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Is it possible to match with decomposed sequences in F#?

开发者 https://www.devze.com 2022-12-11 12:09 出处:网络
I seem to remember an older version of F# allowing structural decomposition when matching sequences just like lists.Is there a way to use the list syntax while keeping the sequence lazy?I\'m hoping to

I seem to remember an older version of F# allowing structural decomposition when matching sequences just like lists. Is there a way to use the list syntax while keeping the sequence lazy? I'm hoping to avoid a lot of calls to Seq.head and Seq.skip 1.

I'm hoping for something like:

let decomposable (xs:seq<'a>) =
  开发者_JAVA百科 match xs with
   | h :: t -> true
   | _ -> false
seq{ 1..100 } |> decomposable

But this only handles lists and gives a type error when using sequences. When using List.of_seq, it seems to evaluate all the elements in the sequence, even if it is infinite.


If you use the LazyList type in the PowerPack, it has Active Patterns called LazyList.Nil and LazyList.Cons that are great for this.

The seq/IEnumerable type is not particulaly amenable to pattern matching; I'd highly recommend LazyList for this. (See also Why is using a sequence so much slower than using a list in this example.)

let s = seq { 1..100 }
let ll = LazyList.ofSeq s
match ll with
| LazyList.Nil -> printfn "empty"
| LazyList.Cons(h,t) -> printfn "head: %d" h


Seq works fine in active patterns! Unless I'm doing something horrible here...

let (|SeqEmpty|SeqCons|) (xs: 'a seq) = 
  if Seq.isEmpty xs then SeqEmpty
  else SeqCons(Seq.head xs, Seq.skip 1 xs)

// Stupid example usage
let a = [1; 2; 3]

let f = function
  | SeqEmpty -> 0
  | SeqCons(x, rest) -> x
  
let result = f a


Remember seq has map reduce functions as well, so you might often be able to get away with only those. In the example, your function is equivalent to "Seq.isEmpty". You might try to launch fsi and just run through the tab completion options (enter "Seq." and hit tab a lot); it might have what you want.

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