Directly using recursion, write a function truesAndLength : bool list -> int * int that
returns both the length of the list (in the first component of the pair), and the number of
elements of the list that are true (in the second component). Your function must only iterate
over the elements of the list once. (Do not use any of the functions from the List module.)
this is my code so far:
let rec length bs =
match bs with
| [] -> 0
| b::bs -> 1 + length bs
let rec trues bs =
match bs with
| [] -> 0
| b::bs -> if b = true then 1 + trues bs else trues bs
let truesandlength bs =
let l = length bs
let t = trues bs
(l, t)
truesandlength [true; true; false]
This works by im iterating through the list 2 times, i can't figure out how to iterate only 1. Any tips?
Based on your comments, here's how I suggest you think about the b::bs case:
Call truesAndLengths recursively on the tail of the list. This gives you tTail (the # of trues) and lTail (the length) of the tail.
Compute t and l for the full list based on the value of b. (E.g. l is one more than lTail.)
Return t, l.
The key is to call truesAndLengths recursively in only one place in your code, passing it the tail of the list.
let rec truesAndLength bs =
let sum a b = (fst a + fst b, snd a + snd b)
match bs with
| [] -> 0, 0
| head::tail ->
if head then sum (1,1) (truesAndLength tail) else sum (1, 0) (truesAndLength tail)
Related
Directly using recursion, write a function truesAndLength : bool list -> int * int that
returns both the length of the list (in the first component of the pair), and the number of
elements of the list that are true (in the second component). Your function must only iterate
over the elements of the list once. (Do not use any of the functions from the List module.)
this is my code so far:
let rec length bs =
match bs with
| [] -> 0
| b::bs -> 1 + length bs
let rec trues bs =
match bs with
| [] -> 0
| b::bs -> if b = true then 1 + trues bs else trues bs
let truesandlength bs =
let l = length bs
let t = trues bs
(l, t)
truesandlength [true; true; false]
This works by im iterating through the list 2 times, i can't figure out how to iterate only 1. Any tips?
Based on your comments, here's how I suggest you think about the b::bs case:
Call truesAndLengths recursively on the tail of the list. This gives you tTail (the # of trues) and lTail (the length) of the tail.
Compute t and l for the full list based on the value of b. (E.g. l is one more than lTail.)
Return t, l.
The key is to call truesAndLengths recursively in only one place in your code, passing it the tail of the list.
let rec truesAndLength bs =
let sum a b = (fst a + fst b, snd a + snd b)
match bs with
| [] -> 0, 0
| head::tail ->
if head then sum (1,1) (truesAndLength tail) else sum (1, 0) (truesAndLength tail)
I'm trying to find the maximum element in a list without using List.Max for a school assignment using the below given template.
let findMax l =
let rec helper(l,m) = failwith "Not implemented"
match l with
| [] -> failwith "Error -- empty list"
| (x::xs) -> helper(xs,x)
The only solution to the problem I can think of, atm is
let rec max_value1 l =
match l with
|[] -> failwith "Empty List"
|[x] -> x
|(x::y::xs) -> if x<y then max_value1 (y::xs)
else max_value1 (x::xs)
max_value1 [1; 17; 3; 6; 1; 8; 3; 11; 6; 5; 9];;
Is there any way I can go from the function I built to one that uses the template? Thanks!
Your helper function should do the work, the outer function just validates that the list is not empty and if it's not, calls the helper, which should be something like this:
let rec helper (l,m) =
match (l, m) with
| [] , m -> m
| x::xs, m -> helper (xs, max m x)
Note, that you since you're matching against the last argument of the function you can remove it and use function instead of match with:
let rec helper = function
| [] , m -> m
| x::xs, m -> helper (xs, max m x)
let findMax l =
let rec helper(l,m) =
match l with
| [] -> m
| (x::xs) -> helper(xs, if (Some x > m) then Some x else m)
helper (l,None)
Example:
[-2;-6;-1;-9;-56;-3] |> findMax
val it : int option = Some -1
An empty list will return None.
You could go for a tuple to pass both, or simply apply the helper function in your main match (instead of the empty list guard clause). I'm including the answer for someone who might find this question in the future and not have a clear answer.
let findMax l =
let rec walk maxValue = function
| [] -> maxValue
| (x::xs) -> walk (if x > maxValue then x else maxValue) xs
match l with
| [] -> failwith "Empty list"
| (head::tail) -> walk head tail
findMax [1; 12; 3; ] //12
Using fold:
let findMax l = l |> List.fold (fun maxValue x -> if x > maxValue then x else maxValue) (List.head l)
I am not sure of what the exact rules of your assigment are but the max of a list is really just List.reduce max. So
let listMax : int list -> int = List.reduce max
You need the type annotation to please the typechecker.
let inline listMax xs = List.reduce max xs
also works and is generic so it works with e.g. floats and strings as well.
I have two snippets of code that tries to convert a float list to a Vector3 or Vector2 list. The idea is to take 2/3 elements at a time from the list and combine them as a vector. The end result is a sequence of vectors.
let rec vec3Seq floatList =
seq {
match floatList with
| x::y::z::tail -> yield Vector3(x,y,z)
yield! vec3Seq tail
| [] -> ()
| _ -> failwith "float array not multiple of 3?"
}
let rec vec2Seq floatList =
seq {
match floatList with
| x::y::tail -> yield Vector2(x,y)
yield! vec2Seq tail
| [] -> ()
| _ -> failwith "float array not multiple of 2?"
}
The code looks very similiar and yet there seems to be no way to extract a common portion. Any ideas?
Here's one approach. I'm not sure how much simpler this really is, but it does abstract some of the repeated logic out.
let rec mkSeq (|P|_|) x =
seq {
match x with
| P(p,tail) ->
yield p
yield! mkSeq (|P|_|) tail
| [] -> ()
| _ -> failwith "List length mismatch" }
let vec3Seq =
mkSeq (function
| x::y::z::tail -> Some(Vector3(x,y,z), tail)
| _ -> None)
As Rex commented, if you want this only for two cases, then you probably won't have any problem if you leave the code as it is. However, if you want to extract a common pattern, then you can write a function that splits a list into sub-list of a specified length (2 or 3 or any other number). Once you do that, you'll only use map to turn each list of the specified length into Vector.
The function for splitting list isn't available in the F# library (as far as I can tell), so you'll have to implement it yourself. It can be done roughly like this:
let divideList n list =
// 'acc' - accumulates the resulting sub-lists (reversed order)
// 'tmp' - stores values of the current sub-list (reversed order)
// 'c' - the length of 'tmp' so far
// 'list' - the remaining elements to process
let rec divideListAux acc tmp c list =
match list with
| x::xs when c = n - 1 ->
// we're adding last element to 'tmp',
// so we reverse it and add it to accumulator
divideListAux ((List.rev (x::tmp))::acc) [] 0 xs
| x::xs ->
// add one more value to 'tmp'
divideListAux acc (x::tmp) (c+1) xs
| [] when c = 0 -> List.rev acc // no more elements and empty 'tmp'
| _ -> failwithf "not multiple of %d" n // non-empty 'tmp'
divideListAux [] [] 0 list
Now, you can use this function to implement your two conversions like this:
seq { for [x; y] in floatList |> divideList 2 -> Vector2(x,y) }
seq { for [x; y; z] in floatList |> divideList 3 -> Vector3(x,y,z) }
This will give a warning, because we're using an incomplete pattern that expects that the returned lists will be of length 2 or 3 respectively, but that's correct expectation, so the code will work fine. I'm also using a brief version of sequence expression the -> does the same thing as do yield, but it can be used only in simple cases like this one.
This is simular to kvb's solution but doesn't use a partial active pattern.
let rec listToSeq convert (list:list<_>) =
seq {
if not(List.isEmpty list) then
let list, vec = convert list
yield vec
yield! listToSeq convert list
}
let vec2Seq = listToSeq (function
| x::y::tail -> tail, Vector2(x,y)
| _ -> failwith "float array not multiple of 2?")
let vec3Seq = listToSeq (function
| x::y::z::tail -> tail, Vector3(x,y,z)
| _ -> failwith "float array not multiple of 3?")
Honestly, what you have is pretty much as good as it can get, although you might be able to make a little more compact using this:
// take 3 [1 .. 5] returns ([1; 2; 3], [4; 5])
let rec take count l =
match count, l with
| 0, xs -> [], xs
| n, x::xs -> let res, xs' = take (count - 1) xs in x::res, xs'
| n, [] -> failwith "Index out of range"
// split 3 [1 .. 6] returns [[1;2;3]; [4;5;6]]
let rec split count l =
seq { match take count l with
| xs, ys -> yield xs; if ys <> [] then yield! split count ys }
let vec3Seq l = split 3 l |> Seq.map (fun [x;y;z] -> Vector3(x, y, z))
let vec2Seq l = split 2 l |> Seq.map (fun [x;y] -> Vector2(x, y))
Now the process of breaking up your lists is moved into its own generic "take" and "split" functions, its much easier to map it to your desired type.
Background:
I have a sequence of contiguous, time-stamped data. The data-sequence has gaps in it where the data is not contiguous. I want create a method to split the sequence up into a sequence of sequences so that each subsequence contains contiguous data (split the input-sequence at the gaps).
Constraints:
The return value must be a sequence of sequences to ensure that elements are only produced as needed (cannot use list/array/cacheing)
The solution must NOT be O(n^2), probably ruling out a Seq.take - Seq.skip pattern (cf. Brian's post)
Bonus points for a functionally idiomatic approach (since I want to become more proficient at functional programming), but it's not a requirement.
Method signature
let groupContiguousDataPoints (timeBetweenContiguousDataPoints : TimeSpan) (dataPointsWithHoles : seq<DateTime * float>) : (seq<seq< DateTime * float >>)= ...
On the face of it the problem looked trivial to me, but even employing Seq.pairwise, IEnumerator<_>, sequence comprehensions and yield statements, the solution eludes me. I am sure that this is because I still lack experience with combining F#-idioms, or possibly because there are some language-constructs that I have not yet been exposed to.
// Test data
let numbers = {1.0..1000.0}
let baseTime = DateTime.Now
let contiguousTimeStamps = seq { for n in numbers ->baseTime.AddMinutes(n)}
let dataWithOccationalHoles = Seq.zip contiguousTimeStamps numbers |> Seq.filter (fun (dateTime, num) -> num % 77.0 <> 0.0) // Has a gap in the data every 77 items
let timeBetweenContiguousValues = (new TimeSpan(0,1,0))
dataWithOccationalHoles |> groupContiguousDataPoints timeBetweenContiguousValues |> Seq.iteri (fun i sequence -> printfn "Group %d has %d data-points: Head: %f" i (Seq.length sequence) (snd(Seq.hd sequence)))
I think this does what you want
dataWithOccationalHoles
|> Seq.pairwise
|> Seq.map(fun ((time1,elem1),(time2,elem2)) -> if time2-time1 = timeBetweenContiguousValues then 0, ((time1,elem1),(time2,elem2)) else 1, ((time1,elem1),(time2,elem2)) )
|> Seq.scan(fun (indexres,(t1,e1),(t2,e2)) (index,((time1,elem1),(time2,elem2))) -> (index+indexres,(time1,elem1),(time2,elem2)) ) (0,(baseTime,-1.0),(baseTime,-1.0))
|> Seq.map( fun (index,(time1,elem1),(time2,elem2)) -> index,(time2,elem2) )
|> Seq.filter( fun (_,(_,elem)) -> elem <> -1.0)
|> PSeq.groupBy(fst)
|> Seq.map(snd>>Seq.map(snd))
Thanks for asking this cool question
I translated Alexey's Haskell to F#, but it's not pretty in F#, and still one element too eager.
I expect there is a better way, but I'll have to try again later.
let N = 20
let data = // produce some arbitrary data with holes
seq {
for x in 1..N do
if x % 4 <> 0 && x % 7 <> 0 then
printfn "producing %d" x
yield x
}
let rec GroupBy comp (input:LazyList<'a>) : LazyList<LazyList<'a>> =
LazyList.delayed (fun () ->
match input with
| LazyList.Nil -> LazyList.cons (LazyList.empty()) (LazyList.empty())
| LazyList.Cons(x,LazyList.Nil) ->
LazyList.cons (LazyList.cons x (LazyList.empty())) (LazyList.empty())
| LazyList.Cons(x,(LazyList.Cons(y,_) as xs)) ->
let groups = GroupBy comp xs
if comp x y then
LazyList.consf
(LazyList.consf x (fun () ->
let (LazyList.Cons(firstGroup,_)) = groups
firstGroup))
(fun () ->
let (LazyList.Cons(_,otherGroups)) = groups
otherGroups)
else
LazyList.cons (LazyList.cons x (LazyList.empty())) groups)
let result = data |> LazyList.of_seq |> GroupBy (fun x y -> y = x + 1)
printfn "Consuming..."
for group in result do
printfn "about to do a group"
for x in group do
printfn " %d" x
You seem to want a function that has signature
(`a -> bool) -> seq<'a> -> seq<seq<'a>>
I.e. a function and a sequence, then break up the input sequence into a sequence of sequences based on the result of the function.
Caching the values into a collection that implements IEnumerable would likely be simplest (albeit not exactly purist, but avoiding iterating the input multiple times. It will lose much of the laziness of the input):
let groupBy (fun: 'a -> bool) (input: seq) =
seq {
let cache = ref (new System.Collections.Generic.List())
for e in input do
(!cache).Add(e)
if not (fun e) then
yield !cache
cache := new System.Collections.Generic.List()
if cache.Length > 0 then
yield !cache
}
An alternative implementation could pass cache collection (as seq<'a>) to the function so it can see multiple elements to chose the break points.
A Haskell solution, because I don't know F# syntax well, but it should be easy enough to translate:
type TimeStamp = Integer -- ticks
type TimeSpan = Integer -- difference between TimeStamps
groupContiguousDataPoints :: TimeSpan -> [(TimeStamp, a)] -> [[(TimeStamp, a)]]
There is a function groupBy :: (a -> a -> Bool) -> [a] -> [[a]] in the Prelude:
The group function takes a list and returns a list of lists such that the concatenation of the result is equal to the argument. Moreover, each sublist in the result contains only equal elements. For example,
group "Mississippi" = ["M","i","ss","i","ss","i","pp","i"]
It is a special case of groupBy, which allows the programmer to supply their own equality test.
It isn't quite what we want, because it compares each element in the list with the first element of the current group, and we need to compare consecutive elements. If we had such a function groupBy1, we could write groupContiguousDataPoints easily:
groupContiguousDataPoints maxTimeDiff list = groupBy1 (\(t1, _) (t2, _) -> t2 - t1 <= maxTimeDiff) list
So let's write it!
groupBy1 :: (a -> a -> Bool) -> [a] -> [[a]]
groupBy1 _ [] = [[]]
groupBy1 _ [x] = [[x]]
groupBy1 comp (x : xs#(y : _))
| comp x y = (x : firstGroup) : otherGroups
| otherwise = [x] : groups
where groups#(firstGroup : otherGroups) = groupBy1 comp xs
UPDATE: it looks like F# doesn't let you pattern match on seq, so it isn't too easy to translate after all. However, this thread on HubFS shows a way to pattern match sequences by converting them to LazyList when needed.
UPDATE2: Haskell lists are lazy and generated as needed, so they correspond to F#'s LazyList (not to seq, because the generated data is cached (and garbage collected, of course, if you no longer hold a reference to it)).
(EDIT: This suffers from a similar problem to Brian's solution, in that iterating the outer sequence without iterating over each inner sequence will mess things up badly!)
Here's a solution that nests sequence expressions. The imperitave nature of .NET's IEnumerable<T> is pretty apparent here, which makes it a bit harder to write idiomatic F# code for this problem, but hopefully it's still clear what's going on.
let groupBy cmp (sq:seq<_>) =
let en = sq.GetEnumerator()
let rec partitions (first:option<_>) =
seq {
match first with
| Some first' -> //'
(* The following value is always overwritten;
it represents the first element of the next subsequence to output, if any *)
let next = ref None
(* This function generates a subsequence to output,
setting next appropriately as it goes *)
let rec iter item =
seq {
yield item
if (en.MoveNext()) then
let curr = en.Current
if (cmp item curr) then
yield! iter curr
else // consumed one too many - pass it on as the start of the next sequence
next := Some curr
else
next := None
}
yield iter first' (* ' generate the first sequence *)
yield! partitions !next (* recursively generate all remaining sequences *)
| None -> () // return an empty sequence if there are no more values
}
let first = if en.MoveNext() then Some en.Current else None
partitions first
let groupContiguousDataPoints (time:TimeSpan) : (seq<DateTime*_> -> _) =
groupBy (fun (t,_) (t',_) -> t' - t <= time)
Okay, trying again. Achieving the optimal amount of laziness turns out to be a bit difficult in F#... On the bright side, this is somewhat more functional than my last attempt, in that it doesn't use any ref cells.
let groupBy cmp (sq:seq<_>) =
let en = sq.GetEnumerator()
let next() = if en.MoveNext() then Some en.Current else None
(* this function returns a pair containing the first sequence and a lazy option indicating the first element in the next sequence (if any) *)
let rec seqStartingWith start =
match next() with
| Some y when cmp start y ->
let rest_next = lazy seqStartingWith y // delay evaluation until forced - stores the rest of this sequence and the start of the next one as a pair
seq { yield start; yield! fst (Lazy.force rest_next) },
lazy Lazy.force (snd (Lazy.force rest_next))
| next -> seq { yield start }, lazy next
let rec iter start =
seq {
match (Lazy.force start) with
| None -> ()
| Some start ->
let (first,next) = seqStartingWith start
yield first
yield! iter next
}
Seq.cache (iter (lazy next()))
Below is some code that does what I think you want. It is not idiomatic F#.
(It may be similar to Brian's answer, though I can't tell because I'm not familiar with the LazyList semantics.)
But it doesn't exactly match your test specification: Seq.length enumerates its entire input. Your "test code" calls Seq.length and then calls Seq.hd. That will generate an enumerator twice, and since there is no caching, things get messed up. I'm not sure if there is any clean way to allow multiple enumerators without caching. Frankly, seq<seq<'a>> may not be the best data structure for this problem.
Anyway, here's the code:
type State<'a> = Unstarted | InnerOkay of 'a | NeedNewInner of 'a | Finished
// f() = true means the neighbors should be kept together
// f() = false means they should be split
let split_up (f : 'a -> 'a -> bool) (input : seq<'a>) =
// simple unfold that assumes f captured a mutable variable
let iter f = Seq.unfold (fun _ ->
match f() with
| Some(x) -> Some(x,())
| None -> None) ()
seq {
let state = ref (Unstarted)
use ie = input.GetEnumerator()
let innerMoveNext() =
match !state with
| Unstarted ->
if ie.MoveNext()
then let cur = ie.Current
state := InnerOkay(cur); Some(cur)
else state := Finished; None
| InnerOkay(last) ->
if ie.MoveNext()
then let cur = ie.Current
if f last cur
then state := InnerOkay(cur); Some(cur)
else state := NeedNewInner(cur); None
else state := Finished; None
| NeedNewInner(last) -> state := InnerOkay(last); Some(last)
| Finished -> None
let outerMoveNext() =
match !state with
| Unstarted | NeedNewInner(_) -> Some(iter innerMoveNext)
| InnerOkay(_) -> failwith "Move to next inner seq when current is active: undefined behavior."
| Finished -> None
yield! iter outerMoveNext }
open System
let groupContigs (contigTime : TimeSpan) (holey : seq<DateTime * int>) =
split_up (fun (t1,_) (t2,_) -> (t2 - t1) <= contigTime) holey
// Test data
let numbers = {1 .. 15}
let contiguousTimeStamps =
let baseTime = DateTime.Now
seq { for n in numbers -> baseTime.AddMinutes(float n)}
let holeyData =
Seq.zip contiguousTimeStamps numbers
|> Seq.filter (fun (dateTime, num) -> num % 7 <> 0)
let grouped_data = groupContigs (new TimeSpan(0,1,0)) holeyData
printfn "Consuming..."
for group in grouped_data do
printfn "about to do a group"
for x in group do
printfn " %A" x
Ok, here's an answer I'm not unhappy with.
(EDIT: I am unhappy - it's wrong! No time to try to fix right now though.)
It uses a bit of imperative state, but it is not too difficult to follow (provided you recall that '!' is the F# dereference operator, and not 'not'). It is as lazy as possible, and takes a seq as input and returns a seq of seqs as output.
let N = 20
let data = // produce some arbitrary data with holes
seq {
for x in 1..N do
if x % 4 <> 0 && x % 7 <> 0 then
printfn "producing %d" x
yield x
}
let rec GroupBy comp (input:seq<_>) = seq {
let doneWithThisGroup = ref false
let areMore = ref true
use e = input.GetEnumerator()
let Next() = areMore := e.MoveNext(); !areMore
// deal with length 0 or 1, seed 'prev'
if not(e.MoveNext()) then () else
let prev = ref e.Current
while !areMore do
yield seq {
while not(!doneWithThisGroup) do
if Next() then
let next = e.Current
doneWithThisGroup := not(comp !prev next)
yield !prev
prev := next
else
// end of list, yield final value
yield !prev
doneWithThisGroup := true }
doneWithThisGroup := false }
let result = data |> GroupBy (fun x y -> y = x + 1)
printfn "Consuming..."
for group in result do
printfn "about to do a group"
for x in group do
printfn " %d" x
Slowly getting the hang of List matching and tail recursion, I needed a function which 'stitches' a list of lists together leaving off intermediate values (easier to show than explain):
merge [[1;2;3];[3;4;5];[5;6;7]] //-> [1;2;3;4;5;6;7]
The code for the List.merge function looks like this:
///Like concat, but removes first value of each inner list except the first one
let merge lst =
let rec loop acc lst =
match lst with
| [] -> acc
| h::t ->
match acc with
| [] -> loop (acc # h) t
| _ -> loop (acc # (List.tl h)) t //first time omit first value
loop [] lst
(OK, it's not quite like concat, because it only handles two levels of list)
Question: How to do this for a Seq of Seqs (without using a mutable flag)?
UPDATE (re comment from Juliet):
My code creates 'paths' composed of 'segments' which are based on an option type:
type SegmentDef = Straight of float | Curve of float * float
let Project sampleinterval segdefs = //('clever' code here)
When I do a List.map (Project 1.) ListOfSegmentDefs, I get back a list where each segment begins on the same point where the previous segment ends. I want to join these lists together to get a Path, keeping only the 'top/tail' of each overlap - but I don't need to do a 'Set', because I know that I don't have any other duplicates.
This is essentially the same as your first solution, but a little more succinct:
let flatten l =
seq {
yield Seq.hd (Seq.hd l) (* first item of first list *)
for a in l do yield! (Seq.skip 1 a) (* other items *)
}
[Edit to add]:
If you need a List version of this code, use append |> Seq.to_list at the end of your method:
let flatten l =
seq {
yield Seq.hd (Seq.hd l) (* first item of first list *)
for a in l do yield! (Seq.skip 1 a) (* other items *)
} |> Seq.to_list
let merge = function
| [] -> []
| xs::xss -> xs # [for _::xs in xss do yield! xs]
or:
let merge = function
| [] -> []
| xs::xss -> xs # List.collect List.tail xss