F#zip实现
我不知道如何在F#中实现Zip函数。谁能告诉我我做错了什么?以下是我在F#zip实现,f#,functional-programming,f#-interactive,F#,Functional Programming,F# Interactive,我不知道如何在F#中实现Zip函数。谁能告诉我我做错了什么?以下是我在fsi.exe中键入的内容: > let rec zip xs ys = - match xs with - | [] -> [] - | head :: tail -> (match ys with - | [] -> [] - | headY :: tailY -> (head, headY) :
fsi.exe中键入的内容:
> let rec zip xs ys =
- match xs with
- | [] -> []
- | head :: tail -> (match ys with
- | [] -> []
- | headY :: tailY -> (head, headY) :: zip tail tailY);;
val zip : xs:'a list -> ys:'b list -> ('a * 'b) list
> zip [1;2;3;4] ["a","b","c","d"];;
val it : (int * (string * string * string * string)) list =
[(1, ("a", "b", "c", "d"))]
在您的示例中,[“a”、“b”、“c”、“d”]
是一个包含一个元素的列表,该元素是四维元组。这就是为什么您会从zip中得到意想不到的结果。
只要使用代码>改为元素分隔符。我认为值得指出的是,让zip
函数尾部递归也是值得的,这样可以避免较大列表上的堆栈溢出
也许是这样的
let zip3 xs ys =
let rec loop r xs ys =
match xs,ys with
| [],[] -> r
| xh::xt,yh::yt -> loop ((xh,yh)::r) xt yt
| _ -> failwith "xs & ys has different length"
loop [] xs ys |> List.rev
备选方案:让rec zip2 xs ys=将xs,ys与| xh::xt,yh::yt->(xh,yh)::(zip2 xt yt)|,|->[]