Suppose I want to run a fold using a function that expects a tuple, f (x, y). This can be done with
List.fold (fun x y -> f (x, y)) x xs
I kind of feel there ought to be a higher-order function that abstracts out that pattern, takes a function that expects a tuple and turns it into a function that expects two separate arguments.
Is there such a function in the F# standard library? If not, easy enough to write but what should it idiomatically be called? untuple?
That function is normally called curry in most functional languages. It's definition is:
let curry f x y = f (x, y)
Unfortunately it's not in the F# core lib, but it's available in many external base libraries, like F#x and F#+.
Related
I have already done some searches, and this question is a duplicate of another post. I am posting this just for future reference.
Is it possible to define SUMPRODUCT without explicitly using variable names x, y?
Original Function:
let SUMPRODUCT x y = List.map2 (*) x y |> List.sum
SUMPRODUCT [1;4] [3;25] // Result: 103
I was hoping to do this:
// CONTAINS ERROR!
let SUMPRODUCT = (List.map2 (*)) >> List.sum
// CONTAINS ERROR!
But F# comes back with an error.
I have already found the solution on another post, but if you have any suggestions please let me know. Thank you.
Function composition only works when the input function takes a single argument. However, in your example, the result of List.map2 (*) is a function that takes two separate arguments and so it cannot be easily composed with List.sum using >>.
There are various ways to work around this if you really want, but I would not do that. I think >> is nice in a few rare cases where it fits nicely, but trying to over-use it leads to unreadable mess.
In some functional languages, the core library defines helpers for turning function with two arguments into a function that takes a tuple and vice versa.
let uncurry f (x, y) = f x y
let curry f x y = f (x, y)
You could use those two to define your sumProduct like this:
let sumProduct = curry ((uncurry (List.map2 (*))) >> List.sum)
Now it is point-free and understanding it is a fun mental challenge, but for all practical purposes, nobody will be able to understand the code and it is also longer than your original explicit version:
let sumProduct x y = List.map2 (*) x y |> List.sum
According to this post:
What am I missing: is function composition with multiple arguments possible?
Sometimes "pointed" style code is better than "pointfree" style code, and there is no good way to unify the type difference of the original function to what I hope to achieve.
Can the function createTuple below be expressed pointfree?
let createTuple = fun v -> (v, v*2)
createTuple 2 |> printfn "%A" // (2,4)
The F# library does not provide many functions for writing code in point-free style (mainly because it is not particularly idiomatic way of writing F#), so you cannot write your createTuple function using just what is available in the core library.
If you really wanted to do this, you could define a couple of helper combinators for working with tuples:
/// Duplicates any given value & returns a tuple with two copies of it
let dup a = a, a
/// Transforms the first element using given function
let mapFst f (a, b) = (f a, b)
/// Transforms the second element (not needed here, but adding for symmetry)
let mapSnd f (a, b) = (a, f b)
With these, you could implement your function in a point-free way:
let createTuple = dup >> mapSnd ((*) 2)
This does the same thing as your function. I think it is significantly harder to decipher what is going on here and I would never actually write that code, but that's another issue :-).
I want to apply a function to both members of a homogenous tuple, resulting in another tuple. Following on from my previous question I defined an operator that seemed to make sense to me:
let (||>>) (a,b) f = f a, f b
However, again I feel like this might be a common use case but couldn't find it in the standard library. Does it exist?
I don't think there is any standard library function that does this.
My personal preference would be to avoid too many custom operators (they make code shorter, but they make it harder to read for people who have not seen the definition before). Applying function to both elements of a tuple is logically close to the map operation on lists (which applies a function to all elements of a list), so I would probably define Tuple2.map:
module Tuple2 =
let map f (a, b) = (f a, f b)
Then you can use the function quite nicely with pipelining:
let nums = (1, 2)
nums |> Tuple2.map (fun x -> x + 1)
This question already has an answer here:
Function Application Operator ($) in F#?
(1 answer)
Closed 8 years ago.
Sometimes I have to write:
myList |> List.iter (fun x -> x)
I would really like to avoid the parentheses. In Haskell there is an operator for this ($)
It would look like this
myList |> List.iter $ fun x -> x
I created a custom operator
let inline (^!) f a = f a
and now I can write it like this
myList |> List.iter ^! fun x -> x
Is there something like this in F#?
There is no way to define custom operator with an explicitly specified associativity in F# - the associativity is determined based on the symbols forming the operator (and you can find it in the MSDN documentation for operators).
In this case, F# does not have any built-in operator that would let you avoid the parentheses and the idiomatic way is to write the code as you write it originally, with parentheses:
myList |> List.iter (fun x -> x)
This is difference in style if you are coming from Haskell, but I do not see any real disadvantage of writing the parentheses - it is just a matter of style that you'll get used to after writing F# for some time. If you want to avoid parentheses (e.g. to write a nice DSL), then you can always named function and write something like:
myList |> List.iter id
(I understand that your example is really just an example, so id would not work for your real use case, but you can always define your own functions if that makes the code more readable).
No, there's nothing like this in a standard F# library. However, you have almost done creating your own operator (by figuring out its name must start with ^).
This snippet by Stephen Swensen demonstrates a high precedence, right associative backward pipe, (^<|).
let inline (^<|) f a = f a
This single-liner from the linked page demonstrates how to use it:
{1..10} |> Seq.map ^<| fun x -> x + 3
And here is an example how to use it for multi-line functions. I find it most useful for real-world multi-liners as you no longer need to keep closing parenthesis at the end:
myList
|> List.map
^<| fun x ->
let ...
returnValue
In F# it's <|
So it would look like:
myList |> List.iter <| fun x -> x
In almost all examples, a y-combinator in ML-type languages is written like this:
let rec y f x = f (y f) x
let factorial = y (fun f -> function 0 -> 1 | n -> n * f(n - 1))
This works as expected, but it feels like cheating to define the y-combinator using let rec ....
I want to define this combinator without using recursion, using the standard definition:
Y = λf·(λx·f (x x)) (λx·f (x x))
A direct translation is as follows:
let y = fun f -> (fun x -> f (x x)) (fun x -> f (x x));;
However, F# complains that it can't figure out the types:
let y = fun f -> (fun x -> f (x x)) (fun x -> f (x x));;
--------------------------------^
C:\Users\Juliet\AppData\Local\Temp\stdin(6,33): error FS0001: Type mismatch. Expecting a
'a
but given a
'a -> 'b
The resulting type would be infinite when unifying ''a' and ''a -> 'b'
How do I write the y-combinator in F# without using let rec ...?
As the compiler points out, there is no type that can be assigned to x so that the expression (x x) is well-typed (this isn't strictly true; you can explicitly type x as obj->_ - see my last paragraph). You can work around this issue by declaring a recursive type so that a very similar expression will work:
type 'a Rec = Rec of ('a Rec -> 'a)
Now the Y-combinator can be written as:
let y f =
let f' (Rec x as rx) = f (x rx)
f' (Rec f')
Unfortunately, you'll find that this isn't very useful because F# is a strict language,
so any function that you try to define using this combinator will cause a stack overflow.
Instead, you need to use the applicative-order version of the Y-combinator (\f.(\x.f(\y.(x x)y))(\x.f(\y.(x x)y))):
let y f =
let f' (Rec x as rx) = f (fun y -> x rx y)
f' (Rec f')
Another option would be to use explicit laziness to define the normal-order Y-combinator:
type 'a Rec = Rec of ('a Rec -> 'a Lazy)
let y f =
let f' (Rec x as rx) = lazy f (x rx)
(f' (Rec f')).Value
This has the disadvantage that recursive function definitions now need an explicit force of the lazy value (using the Value property):
let factorial = y (fun f -> function | 0 -> 1 | n -> n * (f.Value (n - 1)))
However, it has the advantage that you can define non-function recursive values, just as you could in a lazy language:
let ones = y (fun ones -> LazyList.consf 1 (fun () -> ones.Value))
As a final alternative, you can try to better approximate the untyped lambda calculus by using boxing and downcasting. This would give you (again using the applicative-order version of the Y-combinator):
let y f =
let f' (x:obj -> _) = f (fun y -> x x y)
f' (fun x -> f' (x :?> _))
This has the obvious disadvantage that it will cause unneeded boxing and unboxing, but at least this is entirely internal to the implementation and will never actually lead to failure at runtime.
I would say it's impossible, and asked why, I would handwave and invoke the fact that simply typed lambda calculus has the normalization property. In short, all terms of the simply typed lambda calculus terminate (consequently Y can not be defined in the simply typed lambda calculus).
F#'s type system is not exactly the type system of simply typed lambda calculus, but it's close enough. F# without let rec comes really close to the simply typed lambda calculus -- and, to reiterate, in that language you cannot define a term that does not terminate, and that excludes defining Y too.
In other words, in F#, "let rec" needs to be a language primitive at the very least because even if you were able to define it from the other primitives, you would not be able to type this definition. Having it as a primitive allows you, among other things, to give a special type to that primitive.
EDIT: kvb shows in his answer that type definitions (one of the features absent from the simply typed lambda-calculus but present in let-rec-less F#) allow to get some sort of recursion. Very clever.
Case and let statements in ML derivatives are what makes it Turing Complete, I believe they're based on System F and not simply typed but the point is the same.
System F cannot find a type for the any fixed point combinator, if it could, it wasn't strongly normalizing.
What strongly normalizing means is that any expression has exactly one normal form, where a normal form is an expression that cannot be reduced any further, this differs from untyped where every expression has at max one normal form, it can also have no normal form at all.
If typed lambda calculi could construct a fixed point operator in what ever way, it was quite possible for an expression to have no normal form.
Another famous theorem, the Halting Problem, implies that strongly normalizing languages are not Turing complete, it says that's impossible to decide (different than prove) of a turing complete language what subset of its programs will halt on what input. If a language is strongly normalizing, it's decidable if it halts, namely it always halts. Our algorithm to decide this is the program: true;.
To solve this, ML-derivatives extend System-F with case and let (rec) to overcome this. Functions can thus refer to themselves in their definitions again, making them in effect no lambda calculi at all any more, it's no longer possible to rely on anonymous functions alone for all computable functions. They can thus again enter infinite loops and regain their turing-completeness.
Short answer: You can't.
Long answer:
The simply typed lambda calculus is strongly normalizing. This means it's not Turing equivalent. The reason for this basically boils down to the fact that a Y combinator must either be primitive or defined recursively (as you've found). It simply cannot be expressed in System F (or simpler typed calculi). There's no way around this (it's been proven, after all). The Y combinator you can implement works exactly the way you want, though.
I would suggest you try scheme if you want a real Church-style Y combinator. Use the applicative version given above, as other versions won't work, unless you explicitly add laziness, or use a lazy Scheme interpreter. (Scheme technically isn't completely untyped, but it's dynamically typed, which is good enough for this.)
See this for the proof of strong normalization:
http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.127.1794
After thinking some more, I'm pretty sure that adding a primitive Y combinator that behaves exactly the way the letrec defined one does makes System F Turing complete. All you need to do to simulate a Turing machine then is implement the tape as an integer (interpreted in binary) and a shift (to position the head).
Simply define a function taking its own type as a record, like in Swift (there it's a struct) :)
Here, Y (uppercase) is semantically defined as a function that can be called with its own type. In F# terms, it is defined as a record containing a function named call, so for calling a y defined as this type, you have to actually call y.call :)
type Y = { call: Y -> (int -> int) }
let fibonacci n =
let makeF f: int -> int =
fun x ->
if x = 0 then 0 else if x = 1 then 1 else f(x - 1) + f(x - 2)
let y = { call = fun y -> fun x -> (makeF (y.call y)) x }
(y.call y) n
It's not supremely elegant to read but it doesn't resort to recursion for defining a y combinator that is supposed to provide recursion all by itself ^^