转置列表清单

我试图做一个递归函数来获取一个列表的列表的转置n x p到p x n.但我无法这样做.我已经能够创建一个函数来将3 x n列表列表转换为一个列表n x 3:

let rec drop1 list=
    [(match (List.nth list 0) with [] -> [] | a::b -> b);
     (match (List.nth list 1) with [] -> [] | a::b -> b);
     (match (List.nth list 2) with [] -> [] | a::b -> b);]

let rec transpose list=
    if List.length (List.nth list 0) == 0 then []
    else [(match (List.nth list 0) with [] -> 0 | a::b -> a);
          (match (List.nth list 1) with [] -> 0 | a::b -> a);
          (match (List.nth list 2) with [] -> 0 | a::b -> a)]
         :: transpose (drop1 list)

但我无法概括它.我肯定在想错误的方向.这可以推广吗？有更好的解决方案吗？请帮忙.

let rec transpose list = match list with
| []             -> []
| []   :: xss    -> transpose xss
| (x::xs) :: xss ->
    (x :: List.map List.hd xss) :: transpose (xs :: List.map List.tl xss)