1

所以我有这个元组列表(n = 2),女巫我应该“解压缩”并创建一个像这样的新列表:对于像这样的元组列表 val it = (1,"one") :: (2,"two") :: nil : (int,string) alterlist,解压缩函数将创建一个类似 [(1,2),("one", "two")] 的列表。这是我到目前为止得到的:

datatype ('a, 'b) alterlist = nil | :: of ('a*'b) * ('a, 'b) alterlist; 
infixr 5 :: 

fun build4(x, one, y, two) = (x,one)::((y,two)::nil);

fun unzip(alterlist) = 
let 
    fun extract([], i) = []
        | extract((x,y)::xs, i) = if i=0 then x::extract(xs, i)
            else y::extract(xs, i);
in
    (extract(alterlist, 0))::(extract(alterlist, 1))
end;

但我得到一堆错误:

stdIn:48.6-50.26 Error: parameter or result constraints of clauses don't agree [tycon mismatch]
  this clause:      ('Z,'Y) alterlist * 'X -> 'W
  previous clauses:      'V list * 'U -> 'W
  in declaration:
    extract =
      (fn (nil,i) => nil
        | (<pat> :: <pat>,i) =>
            if <exp> = <exp> then <exp> :: <exp> else <exp> :: <exp>)
stdIn:49.41-49.58 Error: operator and operand don't agree [tycon mismatch]
  operator domain: 'Z list * 'Y
  operand:         ('X,'W) alterlist * int
  in expression:
    extract (xs,i)
stdIn:50.9-50.26 Error: operator and operand don't agree [tycon mismatch]
  operator domain: 'Z list * 'Y
  operand:         (_ * _,'X) alterlist * int
  in expression:
    extract (xs,i)
stdIn:48.6-50.26 Error: right-hand-side of clause doesn't agree with function result type [tycon mismatch]
  expression:  (_,_) alterlist
  result type:  'Z list
  in declaration:
    extract =
      (fn (nil,i) => nil
        | (<pat> :: <pat>,i) =>
            if <exp> = <exp> then <exp> :: <exp> else <exp> :: <exp>)

由于我是 ml 新手,我几乎不知道是什么原因造成的。非常感谢帮助!

4

1 回答 1

2

我的建议是首先不使用中缀运算符,因为它有点令人困惑。先解决没有,以后再添加。

这是一个没有中缀的解决方案:

  datatype ('a,'b)alterlist =   Nil 
                            |   element of 'a*('b,'a)alterlist;                         

fun unzip (Nil : ('a,'b)alterlist ) = ([],[])
  | unzip (ablist : ('a,'b)alterlist)  =
    let 
    fun extract  Nil = []
      | extract (element (curr, Nil)) = curr::[]
      | extract (element (curr, element(_,rest))) = curr::(extract rest)
    val element (_, balist) = ablist
    in
    (extract ablist, extract balist)
    end;

因此,例如这样的列表: ["a", 1, "b", 2] 将由以下人员创建:

element ("a", element (1, element ("b", element (2, Nil))));

给你:

val it = element ("a",element (1,element #)) : (string,int) alterlist

* # 只是表示列表比显示的长。

然后,如果您尝试,unzip it; 您会得到:

val it = (["a","b"],[1,2]) : string list * int list

如你所愿。

现在尝试将元素更改为 :: 中缀运算符。

祝你好运!

于 2014-05-23T21:43:50.940 回答