == "TRUE"), how = "list") |>
  lapply(\(x) which(unlist(x)))

[[1]]
integer(0)

[[2]]
[1] 1 2

[[3]]
[1] 1

rrapply::rrapply的另一种紧凑型解决方案：

rrapply::rrapply(mylist, \(x) all(x == "TRUE"), how = "melt")

  L1 L2            value
1  2  1 TRUE, TRUE, TRUE
2  2  2 TRUE, TRUE, TRUE
3  3  1 TRUE, TRUE, TRUE

注意：你的真实的数据中可能有逻辑向量，它是由c(TRUE, FALSE)（没有括号）组成的，在这种情况下，all(x)就足够了。

也许这能帮上忙

which(sapply(mylist, \(x) sapply(x, \(y) all(y == 'TRUE'))), arr.ind = TRUE)

或使用

lapply(mylist, \(x) Filter(\(y) all(y == 'TRUE'), x))
[[1]]
list()

[[2]]
[[2]][[1]]
[1] "TRUE" "TRUE" "TRUE"

[[2]][[2]]
[1] "TRUE" "TRUE" "TRUE"

[[3]]
[[3]][[1]]
[1] "TRUE" "TRUE" "TRUE"

或者可能是

> lapply(mylist, \(x) lapply(x, \(y) all(y == 'TRUE')))
[[1]]
[[1]][[1]]
[1] FALSE

[[1]][[2]]
[1] FALSE

[[1]][[3]]
[1] FALSE

[[2]]
[[2]][[1]]
[1] TRUE

[[2]][[2]]
[1] TRUE

[[2]][[3]]
[1] FALSE

[[3]]
[[3]][[1]]
[1] TRUE

[[3]][[2]]
[1] FALSE

[[3]][[3]]
[1] FALSE

下面是使用lapply的另一个技巧

lapply(
  mylist,
  function(x) {
    as.list(
      colMeans(type.convert(list2DF(x), as.is = TRUE)) == 1
    )
  }
)

它给出了

[[1]]
[[1]][[1]]
[1] FALSE

[[1]][[2]]
[1] FALSE

[[1]][[3]]
[1] FALSE

[[2]]
[[2]][[1]]
[1] TRUE

[[2]][[2]]
[1] TRUE

[[2]][[3]]
[1] FALSE

[[3]]
[[3]][[1]]
[1] TRUE

[[3]][[2]]
[1] FALSE

[[3]][[3]]
[1] FALSE