问题描述
我正在构建合并排序功能,而我的split方法给了我一个值限制错误.我正在使用2个累加参数,这是拆分产生的2个列表,最后将它们打包到一个元组中以供返回.但是,我收到一个值限制错误,我无法弄清楚问题出在哪里.有人有什么想法吗?
I'm building a merge sort function and my split method is giving me a value restriction error. I'm using 2 accumulating parameters, the 2 lists resulting from the split, that I package into a tuple in the end for the return. However I'm getting a value restriction error and I can't figure out what the problem is. Does anyone have any ideas?
let split lst =
let a = []
let b = []
let ctr = 0
let rec helper (lst,l1,l2,ctr) =
match lst with
| [] -> []
| x::xs -> if ctr%2 = 0 then helper(xs, x::l1, l2, ctr+1)
else
helper(xs, l1, x::l2, ctr+1)
helper (lst, a, b, ctr)
(a,b)
任何输入表示赞赏.
推荐答案
如您所写,该代码实际上没有任何意义. F#默认使用不可变值,因此您当前编写的函数可以简化为:
The code, as you have written it, doesn't really make sense. F# uses immutable values by default, therefore your function, as it's currently written, can be simplified to this:
let split lst =
let a = []
let b = []
(a,b)
这可能不是您想要的.实际上,由于绑定是不变的,因此预先声明a, b
和ctr
没有任何价值.
This is probably not what you want. In fact, due to immutable bindings, there is no value in predeclaring a, b
and ctr
.
这里有一个递归函数可以解决这个问题:
Here is a recursive function that will do the trick:
let split lst =
let rec helper lst l1 l2 ctr =
match lst with
| [] -> l1, l2 // return accumulated lists
| x::xs ->
if ctr%2 = 0 then
helper xs (x::l1) l2 (ctr+1) // prepend x to list 1 and increment
else
helper xs l1 (x::l2) (ctr+1) // prepend x to list 2 and increment
helper lst [] [] 0
除了使用递归函数外,您还可以使用List.fold
解决此问题,fold
是一个高阶函数,它概括了我们在上面的递归函数中明确描述的累加过程.
Instead of using a recursive function, you could also solve this problem using List.fold
, fold
is a higher order function which generalises the accumulation process that we described explicitly in the recursive function above.
这种方法更简洁一些,但对于刚开始使用函数式编程的人来说可能不太熟悉,因此我尝试更详细地描述此过程.
This approach is a bit more concise but very likely less familiar to someone new to functional programming, so I've tried to describe this process in more detail.
let split2 lst =
/// Take a running total of each list and a index*value and return a new
/// pair of lists with the supplied value prepended to the correct list
let splitFolder (l1, l2) (i, x) =
match i % 2 = 0 with
|true -> x :: l1, l2 // return list 1 with x prepended and list2
|false -> l1, x :: l2 // return list 1 and list 2 with x prepended
lst
|> List.mapi (fun i x -> i, x) // map list of values to list of index*values
|> List.fold (splitFolder) ([],[]) // fold over the list using the splitFolder function
这篇关于F#拆分功能的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!