Given a simple function, where we do pattern matching on the result of a recursive call, such as:
let rec sumProd = function
| [] -> (0,1)
| x::rest -> let (rSum,rProd) = sumProd rest
(x + rSum,x * rProd)
sumProd [2;5] //Expected (7, 10)
How would I go about changing it into something using higher order functions, e.g. foldBack?
let sumProdHigherOrder lst =
List.foldBack (fun x acc -> (acc + x, acc * x)) lst (0,0)
The above seemed almost like the way to do it, but calling it gives the error: The type 'int' does not match the type 'int * int'
sumProdHigherOrder [2;5] //Expected (7, 10)
What am I missing?
Your missing the tuple functions fst and snd:
List.foldBack (fun x acc -> (fst acc + x, snd acc * x)) [2;5] (0,1)
// val it : int * int = (7, 10)
Or even better, decomposing the tuple at the lambda. I see you just found it:
List.foldBack (fun x (s, m) -> (s + x, m * x)) [2;5] (0,1)
Also note that since the operations are commutative you can do a straight fold:
List.fold (fun (s, m) x -> (s + x, m * x)) (0,1) [2;5]
It will be more efficient.
Right! Of course it shouldn't be the same accumulator that gets passed through the list. After staring intensely at the code for some minutes, I figured it out:
let sumProdHigherOrder lst =
List.foldBack (fun x (acc,acc') -> (acc + x, acc' * x)) lst (0,1)
I am interested to implement fold3, fold4 etc., similar to List.fold and List.fold2. e.g.
// TESTCASE
let polynomial (x:double) a b c = a*x + b*x*x + c*x*x*x
let A = [2.0; 3.0; 4.0; 5.0]
let B = [1.5; 1.0; 0.5; 0.2]
let C = [0.8; 0.01; 0.001; 0.0001]
let result = fold3 polynomial 0.7 A B C
// 2.0 * (0.7 ) + 1.5 * (0.7 )^2 + 0.8 * (0.7 )^3 -> 2.4094
// 3.0 * (2.4094) + 1.0 * (2.4094)^2 + 0.01 * (2.4094)^3 -> 13.173
// 4.0 * (13.173) + 0.5 * (13.173)^2 + 0.001 * (13.173)^3 -> 141.75
// 5.0 * (141.75) + 0.2 * (141.75)^2 + 0.0001 * (141.75)^3 -> 5011.964
//
// Output: result = 5011.964
My first method is grouping the 3 lists A, B, C, into a list of tuples, and then apply list.fold
let fold3 f x A B C =
List.map3 (fun a b c -> (a,b,c)) A B C
|> List.fold (fun acc (a,b,c) -> f acc a b c) x
// e.g. creates [(2.0,1.5,0.8); (3.0,1.0,0.01); ......]
My second method is to declare a mutable data, and use List.map3
let mutable result = 0.7
List.map3 (fun a b c ->
result <- polynomial result a b c // Change mutable data
// Output intermediate data
result) A B C
// Output from List.map3: [2.4094; 13.17327905; 141.7467853; 5011.963942]
// result mutable: 5011.963942
I would like to know if there are other ways to solve this problem. Thank you.
For fold3, you could just do zip3 and then fold:
let polynomial (x:double) (a, b, c) = a*x + b*x*x + c*x*x*x
List.zip3 A B C |> List.fold polynomial 0.7
But if you want this for the general case, then you need what we call "applicative functors".
First, imagine you have a list of functions and a list of values. Let's assume for now they're of the same size:
let fs = [ (fun x -> x+1); (fun x -> x+2); (fun x -> x+3) ]
let xs = [3;5;7]
And what you'd like to do (only natural) is to apply each function to each value. This is easily done with List.map2:
let apply fs xs = List.map2 (fun f x -> f x) fs xs
apply fs xs // Result = [4;7;10]
This operation "apply" is why these are called "applicative functors". Not just any ol' functors, but applicative ones. (the reason for why they're "functors" is a tad more complicated)
So far so good. But wait! What if each function in my list of functions returned another function?
let f1s = [ (fun x -> fun y -> x+y); (fun x -> fun y -> x-y); (fun x -> fun y -> x*y) ]
Or, if I remember that fun x -> fun y -> ... can be written in the short form of fun x y -> ...
let f1s = [ (fun x y -> x+y); (fun x y -> x-y); (fun x y -> x*y) ]
What if I apply such list of functions to my values? Well, naturally, I'll get another list of functions:
let f2s = apply f1s xs
// f2s = [ (fun y -> 3+y); (fun y -> 5+y); (fun y -> 7+y) ]
Hey, here's an idea! Since f2s is also a list of functions, can I apply it again? Well of course I can!
let ys = [1;2;3]
apply f2s ys // Result: [4;7;10]
Wait, what? What just happened?
I first applied the first list of functions to xs, and got another list of functions as a result. And then I applied that result to ys, and got a list of numbers.
We could rewrite that without intermediate variable f2s:
let f1s = [ (fun x y -> x+y); (fun x y -> x-y); (fun x y -> x*y) ]
let xs = [3;5;7]
let ys = [1;2;3]
apply (apply f1s xs) ys // Result: [4;7;10]
For extra convenience, this operation apply is usually expressed as an operator:
let (<*>) = apply
f1s <*> xs <*> ys
See what I did there? With this operator, it now looks very similar to just calling the function with two arguments. Neat.
But wait. What about our original task? In the original requirements we don't have a list of functions, we only have one single function.
Well, that can be easily fixed with another operation, let's call it "apply first". This operation will take a single function (not a list) plus a list of values, and apply this function to each value in the list:
let applyFirst f xs = List.map f xs
Oh, wait. That's just map. Silly me :-)
For extra convenience, this operation is usually also given an operator name:
let (<|>) = List.map
And now, I can do things like this:
let f x y = x + y
let xs = [3;5;7]
let ys = [1;2;3]
f <|> xs <*> ys // Result: [4;7;10]
Or this:
let f x y z = (x + y)*z
let xs = [3;5;7]
let ys = [1;2;3]
let zs = [1;-1;100]
f <|> xs <*> ys <*> zs // Result: [4;-7;1000]
Neat! I made it so I can apply arbitrary functions to lists of arguments at once!
Now, finally, you can apply this to your original problem:
let polynomial a b c (x:double) = a*x + b*x*x + c*x*x*x
let A = [2.0; 3.0; 4.0; 5.0]
let B = [1.5; 1.0; 0.5; 0.2]
let C = [0.8; 0.01; 0.001; 0.0001]
let ps = polynomial <|> A <*> B <*> C
let result = ps |> List.fold (fun x f -> f x) 0.7
The list ps consists of polynomial instances that are partially applied to corresponding elements of A, B, and C, and still expecting the final argument x. And on the next line, I simply fold over this list of functions, applying each of them to the result of the previous.
You could check the implementation for ideas:
https://github.com/fsharp/fsharp/blob/master/src/fsharp/FSharp.Core/array.fs
let fold<'T,'State> (f : 'State -> 'T -> 'State) (acc: 'State) (array:'T[]) =
checkNonNull "array" array
let f = OptimizedClosures.FSharpFunc<_,_,_>.Adapt(f)
let mutable state = acc
for i = 0 to array.Length-1 do
state <- f.Invoke(state,array.[i])
state
here's a few implementations for you:
let fold2<'a,'b,'State> (f : 'State -> 'a -> 'b -> 'State) (acc: 'State) (a:'a array) (b:'b array) =
let mutable state = acc
Array.iter2 (fun x y->state<-f state x y) a b
state
let iter3 f (a: 'a[]) (b: 'b[]) (c: 'c[]) =
let f = OptimizedClosures.FSharpFunc<_,_,_,_>.Adapt(f)
if a.Length <> b.Length || a.Length <> c.Length then failwithf "length"
for i = 0 to a.Length-1 do
f.Invoke(a.[i], b.[i], c.[i])
let altIter3 f (a: 'a[]) (b: 'b[]) (c: 'c[]) =
if a.Length <> b.Length || a.Length <> c.Length then failwithf "length"
for i = 0 to a.Length-1 do
f (a.[i]) (b.[i]) (c.[i])
let fold3<'a,'b,'State> (f : 'State -> 'a -> 'b -> 'c -> 'State) (acc: 'State) (a:'a array) (b:'b array) (c:'c array) =
let mutable state = acc
iter3 (fun x y z->state<-f state x y z) a b c
state
NB. we don't have an iter3, so, implement that. OptimizedClosures.FSharpFunc only allow up to 5 (or is it 7?) params. There are a finite number of type slots available. It makes sense. You can go higher than this, of course, without using the OptimizedClosures stuff.
... anyway, generally, you don't want to be iterating too many lists / arrays / sequences at once. So I'd caution against going too high.
... the better way forward in such cases may be to construct a record or tuple from said lists / arrays, first. Then, you can just use map and iter, which are already baked in. This is what zip / zip3 are all about (see: "(array1.[i],array2.[i],array3.[i])")
let zip3 (array1: _[]) (array2: _[]) (array3: _[]) =
checkNonNull "array1" array1
checkNonNull "array2" array2
checkNonNull "array3" array3
let len1 = array1.Length
if len1 <> array2.Length || len1 <> array3.Length then invalidArg3ArraysDifferent "array1" "array2" "array3" len1 array2.Length array3.Length
let res = Microsoft.FSharp.Primitives.Basics.Array.zeroCreateUnchecked len1
for i = 0 to res.Length-1 do
res.[i] <- (array1.[i],array2.[i],array3.[i])
res
I'm working with arrays at the moment, so my solution pertained to those. Sorry about that. Here's a recursive version for lists.
let fold3 f acc a b c =
let mutable state = acc
let rec fold3 f a b c =
match a,b,c with
| [],[],[] -> ()
| [],_,_
| _,[],_
| _,_,[] -> failwith "length"
| ahead::atail, bhead::btail, chead::ctail ->
state <- f state ahead bhead chead
fold3 f atail btail ctail
fold3 f a b c
i.e. we define a recursive function within a function which acts upon/mutates/changes the outer scoped mutable acc variable (a closure in functional speak). Finally, this gets returned.
It's pretty cool how much type information gets inferred about these functions. In the array examples above, mostly I was explicit with 'a 'b 'c. This time, we let type inference kick in. It knows we're dealing with lists from the :: operator. That's kind of neat.
NB. the compiler will probably unwind this tail-recursive approach so that it is just a loop behind-the-scenes. Generally, get a correct answer before optimising. Just mentioning this, though, as food for later thought.
I think the existing answers provide great options if you want to generalize folding, which was your original question. However, if I simply wanted to call the polynomial function on inputs specified in A, B and C, then I would probably do not want to introduce fairly complex constructs like applicative functors with fancy operators to my code base.
The problem becomes a lot easier if you transpose the input data, so that rather than having a list [A; B; C] with lists for individual variables, you have a transposed list with inputs for calculating each polynomial. To do this, we'll need the transpose function:
let rec transpose = function
| (_::_)::_ as M -> List.map List.head M :: transpose (List.map List.tail M)
| _ -> []
Now you can create a list with inputs, transpose it and calculate all polynomials simply using List.map:
transpose [A; B; C]
|> List.map (function
| [a; b; c] -> polynomial 0.7 a b c
| _ -> failwith "wrong number of arguments")
There are many ways to solve this problem. Few are mentioned like first zip3 all three list, then run over it. Using Applicate Functors like Fyodor Soikin describes means you can turn any function with any amount of arguments into a function that expects list instead of single arguments. This is a good general solution that works with any numbers of lists.
While this is a general good idea, i'm sometimes shocked that so few use more low-level tools. In this case it is a good idea to use recursion and learn more about recursion.
Recursion here is the right-tool because we have immutable data-types. But you could consider how you would implement it with mutable lists and looping first, if that helps. The steps would be:
You loop over an index from 0 to the amount of elements in the lists.
You check if every list has an element for the index
If every list has an element then you pass this to your "folder" function
If at least one list don't have an element, then you abort the loop
The recursive version works exactly the same. Only that you don't use an index to access the elements. You would chop of the first element from every list and then recurse on the remaining list.
Otherwise List.isEmpty is the function to check if a List is empty. You can chop off the first element with List.head and you get the remaining list with the first element removed by List.tail. This way you can just write:
let rec fold3 f acc l1 l2 l3 =
let h = List.head
let t = List.tail
let empty = List.isEmpty
if (empty l1) || (empty l2) && (empty l3)
then acc
else fold3 f (f acc (h l1) (h l2) (h l3)) (t l1) (t l2) (t l3)
The if line checks if every list has at least one element. If that is true
it executes: f acc (h l1) (h l2) (h l3). So it executes f and passes it the first element of every list as an argument. The result is the new accumulator of
the next fold3 call.
Now that you worked on the first element of every list, you must chop off the first element of every list, and continue with the remaining lists. You achieve that with List.tail or in the above example (t l1) (t l2) (t l3). Those are the next remaining lists for the next fold3 call.
Creating a fold4, fold5, fold6 and so on isn't really hard, and I think it is self-explanatory. My general advice is to learn a little bit more about recursion and try to write recursive List functions without Pattern Matching. Pattern Matching is not always easier.
Some code examples:
fold3 (fun acc x y z -> x + y + z :: acc) [] [1;2;3] [10;20;30] [100;200;300] // [333;222;111]
fold3 (fun acc x y z -> x :: y :: z :: acc) [] [1;2;3] [10;20;30] [100;200;300] // [3; 30; 300; 2; 20; 200; 1; 10; 100]
I have to write a program which give in output a tuple with: min and max of a not-empty list and the value that appears most often.
In particular:
min_max [1;0;-1;2;0;-4] ==> (-4; 2)
min_max: int list -> (int * int)
mode [-1;2;1;2;5;-1;5;5;2] ==> 2
mode: int list -> int
This is the code that I wrote for max (min is almost equal) but how can I do to receive as output a tuple with the two values?
let rec max_list xs =
match xs with
| [] -> failwith "xs" "Empty list"
| [x] -> x
| x1::x2::xs' -> max_list((max2 x1 x2)::xs');;
I'll take the first suggestion from #Mark Seemann's answer and run with it, in order to make it generic, working with any collection type, and handle the case of the empty collection sensibly.
let tryMinMax xs =
Seq.fold (function
| Some(mn, mx) -> fun i -> Some(min mn i, max mx i)
| None -> fun i -> Some(i, i) ) None xs
[1;0;-1;2;0;-4]
|> tryMinMax
// val it : (int * int) option = Some (-4, 2)
For the most frequent part of the question:
let mostFrequent xs =
xs
|> Seq.countBy id
|> Seq.maxBy snd
|> fst
[1;0;-1;2;0;-4]
|> mostFrequent
// val it : int = 0
let minMax xs =
xs
|> List.fold
(fun (mn, mx) i -> min mn i, max mx i)
(System.Int32.MaxValue, System.Int32.MinValue)
Not particularly efficient, but fun to write:
let mode xs =
xs
|> List.groupBy id
|> List.map (fun (i, is) -> i, Seq.length is)
|> List.maxBy snd
|> fst
Option without the use of standard modules:
open System
let tryMinMax xs =
let rec minMax xs mn mx =
match xs with | [] -> mn, mx | h::t -> minMax t (min mn h) (max mx h)
match xs with | [] -> None | _ -> Some(minMax xs Int32.MaxValue Int32.MinValue)
dotnetfiddle
On the second question - show their attempts to solve.
I am new to F# and writing some simple algorithm to get used to the language, which needs argMax. Does the standard library come with a function for searching for a list element that maximizes a function? That is, if there's an existing function that behaves like this one:
let argMax f xs =
let rec go a fa zs =
match zs with
| [] -> a
| z :: zs' ->
let fz = f z
if fz > fa
then go z fz zs'
else go a fa zs'
match xs with
| [] -> invalidArg "xs" "empty"
| x :: xs' -> go x (f x) xs'
Yes, but it's called List.maxBy.
Here's an example:
let f x = -(x * x) + 100 * x + 1000
List.maxBy f [0..1000]
// val it : int = 50
f 50
// val it : int = 3500
There is also List.minBy and the same functions are available for Seq and Array.
I have been trying to write a curried function "multvec" which uses u=(u1, u2, ..., un) and v=(v1, v2, ..., vn) and outputs u1*v1 + u2*v2 + ... + un*vn. I think I have the logic mostly correct (at least it would be in other languages...) but I keep getting an:
stdin(11,57): error FS0001: This expression was expected to have type
'a list
but here has type
'c list * 'd list -> 'b list
The code is below: The problem is clearly in the call to product in the last line. However, I was under the impression that the base case ( x * y )::[ ] would just product an 'a list, instead of what it actually produces.
let rec multvec xs ys = function
| [ ], [ ] -> failwith "Both lists cannot be empty"
| x::[ ], y::[ ] -> ( x * y )::[ ]
| x::xs, y::ys -> let product = multvec xs ys
( x * y ) + ( List.reduce (+) product )
Any clarity on this error would be greatly appreciated! Thank you in advance.
Your code honestly has more wrong with it than right: ;-]
It's not tail-recursive
Use of function when xs and ys are separate parameters
Incomplete pattern match
Having one branch evaluate to a list and another evaluate to a scalar
All products are manually added together except for the last, which has a List.reduce run over a single-element list – slightly backwards ;-]
Here's a sensible implementation that fixes all of the above:
let inline multvec xs ys = List.map2 (*) xs ys |> List.sum
Note that if performance is a primary concern then it may be worth avoiding List.sum, as it uses checked arithmetic. One could do the following instead if using unchecked arithmetic is okay:
let inline multvec xs ys = List.map2 (*) xs ys |> List.reduce (+)
If you really want to do this manually, here's one approach:
let inline multvec xs ys =
let rec impl acc = function
| [], [] -> acc
| x::xs', y::ys' -> impl (x * y + acc) (xs', ys')
| _ -> failwith "lists must be of equal length"
impl LanguagePrimitives.GenericZero (xs, ys)
To add to ildjarn's answer, you can fuse map2 and reduce into a single function call using fold2 to compute the dot product ("multvec") of two vectors:
let inline dot xs ys =
let zero = LanguagePrimitives.GenericZero
List.fold2 (fun acc x y -> acc + x * y) zero xs ys
This would save you from creating an unnecessary temporary list.