I would like to generate a sequence of sequences of sets of tuple
It's for generating coordinates of lines in board game (squares);horizontal lines,vertical lines, and the main 2 diagonal lines
The one I created using list (using the generateLines function) works well. It generates the coordinates of lines in a list of lists of tuples. However, after I convert it into sequence of sequences, I don't have any idea how to check whether it's correct
I've tried this :
CoordinateLines 3 |> Seq.map (fun x ->Seq.toList) |> Seq.toList
But it doesn't work, with error : error FS0030: Value restriction. The value 'it' has been inferred to have generic type
val it : ('_a -> '_b list) list when '_a :> seq<'_b>
Either define 'it' as a simple data term, make it a function with explicit arguments or, if you do not intend for it to be generic, add a type annotation.
let CoordinateLines size : seq<seq<int*int>> =
let generateLines size =
[for a in 0..(size-1) do
yield [for b in 0..(size-1) do
yield(a,b)
]
]
#
[for a in 0..(size-1) do
yield [for b in 0..(size-1) do
yield(b,a)
]
]
#
[[for a = 0 to (size-1) do
yield (a,(size-1)-a)
]]
#
[[for a = 0 to (size-1) do
yield (a,a)
]]
generateLines size
|> List.toSeq
|> Seq.map (fun x -> List.toSeq x)
You are getting this error, because you forgot to pass x as the argument to Seq.toList in the lambda function passed to Seq.map. The following works fine:
CoordinateLines 3 |> Seq.map (fun x ->Seq.toList x) |> Seq.toList
Alternatively, you can also use Seq.toList directly as the argument to Seq.map:
CoordinateLines 3 |> Seq.map Seq.toList |> Seq.toList
That said, I don't see any reason for converting the lists to seq<'a> - the list type implements the seq<'a> interface already, so anything that you want to do with sequences will also work with lists. And since you are generating fully evaluated lists already, it might actually turn out that you can write some subsequent logic more nicely using lists.
If I was doing this, I'd just write a function that returns list of lists. You can also use the
[ for <var> in <range> -> <expr> ] notation (which lets you omit some yield keywords), and you can avoid using # by putting all the code that generates nested list in one big [ .. ] block with multiple nested fors and yields. This makes your code a bit shorter:
let CoordinateLines size =
[ for a in 0..(size-1) ->
[ for b in 0..(size-1) -> (a,b) ]
for a in 0..(size-1) ->
[ for b in 0..(size-1) -> (b,a) ]
yield [ for a in 0..(size-1) -> (a,(size-1)-a) ]
yield [ for a in 0..(size-1) -> (a,a) ] ]
Related
I am learning F# and the use cases of the |>, >>, and << operators confuse me. I get that everything if statements, functions, etc. act like variables but how do these work?
Usually we (community) say the Pipe Operator |> is just a way, to write the last argument of a function before the function call. For example
f x y
can be written
y |> f x
but for correctness, this is not true. It just pass the next argument to a function. So you could even write.
y |> (x |> f)
All of this, and all other kind of operators works, because in F# all functions are curried by default. This means, there exists only functions with one argument. Functions with many arguments, are implemented that a functions return another function.
You could also write
(f x) y
for example. The function f is a function that takes x as argument and returns another function. This then gets y passed as an argument.
This process is automatically done by the language. So if you write
let f x y z = x + y + z
it is the same as:
let f = fun x -> fun y -> fun z -> x + y + z
Currying is by the way the reason why parenthesis in a ML-like language are not enforced compared to a LISP like language. Otherwise you would have needded to write:
(((f 1) 2) 3)
to execute a function f with three arguments.
The pipe operator itself is just another function, it is defined as
let (|>) x f = f x
It takes a value x as its first argument. And a function f as its second argument. Because operators a written "infix" (this means between two operands) instead of "prefix" (before arguments, the normal way), this means its left argument to the operator is the first argument.
In my opinion, |> is used too much by most F# people. It makes sense to use piping if you have a chain of operations, one after another. Typically for example if you have multiple list operations.
Let's say, you want to square all numbers in a list and then filter only the even ones. Without piping you would write.
List.filter isEven (List.map square [1..10])
Here the second argument to List.filter is a list that is returned by List.map. You can also write it as
List.map square [1..10]
|> List.filter isEven
Piping is Function application, this means, you will execute/run a function, so it computes and returns a value as its result.
In the above example List.map is first executed, and the result is passed to List.filter. That's true with piping and without piping. But sometimes, you want to create another function, instead of executing/running a function. Let's say you want to create a function, from the above. The two versions you could write are
let evenSquares xs = List.filter isEven (List.map square xs)
let evenSquares xs = List.map square xs |> List.filter isEven
You could also write it as function composition.
let evenSquares = List.filter isEven << List.map square
let evenSquares = List.map square >> List.filter isEven
The << operator resembles function composition in the "normal" way, how you would write a function with parenthesis. And >> is the "backwards" compositon, how it would be written with |>.
The F# documentation writes it the other way, what is backward and forward. But i think the F# language creators are wrong.
The function composition operators are defined as:
let (<<) f g x = f (g x)
let (>>) f g x = g (f x)
As you see, the operator has technically three arguments. But remember currying. When you write f << g, then the result is another functions, that expects the last argument x. Passing less arguments then needed is also often called Partial Application.
Function composition is less often used in F#, because the compiler sometimes have problems with type inference if the function arguments are generic.
Theoretically you could write a program without ever defining a variable, just through function composition. This is also named Point-Free style.
I would not recommend it, it often makes code harder to read and/or understand. But it is sometimes used if you want to pass a function to another
Higher-Order function. This means, a functions that take another function as an argument. Like List.map, List.filter and so on.
Pipes and composition operators have simple definition but are difficult to grasp. But once we have understand them, they are super useful and we miss them when we get back to C#.
Here some explanations but you get the best feedbacks from your own experiments. Have fun!
Pipe right operator |>
val |> fn ≡ fn val
Utility:
Building a pipeline, to chain calls to functions: x |> f |> g ≡ g (f x).
Easier to read: just follow the data flow
No intermediary variables
Natural language in english: Subject Verb.
It's regular in object-oriented code : myObject.do()
In F#, the "subject" is usually the last parameter: List.map f list. Using |>, we get back the natural "Subject Verb" order: list |> List.map f
Final benefit but not the least: help type inference:
let items = ["a"; "bb"; "ccc"]
let longestKo = List.maxBy (fun x -> x.Length) items // ❌ Error FS0072
// ~~~~~~~~
let longest = items |> List.maxBy (fun x -> x.Length) // ✅ return "ccc"
Pipe left operator <|
fn <| expression ≡ fn (expression)
Less used than |>
✅ Small benefit: avoiding parentheses
❌ Major drawback: inverse of the english natural "left to right" reading order and inverse of execution order (because of left-associativity)
printf "%i" 1+2 // 💥 Error
printf "%i" (1+2) // With parentheses
printf "%i" <| 1+2 // With pipe left
What about this kind of expression: x |> fn <| y ❓
In theory, allow using fn in infix position, equivalent of fn x y
In practice, it can be very confusing for some readers not used to it.
👉 It's probably better to avoid using <|
Forward composition operator >>
Binary operator placed between 2 functions:
f >> g ≡ fun x -> g (f x) ≡ fun x -> x |> f |> g
Result of the 1st function is used as argument for the 2nd function
→ types must match: f: 'T -> 'U and g: 'U -> 'V → f >> g :'T -> 'V
let add1 x = x + 1
let times2 x = x * 2
let add1Times2 x = times2(add1 x) // 😕 Style explicit but heavy
let add1Times2' = add1 >> times2 // 👍 Style concise
Backward composition operator <<
f >> g ≡ g << f
Less used than >>, except to get terms in english order:
let even x = x % 2 = 0
// even not 😕
let odd x = x |> even |> not
// "not even" is easier to read 👍
let odd = not << even
☝ Note: << is the mathematical function composition ∘: g ∘ f ≡ fun x -> g (f x) ≡ g << f.
It's confusing in F# because it's >> that is usually called the "composition operator" ("forward" being usually omitted).
On the other hand, the symbols used for these operators are super useful to remember the order of execution of the functions: f >> g means apply f then apply g. Even if argument is implicit, we get the data flow direction:
>> : from left to right → f >> g ≡ fun x -> x |> f |> g
<< : from right to left → f << g ≡ fun x -> f <| (g <| x)
(Edited after good advices from David)
Does Function Composition rely on Partial Application?
Here's my understanding:
Observe the following functions that have some duplication of function calls:
let updateCells (grid:Map<(int * int), Cell>) =
grid |> Map.toSeq
|> Seq.map snd
|> Seq.fold (fun grid c -> grid |> setReaction (c.X, c.Y)) grid
let mutable _cells = ObservableCollection<Cell>( grid |> Map.toSeq
|> Seq.map snd
|> Seq.toList )
let cycleHandler _ =
self.Cells <- ObservableCollection<Cell>( grid |> cycleThroughCells
|> Map.toSeq
|> Seq.map snd
|> Seq.toList )
If you’ve noticed, the following code appears in all three functions:
grid |> Map.toSeq
|> Seq.map snd
Function Composition
Within functional programming, we can fuse functions together so that they can become one function.
To do this, let’s create a new function from the duplicated sequence of functions:
let getCells = Map.toSeq >> Seq.map snd >> Seq.toList
Now if you’re attentive, you will have noticed that we don’t use any arguments when using Function Composition. Hence, the grid value is not used. The reason behind this is because of Partial Application.
Partial Application
I’m still learning all these functional programming techniques. However, my understanding is that partial application is a technique within functional programming that postpones the need to accept a complete set of arguments for a given function. In other words, partial application is the act of deferring the acceptance of a complete set of arguments for a given function in which there is an expectation that the end-client will provide the rest of the arguments later. At least, that’s my understanding.
We can now take a function like:
let updateCells (grid:Map<(int * int), Cell>) =
grid |> Map.toSeq
|> Seq.map snd
|> Seq.fold (fun grid c -> grid |> setReaction (c.X, c.Y)) grid
And refactor it to something like:
let updateCells (grid:Map<(int * int), Cell>) =
grid |> getCells
|> Seq.fold (fun grid c -> grid |> setReaction (c.X, c.Y)) grid
Are my thoughts regarding Function Composition being coupled with Partial Application accurate?
Generics
Actually, if you take the expression
let getCells = Map.toSeq >> Seq.map snd >> Seq.toList
and attempt to compile it as a stand-alone expression, you'll get a compiler error:
error FS0030: Value restriction. The value 'getCells' has been inferred to have generic type
val getCells : (Map<'_a,'_b> -> '_b list) when '_a : comparison
Either make the arguments to 'getCells' explicit or, if you do not intend for it to be generic, add a type annotation.
The reason it works in your case is because you're using the getCells function with grid, which means that the compiler infers it to have a constrained type.
In order to keep it generic, you can rephrase it using an explicit argument:
let getCells xs = xs |> Map.toSeq |> Seq.map snd |> Seq.toList
This expression is a valid stand-alone expression of the type Map<'a,'b> -> 'b list when 'a : comparison.
Point-free
The style used with the >> function composition operator is called point-free. It works well with partial application, but isn't quite the same.
Application
There is, however, an example of partial function application in this example:
let getCells xs = xs |> Map.toSeq |> Seq.map snd |> Seq.toList
The function snd has the following type:
'a * 'b -> 'b
It's function that takes a single argument.
You could also write the above getCells function without partial application of the snd function:
let getCells xs = xs |> Map.toSeq |> Seq.map (fun x -> snd x) |> Seq.toList
Notice that instead of a partially applied function passed to Seq.map, you can pass a lambda expression. The getCells function is still a function composed from other functions, but it no longer relies on partial application of snd.
Thus, to partially (pun intended) answer your question: function composition doesn't have to rely on partial function composition.
Currying
In F#, functions are curried by default. This means that all functions take exactly one argument, and returns a value. Sometimes (often), the return value is another function.
Consider, as an example, the Seq.map function. If you call it with one argument, the return value is another function:
Seq.map snd
This expression has the type seq<'a * 'b> -> seq<'b>, because the return value of Seq.map snd is another function.
Eta reduction
This means that you can perform an Eta reduction on the above lambda expression fun x -> snd x, because x appears on both sides of the expression. The result is simply snd, and the entire expression becomes
let getCells xs = xs |> Map.toSeq |> Seq.map snd |> Seq.toList
As you can see, partial application isn't necessary for function composition, but it does make it much easier.
Impartial
The above composition using the pipe operator (|>) still relies on partial application of the functions Map.toSeq, Seq.map, etcetera. In order to demonstrate that composition doesn't rely on partial application, here's an 'impartial' (the opposite of partial? (pun)) alternative:
let getCells xs =
xs
|> (fun xs' -> Map.toSeq xs')
|> (fun xs' -> Seq.map (fun x -> snd x) xs')
|> (fun xs' -> Seq.toList xs')
Notice that this version makes extensive use of lambda expressions instead of partial application.
I wouldn't compose functions in this way; I only included this alternative to demonstrate that it can be done.
Composition depends on first-class functions, not really on partial applications.
What is required to implement composition is that:
Functions must be able to be taken as arguments and returned as return values
Function signatures must be valid types (if you want the composition to be strongly-typed)
Partial application creates more opportunities for composition, but in principle you can easily define function composition without it.
For example, C# doesn't have partial application*, but you can still compose two functions together, as long as the signatures match:
Func<a, c> Compose<a, b, c>(this Func<a, b> f,
Func<b, c> g)
{
return x => g(f(x));
}
which is just >> with an uglier syntax: f.Compose(g).
However, there is one interesting connection between composition and partial application. The definition of the >> operator is:
let (>>) f g x = g(f(x))
and so, when you write foo >> bar, you are indeed partially applying the (>>) function, ie omitting the x argument to get the fun x = g(f(x)) partial result.
But, as I said above, this isn't strictly necessary. The Compose function above is equivalent to F#'s >> operator and doesn't involve any partial application; lambdas perform the same role in a slightly more verbose way.
* Unless you manually implement it, which nobody does. I.e. instead of writing
string foo(int a, int b)
{
return (a + b).ToString();
}
you'd have to write
Func<int, string> foo(int a)
{
return b => (a + b).ToString();
}
and then you'd be able to pass each argument separately just like in F#.
I have a code which looks like this:
this.GetItemTypeIdsAsListForOneItemTypeIdTreeUpIncludeItemType itemType.AutoincrementedId
|> Array.map (fun i -> i.AutoincrementedId)
|> Array.map (BusinessLogic.EntityTypes.getFullSetOfEntityTypeFieldValuesForItemTypeAid item.Autoincrementedid)
|> Array.fold Array.append [||]
|> Array.map (fun fv -> { fv with ReferenceAutoId = aid } )
|> Array.toSeq
|> Seq.distinctBy (fun fv -> fv.Fieldname)
|> Seq.toArray
Sometimes such code gets the unusual result which I need to explain. Usually there is not error in the code. There is an error in the data. And I need to explain why this backet of data is incorrect. What is the best way to do it ?
I just want to look at the list on each step of this expression.
Something like:
func data
|> func2 && Console.WriteLine
|> func3 && Console.WriteLine
....
Get input, split it on two. Pass one of the output to the next function, and second output to Console.
For a quick and dirty solution, you can always add a function like this one:
// ('a -> unit) -> 'a -> 'a
let tee f x = f x; x
If, for example, you have a composition like this:
[1..10]
|> List.map string
|> String.concat "|"
you can insert tee in order to achieve a side-effect:
[1..10]
|> List.map string
|> tee (printfn "%A")
|> String.concat "|"
That's not functional, but can be used in a pinch if you just need to look at some intermediate values; e.g. for troubleshooting.
Otherwise, for a 'proper' functional solution, perhaps application of the State monad might be appropriate. That will enable you to carry around state while performing the computation. The state could, for example, contain custom messages collected along the way...
If you just want to 'exit' as soon as you discover that something is wrong, though, then the Either monad is the appropriate way to go.
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Linked list partition function and reversed results
Actually I don't care about the input type or the output type, any of seq, array, list will do. (It doesn't have to be generic) Currently my code takes list as input and (list * list) as output
let takeWhile predicator list =
let rec takeWhileRec newList remain =
match remain with
| [] -> (newList |> List.rev, remain)
| x::xs -> if predicator x then
takeWhileRec (x::newList) xs
else
(newList |> List.rev, remain)
takeWhileRec [] list
However, there is a pitfall. As fas as I see, List.rev is O(n^2), which would likely to dominate the overall speed? I think it is even slower than the ugly solution: Seq.takeWhile, then count, and then take tail n times... which is still O(n)
(If there is a C# List, then i would use that without having to reverse it...)
A side question, what's difference between Array.ofList and List.toArray , or more generally, A.ofB and B.ofA in List, Seq, Array?
is seq myList identical to List.toSeq myList?
Another side question, is nested Seq.append have same complexity as Seq.concat?
e.g.
Seq.append (Seq.append (Seq.append a b) c) d // looks aweful
Seq.concat [a;b;c;d]
1)The relevant implementation of List.rev is in local.fs in the compiler - it is
// optimized mutation-based implementation. This code is only valid in fslib, where mutation of private
// tail cons cells is permitted in carefully written library code.
let rec revAcc xs acc =
match xs with
| [] -> acc
| h::t -> revAcc t (h::acc)
let rev xs =
match xs with
| [] -> xs
| [_] -> xs
| h1::h2::t -> revAcc t [h2;h1]
The comment does seem odd as there is no obvious mutation. Note that this is in fact O(n) not O(n^2)
2) As pad said there is no difference - I prefer to use the to.. as I think
A
|> List.map ...
|> List.toArray
looks nicer than
A
|> List.map ...
|> Array.ofList
but that is just me.
3)
Append (compiler source):
[<CompiledName("Append")>]
let append (source1: seq<'T>) (source2: seq<'T>) =
checkNonNull "source1" source1
checkNonNull "source2" source2
fromGenerator(fun () -> Generator.bindG (toGenerator source1) (fun () -> toGenerator source2))
Note that for each append we get an extra generator that has to be walked through. In comparison, the concat implementation will just have 1 single extra function rather than n so using concat is probably better.
To answer your questions:
1) Time complexity of List.rev is O(n) and worst-case complexity of takeWhile is also O(n). So using List.rev doesn't increase complexity of the function. Using ResizeArray could help you avoid List.rev, but you have to tolerate a bit of mutation.
let takeWhile predicate list =
let rec loop (acc: ResizeArray<_>) rest =
match rest with
| x::xs when predicate x -> acc.Add(x); loop acc xs
| _ -> (acc |> Seq.toList, rest)
loop (ResizeArray()) list
2) There is no difference. Array.ofList and List.toArray uses the same function internally (see here and here).
3). I think Seq.concat has the same complexity with a bunch of Seq.append. In the context of List andArray, concat is more efficient than append because you have more information to pre-allocate space for outputs.
how about this:
let takeWhile pred =
let cont = ref true
List.partition (pred >> fun r -> !cont && (cont := r; r))
It uses a single library function, List.partition, which is efficiently implemented.
Hope this is what you meant :)
Hey guys, I'm trying to get cozy with functional programming (particularly with F#), and I've hit a wall when it comes to building tail-recursive functions. I'm pretty good with turning basic recursion (where the function basically calls itself once per invocation), into tail recursion, but I now have a slightly more complicated situation.
In my case, the function must accept a single list as a parameter. When the function is called, I have to remove the first element from the list, and then recur using the remainder of the list. Then I need to apply the first element which I removed in some way to the result of the recursion. Next, I remove the second element and do the same thing (Note: when I say "remove the seond element", that is from the original list, so the list passed at the recursion includes the first element as well). I do the same for the third, fourth, etc. elements of the list.
Is there a way to convert the above situation into a tail-recursive function? Maybe nested tail-recursive functions??? Thank you for any answers.
Okay, so here's my basic code. This particular one is a permutation generator (I'm not too concern with the permutation part, though - it's the recursion I'd like to focusing on):
let permutationsOther str =
match str with
| value :: [] ->
[[value]]
| _ ->
let list = (List.map (fun a -> // This applies the remove part for every element a
let lst = (List.filter (fun b -> b <> a) str) // This part removes element a from the list
let permutedLst = permutations lst // recursive call
consToAll a permutedLst // constToAll this is my own function which performs "cons" operation with a and every element in the list permutedLst
) str)
List.reduce (fun acc elem -> elem # acc) list // flatten list of lists produce by map into a single list
I hope this is clear enough - I'll be happy to provide clarifications if needed.
By the way, I have found just a way to rewrite this particular function so that it only uses a single recursion, but it was a fluke more than an informed decision. However, this has encouraged me that there may be a general method of turning multiple recursion into single recursion, but I have not yet found it.
Conversion to CPS should do the trick:
NOTE 1: Source of the sample is typed directly in browser, so may contain errors :(. But I hope it can demonstrate the general idea.
NOTE 2: consToAll function should be converted to CPS too: consToAll: 'T -> 'T list list -> ('T list list -> 'R) -> 'R
let remove x l = List.filter ((<>) x) l // from original post: should duplicates also be removed ???
let permute l =
let rec loop k l =
match l with
| [] -> k []
| [value] -> k [[value]]
| _ -> filter l [] l (fun r -> r |> List.reduce (fun acc elem -> elem # acc) |> k )
and filter l acc orig fk =
match l with
| [] -> fk acc
| x::xs ->
remove x orig
|> loop (fun res ->
consToAll x res (fun rs -> filter xs (rs::acc) orig fk)
)
loop id l