Question

我需要“解压缩”一个元组列表，即使我尝试使用unzip函数，它也没有按照我想要的方式执行。事实上，据我所知，unzip函数会返回一个列表元组，但我想要的是这个。

给出一个像这样的元组列表：

L1 = [(("s0","l0"),("l0","s1")),(("s1","l1"),("l1","s2"))]

类型：

(string*string)*(string*string) list

该函数应该返回由解压缩元组组成的单个列表：

L2 = [("s0","l0"),("l0","s1"),("s1","l1"),("l1","s2")]

类型：

(string*string) list

我找到的唯一解决方案是：

fun f(h as (x,y,z),nil) =  [(x,x^y^z)]@[(x^y^z,z)]
|f (h as (x,y,z),t::rest) = 
    let val i = ref 0
        in if !i=0 then (i := !i+1; (x,x^y^z)::f(h,t::rest)) else (i := !i-1;(x^y^z,z)::f(t,rest)) end;

并且我认为它可以工作但是当我运行该功能时系统崩溃。

Answer 1

我不完全确定你发布的功能是什么，但它几乎肯定不会为你解压缩2元组的列表。它似乎期望3元组，并启动一些解引用/连接。你是怎么找到它的？

如果你想要一个带有2元组列表并返回其元素列表的函数，你可以这样做

fun peel lst =
    let fun recur [] acc = List.rev acc
      | recur ((a, b)::rest) acc = 
        recur rest (b::a::acc)
    in 
    recur lst []
    end

即，遍历列表，收集每个元组的两个元素，并在结束时返回累加器的反向。

Standard ML of New Jersey v110.76 [built: Thu Feb 19 00:37:13 2015]
- fun peel lst =
    let fun recur [] acc = List.rev acc
      | recur ((a, b)::rest) acc = 
        recur rest (b::a::acc)
    in 
    recur lst []
    end ;;
= = = = = = [autoloading]
[library $SMLNJ-BASIS/basis.cm is stable]
[autoloading done]
val peel = fn : ('a * 'a) list -> 'a list
- peel [(("s0","l0"),("l0","s1")),(("s1","l1"),("l1","s2"))] ;;
val it = [("s0","l0"),("l0","s1"),("s1","l1"),("l1","s2")]
  : (string * string) list
-

由于其定义的方式，它具有（可能不必要的奖励）它可以类似地处理任何2元组的列表。

- peel (peel [(("s0","l0"),("l0","s1")),(("s1","l1"),("l1","s2"))]) ;;
val it = ["s0","l0","l0","s1","s1","l1","l1","s2"] : string list
-

SML - 在一个列表中解压缩元组

1 个答案: