let rec findMatches str list =
let hd :: tl = list
match list with
| [] -> []
| (s, _) as hd :: tl when s = str -> hd :: findMatches str tl
| _::tl -> findMatches str tl
This is my current function and i am stuck on how to create a new list and returning the list, I would want to test my function with this
matchs "A" [("A",5); ("BB",6); ("AA",9); ("A",0)];;
and i want it to reutrn
val it : int list = [0; 5]
so i know that i need a int list returned
It is easy to achieve your goal using a recursive inner function with an accumulator argument to collect the results one by one:
let findMatches str list =
let rec inner acc = function
| [] -> acc
| (s, n) :: tl ->
inner (if s = str then n :: acc else acc) tl
inner [] list
This is a perfect candidate to use List.fold from the F# library
let toMatch = "A"
let test =
[ ("A", 5)
("BB", 6)
("AA", 9)
("A", 0) ]
let findMatches toMatch items =
List.fold
(fun output item ->
if toMatch = (fst item) then
(snd item) :: output //Append if we find a match otherwise just return the same list
else
output)
[] //Set initial output to the empty list
items
findMatches toMatch test
Related
I am trying to distribute a single item across every element in a list. For example, if I had a list [1,2,3] and an item "DOG". I would like to get the result [("DOG", 1); ("DOG", 2); ("DOG", 3)].
I am trying to use a match expression to complete this, but I am not sure how to start. I currently have:
let rec con (x: 'a) (l: 'b list) =
match l with
| [] -> []
A way to transform every item of a list in some way is the List.map function. It takes two parameters: a function for transforming every item and the list itself.
let l = [1; 2; 3]
let dogs = List.map (fun x -> ("DOG", x)) l
Fyodor's answer is the best way to do this, but if you want a recursive solution with pattern matching, it would look like this:
let rec con (x: 'a) (l: 'b list) =
match l with
| [] -> []
| h :: t -> (x, h) :: (con x t)
Or, more succinctly:
let rec con x = function
| [] -> []
| h :: t -> (x, h) :: (con x t)
Test code:
con "DOG" [1; 2; 3]
|> printfn "%A" // [DOG,1; DOG,2; DOG,3]
I've got this function, what I want to do is take a list, split every element in the list on '.' and place the separated elements in 2 different lists, my problem now is I'm getting an error that says the let expression is unfinished and I think it has to do with not having a return value, is there a way to bypass this or am I doing something completely wrong?
let klist = []
let olist = []
let listSplit list =
match list.Split '.' with
| [| x;y |] -> x :: klist, y :: olist
| [| x |] -> x :: klist
| _ -> None;;
It looks like you want something like
let listSplit (list: string list) =
let acc (kl, ol) = function
| [| x; y |] -> (x::kl, y::ol)
| [| x |] -> (x::kl, ol)
list |> List.map (fun s -> s.Split('.'))
|> List.filter(fun a -> a.Length = 1 || a.Length = 2)
|> List.fold acc ([],[])
Your existing code has a few problems:
list has no split method. Strings do have a Split method, so you probably want to split each string in the input list, which you can do with List.map.
Your match expression does not type check, and each branch has a different type - the first returns string list * string list, the second string list and the third a' option.
I have a pretty trivial task but I can't figure out how to make the solution prettier.
The goal is taking a List and returning results, based on whether they passed a predicate. The results should be grouped. Here's a simplified example:
Predicate: isEven
Inp : [2; 4; 3; 7; 6; 10; 4; 5]
Out: [[^^^^]......[^^^^^^^^]..]
Here's the code I have so far:
let f p ls =
List.foldBack
(fun el (xs, ys) -> if p el then (el::xs, ys) else ([], xs::ys))
ls ([], [])
|> List.Cons // (1)
|> List.filter (not << List.isEmpty) // (2)
let even x = x % 2 = 0
let ret =
[2; 4; 3; 7; 6; 10; 4; 5]
|> f even
// expected [[2; 4]; [6; 10; 4]]
This code does not seem to be readable that much. Also, I don't like lines (1) and (2). Is there any better solution?
Here is my take. you need a few helper functions first:
// active pattern to choose between even and odd intengers
let (|Even|Odd|) x = if (x % 2) = 0 then Even x else Odd x
// fold function to generate a state tupple of current values and accumulated values
let folder (current, result) x =
match x, current with
| Even x, _ -> x::current, result // even members a added to current list
| Odd x, [] -> current, result // odd members are ignored when current is empty
| Odd x, _ -> [], current::result // odd members starts a new current
// test on data
[2; 4; 3; 7; 6; 10; 4; 5]
|> List.rev // reverse list since numbers are added to start of current
|> List.fold folder ([], []) // perform fold over list
|> function | [],x -> x | y,x -> y::x // check that current is List.empty, otherwise add to result
How about this one?
let folder p l = function
| h::t when p(l) -> (l::h)::t
| []::_ as a -> a
| _ as a -> []::a
let f p ls =
ls
|> List.rev
|> List.fold (fun a l -> folder p l a) [[]]
|> List.filter ((<>) [])
At least the folder is crystal clear and effective, but then you pay the price for this by list reversing.
Here is a recursive solution based on a recursive List.filter
let rec _f p ls =
match ls with
|h::t -> if p(h) then
match f p t with
|rh::rt -> (h::rh)::rt
|[] -> (h::[])::[]
else []::f p t
|[] -> [[]]
let f p ls = _f p ls |> List.filter (fun t -> t <> [])
Having to filter at the end does seem inelegant though.
Here you go. This function should also have fairly good performance.
let groupedFilter (predicate : 'T -> bool) (list : 'T list) =
(([], []), list)
||> List.fold (fun (currentGroup, finishedGroups) el ->
if predicate el then
(el :: currentGroup), finishedGroups
else
match currentGroup with
| [] ->
[], finishedGroups
| _ ->
// This is the first non-matching element
// following a matching element.
// Finish processing the previous group then
// add it to the finished groups list.
[], ((List.rev currentGroup) :: finishedGroups))
// Need to do a little clean-up after the fold.
|> fun (currentGroup, finishedGroups) ->
// If the current group is non-empty, finish it
// and add it to the list of finished groups.
let finishedGroups =
match currentGroup with
| [] -> finishedGroups
| _ ->
(List.rev currentGroup) :: finishedGroups
// Reverse the finished groups list so the grouped
// elements will be in their original order.
List.rev finishedGroups;;
With the list reversing, I would like to go to #seq instead of list.
This version uses mutation (gasp!) internally for efficiency, but may also be a little slower with the overhead of seq. I think it is quite readable though.
let f p (ls) = seq {
let l = System.Collections.Generic.List<'a>()
for el in ls do
if p el then
l.Add el
else
if l.Count > 0 then yield l |> List.ofSeq
l.Clear()
if l.Count > 0 then yield l |> List.ofSeq
}
I can't think of a way to do this elegantly using higher order functions, but here's a solution using a list comprehension. I think it's fairly straightforward to read.
let f p ls =
let rec loop xs =
[ match xs with
| [] -> ()
| x::xs when p x ->
let group, rest = collectGroup [x] xs
yield group
yield! loop rest
| _::xs -> yield! loop xs ]
and collectGroup acc = function
| x::xs when p x -> collectGroup (x::acc) xs
| xs -> List.rev acc, xs
loop ls
(I am aware of this question, but it relates to sequences, which is not my problem here)
Given this input (for example):
let testlist =
[
"*text1";
"*text2";
"text3";
"text4";
"*text5";
"*text6";
"*text7"
]
let pred (s:string) = s.StartsWith("*")
I would like to be able to call MyFunc pred testlist and get this output:
[
["*text1";"*text2"];
["*text5";"*text6";"*text7"]
]
This is my current solution, but I don't really like the nested List.revs (ignore the fact that it takes Seq as input)
let shunt pred sq =
let shunter (prevpick, acc) (pick, a) =
match pick, prevpick with
| (true, true) -> (true, (a :: (List.hd acc)) :: (List.tl acc))
| (false, _) -> (false, acc)
| (true, _) -> (true, [a] :: acc)
sq
|> Seq.map (fun a -> (pred a, a))
|> Seq.fold shunter (false, [])
|> snd
|> List.map List.rev
|> List.rev
there is a List.partition function in the F# core library (in case you wanted to implement this just to have it working and not to learn how to write recursive functions yourself). Using this function, you can write this:
> testlist |> List.partition (fun s -> s.StartsWith("*"))
val it : string list * string list =
(["*text1"; "*text2"; "*text5"; "*text6"; "*text7"], ["text3"; "text4"])
Note that this function returns a tuple instead of returning a list of lists. This is a bit different to what you wanted, but if the predicate returns just true or false, then this makes more sense.
The implementation of partition function that returns tuples is also a bit simpler, so it may be useful for learning purposes:
let partition pred list =
// Helper function, which keeps results collected so
// far in 'accumulator' arguments outTrue and outFalse
let rec partitionAux list outTrue outFalse =
match list with
| [] ->
// We need to reverse the results (as we collected
// them in the opposite order!)
List.rev outTrue, List.rev outFalse
// Append element to one of the lists, depending on 'pred'
| x::xs when pred x -> partitionAux xs (x::outTrue) outFalse
| x::xs -> partitionAux xs outTrue (x::outFalse)
// Run the helper function
partitionAux list [] []
Edit: rev-less version using foldBack added below.
Here's some code that uses lists and tail-recursion:
//divides a list L into chunks for which all elements match pred
let divide pred L =
let rec aux buf acc L =
match L,buf with
//no more input and an empty buffer -> return acc
| [],[] -> List.rev acc
//no more input and a non-empty buffer -> return acc + rest of buffer
| [],buf -> List.rev (List.rev buf :: acc)
//found something that matches pred: put it in the buffer and go to next in list
| h::t,buf when pred h -> aux (h::buf) acc t
//found something that doesn't match pred. Continue but don't add an empty buffer to acc
| h::t,[] -> aux [] acc t
//found input that doesn't match pred. Add buffer to acc and continue with an empty buffer
| h::t,buf -> aux [] (List.rev buf :: acc) t
aux [] [] L
usage:
> divide pred testlist;;
val it : string list list =
[["*text1"; "*text2"]; ["*text5"; "*text6"; "*text7"]]
Using a list as data structure for a buffer means that it always needs to be reversed when outputting the contents. This may not be a problem if individual chunks are modestly sized. If speed/efficiency becomes an issue, you could use a Queue<'a> or a `List<'a>' for the buffers, for which appending is fast. But using these data structures instead of lists also means that you lose the powerful list pattern matching. In my opinion, being able to pattern match lists outweighs the presence of a few List.rev calls.
Here's a streaming version that outputs the result one block at a time. This avoids the List.rev on the accumulator in the previous example:
let dividestream pred L =
let rec aux buf L =
seq { match L, buf with
| [],[] -> ()
| [],buf -> yield List.rev buf
| h::t,buf when pred h -> yield! aux (h::buf) t
| h::t,[] -> yield! aux [] t
| h::t,buf -> yield List.rev buf
yield! aux [] t }
aux [] L
This streaming version avoids the List.rev on the accumulator. Using List.foldBack can be used to avoid reversing the accumulated chunks as well.
update: here's a version using foldBack
//divides a list L into chunks for which all elements match pred
let divide2 pred L =
let f x (acc,buf) =
match pred x,buf with
| true,buf -> (acc,x::buf)
| false,[] -> (acc,[])
| false,buf -> (buf::acc,[])
let rest,remainingBuffer = List.foldBack f L ([],[])
match remainingBuffer with
| [] -> rest
| buf -> buf :: rest
Just reverse the list once up front, and then build the structure in order easily:
let Shunt p l =
let mutable r = List.rev l
let mutable result = []
while not r.IsEmpty do
let mutable thisBatch = []
while not r.IsEmpty && not(p r.Head) do
r <- r.Tail
while not r.IsEmpty && p r.Head do
thisBatch <- r.Head :: thisBatch
r <- r.Tail
if not thisBatch.IsEmpty then
result <- thisBatch :: result
result
The outer while deals with each 'batch', and the first inner while skips over any that don't match the predicate, followed by another while that grabs all those that do and stores them in the current batch. If there was anything in this batch (the final one may be empty), prepend it to the final result.
This is an example where I think locally imperative code is simply superior to a purely functional counterpart. The code above is so easy to write and to reason about.
Another version of shunt:
let shunt pred lst =
let rec tWhile pred lst =
match lst with
| [] -> [], []
| hd :: tl when pred hd -> let taken, rest = tWhile pred tl
(hd :: taken), rest
| lst -> [], lst
let rec collect = function
| [] -> []
| lst -> let taken, rest = tWhile pred lst
taken :: (collect (snd (tWhile (fun x -> not (pred x)) rest)))
collect lst
This one avoids List.rev but it's not tail recursive - so only suitable for small lists.
yet another one...
let partition pred lst =
let rec trec xs cont =
match xs with
| [] -> ([],[]) |> cont
| h::t when pred h -> (fun (y,n) -> h::y,n) >> cont |> trec t
| h::t -> (fun (y,n) -> y,h::n) >> cont |> trec t
trec lst id
then we can define shunt:
let shunt pred lst = lst |> partition pred |> (fun (x,y) -> [x;y])
I was doing an exercise on F# Wiki Book on List (scroll to the bottom) to create a Pair method.
I was able to pair a integer list without problem but an F# exception was thrown for a string list. It is just too cryptic for me to decipher what the exception means for an F# beginner like me.
Here is my initial attempt to implementing Pair on fsi.exe
> let pair l =
- let rec loop acc = function
- | [] -> acc
- | (hd1 :: hd2 :: tl) -> loop ((hd1, hd2) :: acc) tl
- List.rev(loop [] l)
-
- printfn "%A" ([1..10] |> pair)
- printfn "%A" ([ "one"; "two"; "three"; "four"; "five" ] |> pair);;
let rec loop acc = function
-----------------------^
stdin(2,24): warning FS0025: Incomplete pattern matches on this expression.
For example, the value '[_]' will not be matched
val pair : 'a list -> ('a * 'a) list
[(1, 2); (3, 4); (5, 6); (7, 8); (9, 10)]
Microsoft.FSharp.Core.MatchFailureException:
Exception of type 'Microsoft.FSharp.Core.MatchFailureException' was thrown.
at FSI_0002.clo#2T.Invoke(List`1 acc, List`1 _arg1)
at FSI_0002.pair[T](List`1 l)
at <StartupCode$FSI_0002>.$FSI_0002._main()
stopped due to error
So Pair does work on integer version
and the function signature
val pair : 'a list -> ('a * 'a) list
indicates that Pair operates on a generic list.
Question: Then why would Pair not work on a string list?
[ANSWER] (my version)
Simply returning accumulated list for else case (_) did the trick.
And the warning is taken care of, as well.
let pair l =
let rec loop acc = function
// | [] -> acc
| (hd1 :: hd2 :: tl) -> loop ((hd1, hd2) :: acc) tl
| _ -> acc
List.rev(loop [] l)
printfn "%A" ([1..10] |> pair)
printfn "%A" ([ "one"; "two"; "three"; "four"; "five" ] |> pair)
[EDIT2] Well, I will also post my version of Unpair for completeness.
let unpair l = [for (a,b) in l do yield! a :: b :: []]
Here is somewhat flawed benchmarking using solution version against that of mine for 1 million item lists
#light
open System;
let pn l = printfn "%A" l
let duration f =
let startTime = DateTime.Now;
let returnValue = f()
let endTime = DateTime.Now;
printfn "Duration (ms): %f" (endTime - startTime).TotalMilliseconds
returnValue
let ll = [for a in 1..1000000 do yield (a)]
let tl = [for a in 1..1000000 do yield (a,a)]
let pair1 l =
let rec loop acc = function
| [] | [_] -> List.rev acc
| h1 :: h2 :: tl -> loop ((h1, h2) :: acc) tl
loop [] l
let unpair1 l =
let rec loop acc = function
| [] -> List.rev acc
| (h1, h2) :: tl -> loop (h2 :: h1 :: acc) tl
loop [] l
let pair2 l =
let rec loop acc = function
| (hd1 :: hd2 :: tl) -> loop ((hd1, hd2) :: acc) tl
| _ | [_] -> acc
List.rev(loop [] l)
let unpair2 l = [for (a,b) in l do yield! a :: b :: []]
pn(duration (fun() -> ll |> pair1))
pn(duration (fun() -> tl |> unpair1))
pn(duration (fun() -> ll |> pair2))
pn(duration (fun() -> tl |> unpair2))
Benchmark Result:
Solution version
PAIR -> Duration (ms): 255.000000
UNPAIR -> Duration (ms): 840.000000
My version
PAIR -> Duration (ms): 220.000000
UNPAIR -> Duration (ms): 1624.000000
I don't think your version of Pair would work on a list of an odd number of anything. You happen to test an even number of ints and a odd number of strings. I think your second argument to match implies a list of at least two members. So you break off 2 break off 2 and get to a list with 1 element and none of your conditions match.
[_] is a 1 item list with anything in it. You must provide a predicate that matches it.