List.Sort
sorts a list from low to high - How does one sort from high to low? Is there some kind of library function for this?
For a list of numbers:
list
|> List.sortBy (fun x -> -x)
The function (fun x -> -x) negates the number, therefore reversing the order.
For comparables in general, use List.sortWith with compare. Observe the ordering of a b in compare:
> List.sortWith (fun a b -> compare a b) ["a";"s";"d";"f"];;
val it : string list = ["a"; "d"; "f"; "s"]
> List.sortWith (fun a b -> compare b a) ["a";"s";"d";"f"];;
val it : string list = ["s"; "f"; "d"; "a"]
If you looked at the linked thread F# Seq.sortBy in descending order, there is a chance of overflow when you use List.sortBy (fun x -> -x). To be correct, it should be:
List.sortBy (fun x -> -x-1)
In F# 4.0 (that comes with Visual Studio 2015 Preview), there are sortDescending/sortByDescending functions for this exact purpose.
You can use
list
|> List.sortDescending
or
list
|> List.sortByDescending id
See the comprehensive list of new core library functions at https://github.com/fsharp/FSharpLangDesign/blob/master/FSharp-4.0/ListSeqArrayAdditions.md.
You can use List.sortBy to sort by a custom function, and use the unary minus operator ~- as such function in a compact notation:
let list = [1..10]
list |> List.sortBy (~-)
Related
I'd like to implement something akin to imaginary Array.multipick:
Array.multipick : choosers:('a -> bool) [] -> array:'a [] -> 'a []
Internally, we test each array's element with all choosers, the first chooser to return true is removed from choosers array, and we add that chooser's argument to the result. After that, we continue interation while choosers array has elements left.
The last part is important, because without early exit requirement this could be solved with just Array.fold.
This could be easily implemented with something like:
let rec impl currentIndex currentChoosers results
But it's too procedural for my taste. Maybe there's more elegant solution?
It's quite difficult to write elegant code using arrays of changing size. Here is some code that works on lists instead and does not mutate any values.
let rec pick accum elem tried = function
| [] -> (accum, List.rev tried)
| chooser :: rest ->
if chooser elem then (elem :: accum, List.rev_append tried rest)
else pick accum elem (chooser :: tried) rest
let rec multipick_l accum choosers list =
match choosers, list with
| [], _
| _, [] -> List.rev accum
| _, elem :: elems ->
let (accum', choosers') = pick accum elem [] choosers in
multipick_l accum' choosers' elems
let multipick choosers array =
Array.of_list
(multipick_l [] (Array.to_list choosers) (Array.to_list array))
If you think that Array.fold_left is usable except for the early exit requirement, you can use an exception to exit early.
A fold with an early exit is a good idea, however a production-worthy one specifically targeting arrays would need to be written in a fairly imperative manner. For simplicity, I'll grab the more general sequence one from this answer.
let multipick (choosers: ('a -> bool) array) (arr: 'a array) : 'a array =
let indexed =
choosers
|> Seq.indexed
|> Map.ofSeq
((indexed, []), arr)
||> foldWhile (fun (cs, res) e ->
if Map.isEmpty cs then
None
else
match cs |> Seq.tryFind (fun kvp -> kvp.Value e) with
| Some kvp -> Some (Map.remove kvp.Key cs, e :: res)
| None -> Some (cs, res))
|> snd
|> List.rev
|> Array.ofList
I'm using a Map keyed by array index to keep track of remaining functions - this allows for easy removal of elements, but still retains their order (since map key-value pairs are ordered by keys when iterating).
F# Set wouldn't work with functions due to comparison constraint. System.Collections.Generic.HashSet would work, but it's mutable, and I'm not sure if it would retain ordering.
I'm new to F# and I'm trying to figure out how to return a random string value from a list/array of strings.
I have a list like this:
["win8FF40", "win10Chrome45", "win7IE11"]
How can I randomly select and return one item from the list above?
Here is my first try:
let combos = ["win8FF40";"win10Chrome45";"win7IE11"]
let getrandomitem () =
let rnd = System.Random()
fun (combos : string[]) -> combos.[rnd.Next(combos.Length)]
Both the answers given here by latkin and mydogisbox are good, but I still want to add a third approach that I sometimes use. This approach isn't faster, but it's more flexible and more composable, and fast enough for small sequences. Depending on your needs, you can use one of higher performance options given here, or you can use the following.
Single-argument function using Random
Instead of directly enabling you to select a single element, I often define a shuffleR function like this:
open System
let shuffleR (r : Random) xs = xs |> Seq.sortBy (fun _ -> r.Next())
This function has the type System.Random -> seq<'a> -> seq<'a>, so it works with any sort of sequence: lists, arrays, collections, and lazily evaluated sequences (although not with infinite sequences).
If you want a single random element from a list, you can still do that:
> [1..100] |> shuffleR (Random ()) |> Seq.head;;
val it : int = 85
but you can also take, say, three randomly picked elements:
> [1..100] |> shuffleR (Random ()) |> Seq.take 3;;
val it : seq<int> = seq [95; 92; 12]
No-argument function
Sometimes, I don't care about having to pass in that Random value, so I instead define this alternative version:
let shuffleG xs = xs |> Seq.sortBy (fun _ -> Guid.NewGuid())
It works in the same way:
> [1..100] |> shuffleG |> Seq.head;;
val it : int = 11
> [1..100] |> shuffleG |> Seq.take 3;;
val it : seq<int> = seq [69; 61; 42]
Although the purpose of Guid.NewGuid() isn't to provide random numbers, it's often random enough for my purposes - random, in the sense of being unpredictable.
Generalised function
Neither shuffleR nor shuffleG are truly random. Due to the ways Random and Guid.NewGuid() work, both functions may result in slightly skewed distributions. If this is a concern, you can define an even more general-purpose shuffle function:
let shuffle next xs = xs |> Seq.sortBy (fun _ -> next())
This function has the type (unit -> 'a) -> seq<'b> -> seq<'b> when 'a : comparison. It can still be used with Random:
> let r = Random();;
val r : Random
> [1..100] |> shuffle (fun _ -> r.Next()) |> Seq.take 3;;
val it : seq<int> = seq [68; 99; 54]
> [1..100] |> shuffle (fun _ -> r.Next()) |> Seq.take 3;;
val it : seq<int> = seq [99; 63; 11]
but you can also use it with some of the cryptographically secure random number generators provided by the Base Class Library:
open System.Security.Cryptography
open System.Collections.Generic
let rng = new RNGCryptoServiceProvider ()
let bytes = Array.zeroCreate<byte> 100
rng.GetBytes bytes
let q = bytes |> Queue
FSI:
> [1..100] |> shuffle (fun _ -> q.Dequeue()) |> Seq.take 3;;
val it : seq<int> = seq [74; 82; 61]
Unfortunately, as you can see from this code, it's quite cumbersome and brittle. You have to know the length of the sequence up front; RNGCryptoServiceProvider implements IDisposable, so you should make sure to dispose of rng after use; and items will be removed from q after use, which means it's not reusable.
Cryptographically random sort or selection
Instead, if you really need a cryptographically correct sort or selection, it'd be easier to do it like this:
let shuffleCrypto xs =
let a = xs |> Seq.toArray
use rng = new RNGCryptoServiceProvider ()
let bytes = Array.zeroCreate a.Length
rng.GetBytes bytes
Array.zip bytes a |> Array.sortBy fst |> Array.map snd
Usage:
> [1..100] |> shuffleCrypto |> Array.head;;
val it : int = 37
> [1..100] |> shuffleCrypto |> Array.take 3;;
val it : int [] = [|35; 67; 36|]
This isn't something I've ever had to do, though, but I thought I'd include it here for the sake of completeness. While I haven't measured it, it's most likely not the fastest implementation, but it should be cryptographically random.
Your problem is that you are mixing Arrays and F# Lists (*type*[] is a type notation for Array). You could modify it like this to use lists:
let getrandomitem () =
let rnd = System.Random()
fun (combos : string list) -> List.nth combos (rnd.Next(combos.Length))
That being said, indexing into a List is usually a bad idea since it has O(n) performance since an F# list is basically a linked-list. You would be better off making combos into an array if possible like this:
let combos = [|"win8FF40";"win10Chrome45";"win7IE11"|]
I wrote a blog post on exactly this topic a while ago: http://latkin.org/blog/2013/11/16/selecting-a-random-element-from-a-linked-list-3-approaches-in-f/
3 approaches are given there, with discussion of performance and tradeoffs of each.
To summarize:
// pro: simple, fast in practice
// con: 2-pass (once to get length, once to select nth element)
let method1 lst (rng : Random) =
List.nth lst (rng.Next(List.length lst))
// pro: ~1 pass, list length is not bound by int32
// con: more complex, slower in practice
let method2 lst (rng : Random) =
let rec step remaining picks top =
match (remaining, picks) with
| ([], []) -> failwith "Don't pass empty list"
// if only 1 element is picked, this is the result
| ([], [p]) -> p
// if multiple elements are picked, select randomly from them
| ([], ps) -> step ps [] -1
| (h :: t, ps) ->
match rng.Next() with
// if RNG makes new top number, picks list is reset
| n when n > top -> step t [h] n
// if RNG ties top number, add current element to picks list
| n when n = top -> step t (h::ps) top
// otherwise ignore and move to next element
| _ -> step t ps top
step lst [] -1
// pro: exactly 1 pass
// con: more complex, slowest in practice due to tuple allocations
let method3 lst (rng : Random) =
snd <| List.fold (fun (i, pick) elem ->
if rng.Next(i) = 0 then (i + 1, elem)
else (i + 1, pick)
) (0, List.head lst) lst
Edit: I should clarify that above shows a few ways to get a random element from a list, assuming you must use a list. If it fits with the rest of your program's design, it is definitely more efficient to take a random element from an array.
I need a very efficient way to find duplicates in an unsorted sequence. This is what I came up with, but it has a few shortcomings, namely it
unnecessarily counts occurrences beyond 2
consumes the entire sequence before yielding duplicates
creates several intermediate sequences
module Seq =
let duplicates items =
items
|> Seq.countBy id
|> Seq.filter (snd >> ((<) 1))
|> Seq.map fst
Regardless of the shortcomings, I don't see a reason to replace this with twice the code. Is it possible to improve this with comparably concise code?
A more elegant functional solution:
let duplicates xs =
Seq.scan (fun xs x -> Set.add x xs) Set.empty xs
|> Seq.zip xs
|> Seq.choose (fun (x, xs) -> if Set.contains x xs then Some x else None)
Uses scan to accumulate sets of all elements seen so far. Then uses zip to combine each element with the set of elements before it. Finally, uses choose to filter out the elements that are in the set of previously-seen elements, i.e. the duplicates.
EDIT
Actually my original answer was completely wrong. Firstly, you don't want duplicates in your outputs. Secondly, you want performance.
Here is a purely functional solution that implements the algorithm you're after:
let duplicates xs =
(Map.empty, xs)
||> Seq.scan (fun xs x ->
match Map.tryFind x xs with
| None -> Map.add x false xs
| Some false -> Map.add x true xs
| Some true -> xs)
|> Seq.zip xs
|> Seq.choose (fun (x, xs) ->
match Map.tryFind x xs with
| Some false -> Some x
| None | Some true -> None)
This uses a map to track whether each element has been seen before once or many times and then emits the element if it is seen having only been seen once before, i.e. the first time it is duplicated.
Here is a faster imperative version:
let duplicates (xs: _ seq) =
seq { let d = System.Collections.Generic.Dictionary(HashIdentity.Structural)
let e = xs.GetEnumerator()
while e.MoveNext() do
let x = e.Current
let mutable seen = false
if d.TryGetValue(x, &seen) then
if not seen then
d.[x] <- true
yield x
else
d.[x] <- false }
This is around 2× faster than any of your other answers (at the time of writing).
Using a for x in xs do loop to enumerate the elements in a sequence is substantially slower than using GetEnumerator directly but generating your own Enumerator is not significantly faster than using a computation expression with yield.
Note that the TryGetValue member of Dictionary allows me to avoid allocation in the inner loop by mutating a stack allocated value whereas the TryGetValue extension member offered by F# (and used by kvb in his/her answer) allocates its return tuple.
Here's an imperative solution (which is admittedly slightly longer):
let duplicates items =
seq {
let d = System.Collections.Generic.Dictionary()
for i in items do
match d.TryGetValue(i) with
| false,_ -> d.[i] <- false // first observance
| true,false -> d.[i] <- true; yield i // second observance
| true,true -> () // already seen at least twice
}
This is the best "functional" solution I could come up with that doesn't consume the entire sequence up front.
let duplicates =
Seq.scan (fun (out, yielded:Set<_>, seen:Set<_>) item ->
if yielded.Contains item then
(None, yielded, seen)
else
if seen.Contains item then
(Some(item), yielded.Add item, seen.Remove item)
else
(None, yielded, seen.Add item)
) (None, Set.empty, Set.empty)
>> Seq.Choose (fun (x,_,_) -> x)
Assuming your sequence is finite, this solution requires one run on the sequence:
open System.Collections.Generic
let duplicates items =
let dict = Dictionary()
items |> Seq.fold (fun acc item ->
match dict.TryGetValue item with
| true, 2 -> acc
| true, 1 -> dict.[item] <- 2; item::acc
| _ -> dict.[item] <- 1; acc) []
|> List.rev
You can provide length of the sequence as the capacity of Dictionary, but it requires to enumerate the whole sequence once more.
EDIT:
To resolve 2nd problem, one could generate duplicates on demand:
open System.Collections.Generic
let duplicates items =
seq {
let dict = Dictionary()
for item in items do
match dict.TryGetValue item with
| true, 2 -> ()
| true, 1 -> dict.[item] <- 2; yield item
| _ -> dict.[item] <- 1
}
Functional solution:
let duplicates items =
let test (unique, result) v =
if not(unique |> Set.contains v) then (unique |> Set.add v ,result)
elif not(result |> Set.contains v) then (unique,result |> Set.add v)
else (unique, result)
items |> Seq.fold test (Set.empty, Set.empty) |> snd |> Set.toSeq
Here is my failed attempt at the problem any help would be appreciated.
I tried to come up with the best algo for the power set that worked on eager lists. This part seems to be working fine. The part I'm having trouble with is translating it to work with Sequences so it can run it on streaming\infinite lists. I really don't like the yield syntax maybe because I don't understand it well but I would rather have it without using the yield syntax as well.
//All Combinations of items in a list
//i.e. the Powerset given each item is unique
//Note: lists are eager so can't be used for infinite
let listCombinations xs =
List.fold (fun acc x ->
List.collect (fun ys -> ys::[x::ys]) acc) [[]] xs
//This works fine (Still interested if it could be faster)
listCombinations [1;2;3;4;5] |> Seq.iter (fun x -> printfn "%A" x)
//All Combinations of items in a sequence
//i.e. the Powerset given each item is unique
//Note: Not working
let seqCombinations xs =
Seq.fold (fun acc x ->
Seq.collect (fun ys ->
seq { yield ys
yield seq { yield x
yield! ys} }) acc) Seq.empty xs
//All Combinations of items in a sequence
//i.e. the Powerset given each item is unique
//Note: Not working (even wrong type signature)
let seqCombinations2 xs =
Seq.fold (fun acc x ->
Seq.collect (fun ys ->
Seq.append ys (Seq.append x ys)) acc) Seq.empty xs
//Sequences to test on
let infiniteSequence = Seq.initInfinite (fun i -> i + 1)
let finiteSequence = Seq.take 5 infiniteSequence
//This should work easy since its in a finite sequence
//But it does not, so their must be a bug in 'seqCombinations' above
for xs in seqCombinations finiteSequence do
for y in xs do
printfn "%A" y
//This one is much more difficult to get to work
//since its the powerset on the infinate sequence
//None the less If someone could help me find a way to make this work
//This is my ultimate goal
let firstFew = Seq.take 20 (seqCombinations infiniteSequence)
for xs in firstFew do
for y in xs do
printfn "%A" y
Your seqCombinations is almost correct, but you didn't translate it from lists to sequences properly. The equivalent of [[]] is not Seq.empty, but Seq.singleton Seq.empty:
let seqCombinations xs =
Seq.fold (fun acc x ->
Seq.collect (fun ys ->
seq { yield ys
yield seq { yield x
yield! ys} }) acc) (Seq.singleton Seq.empty) xs
The code above works for finite sequences. But for infinite ones, it doesn't work, because it first tries to reach the end, which it obviously never does for infinite sequences.
If you want a function that will work with infinite sequences I managed to figure out two ways, but neither of them is particularly nice. One of them uses mutable state:
let seqCombinations xs =
let combs = ref [[]]
seq {
yield! !combs
for x in xs do
let added = List.map (fun ys -> x::ys) !combs
yield! added
combs := !combs # added
}
The other is too much about dealing with details of seq<T>:
open System.Collections.Generic
let seqCombinations (xs : seq<_>) =
let rec combs acc (e : IEnumerator<_>) =
seq {
if (e.MoveNext()) then
let added = List.map (fun ys -> (e.Current)::ys) acc
yield! added
yield! combs (acc # added) e }
use enumerator = xs.GetEnumerator()
seq {
yield []
yield! combs [[]] enumerator
}
I think this would be much easier if you could treat infinite sequences as head and tail, like finite lists in F# or any sequence in Haskell. But it's certainly possible there is a nice way to express this in F#, and I just didn't find it.
I've asked a similar question recently at Generate powerset lazily and got some nice answers.
For powerset of finite sets, the answer by #Daniel in the above link is an efficient solution and probably suits your purpose. You can come up with a test case to compare between his approach and yours.
Regarding powerset of infinite sets, here is a bit of maths. According to Cantor's theorem, the power set of a countably infinite set is uncountably infinite. It means there is no way to enumerate powerset of all integers (which is countably infinite) even in a lazy way. The intuition is the same for real numbers; since real number is uncountably infinite, we can't actually model them using infinite sequences.
Therefore, there is no algorithm to enumerate powerset of a countably infinite set. Or that kind of algorithm just doesn't make sense.
This is sort of a joke, but will actually generate the correct result for an infinite sequence (it's just that it can't be proven--empirically, not mathematically).
let powerset s =
seq {
yield Seq.empty
for x in s -> seq [x]
}
I've recently written a piece of code to read some data from a file, store it in a tuple and sort all the collected data by the first element of the tuple. After some tests I've noticed that using Seq.sortBy (and Array.sortBy) is extremely slower than using IEnumerable.OrderBy.
Below are two snippets of code which should show the behaviour I'm talking about:
(filename
|> File.ReadAllLines
|> Array.Parallel.map(fun ln -> let arr = ln.Split([|' '|], StringSplitOptions.RemoveEmptyEntries)
|> Array.map(double)
|> Array.sort in arr.[0], arr.[1])
).OrderBy(new Func(fun (a,b) -> a))
and
filename
|> File.ReadAllLines
|> Array.Parallel.map(fun ln -> let arr = ln.Split([|' '|], StringSplitOptions.RemoveEmptyEntries) |> Array.map(double) |> Array.sort in arr.[0], arr.[1])
|> Seq.sortBy(fun (a,_) -> a)
On a file containing 100000 lines made of two doubles, on my computer the latter version takes over twice as long as the first one (no improvements are obtained if using Array.sortBy).
Ideas?
the f# implementation uses a structural comparison of the resulting key.
let sortBy keyf seq =
let comparer = ComparisonIdentity.Structural
mkDelayedSeq (fun () ->
(seq
|> to_list
|> List.sortWith (fun x y -> comparer.Compare(keyf x,keyf y))
|> to_array) :> seq<_>)
(also sort)
let sort seq =
mkDelayedSeq (fun () ->
(seq
|> to_list
|> List.sortWith Operators.compare
|> to_array) :> seq<_>)
both Operators.compare and the ComparisonIdentity.Structural.Compare become (eventually)
let inline GenericComparisonFast<'T> (x:'T) (y:'T) : int =
GenericComparisonIntrinsic x y
// lots of other types elided
when 'T : float = if (# "clt" x y : bool #)
then (-1)
else (# "cgt" x y : int #)
but the route to this for the Operator is entirely inline, thus the JIT compiler will end up inserting a direct double comparison instruction with no additional method invocation overhead except for the (required in both cases anyway) delegate invocation.
The sortBy uses a comparer so will go through an additional virtual method call but is basically about the same.
In comparison the OrderBy function also must go through virtual method calls for the equality (Using EqualityComparer<T>.Default) but the significant difference is that it sorts in place and uses the buffer created for this as the result. In comparison if you take a look at the sortBy you will see that it sorts the list (not in place, it uses the StableSortImplementation which appears to be merge sort) and then creates a copy of it as a new array. This additional copy (given the size of your input data) is likely the principle cause of the slow down though the differing sort implementations may also have an effect.
That said this is all guessing. If this area is a concern for you in performance terms then you should simply profile to find out what is taking the time.
If you wish to see what effect the sorting/copying change would have try this alternate:
// these are taken from the f# source so as to be consistent
// beware doing this, the compiler may know about such methods
open System.Collections.Generic
let mkSeq f =
{ new IEnumerable<'b> with
member x.GetEnumerator() = f()
interface System.Collections.IEnumerable with
member x.GetEnumerator() = (f() :> System.Collections.IEnumerator) }
let mkDelayedSeq (f: unit -> IEnumerable<'T>) =
mkSeq (fun () -> f().GetEnumerator())
// the function
let sortByFaster keyf seq =
let comparer = ComparisonIdentity.Structural
mkDelayedSeq (fun () ->
let buffer = Seq.to_array seq
Array.sortInPlaceBy (fun x y -> comparer.Compare(keyf x,keyf y)) buffer
buffer :> seq<_>)
I get some reasonable percentage speedups within the repl with very large (> million) input sequences but nothing like an order of magnitude. Your mileage, as always, may vary.
A difference of x2 is not much when sorts are O(n.log(n)).
Small differences in data structures (e.g. optimising for input being ICollection<T>) could make this scale of difference.
And F# is currently Beta (not so much focus on optimisation vs. getting the language and libraries right), plus the generality of F# functions (supporting partial application etc.) could lead to a slight slow down in calling speed: more than enough to account for the different.