I have a seq<'A>. I want to map this to a seq<(int, 'A)>, where the integer is an auto-generated sequence of values starting at 0. I know I can do this with a mutable counter and a loop, but is there a more elegant way to do this, perhaps using Seq.map?
Check out Seq.mapi: http://msdn.microsoft.com/en-us/library/ee340431.aspx
let a = [ 1; 2; 3 ]
let s = a |> Seq.mapi (fun i x -> i,x)
Related
I am attempting to generate a series of guesses for the second Taxicab number. What I want to do is is call the Attempt function on a series of integers in a finite sequence. I have my two questions about implementation in the comments.
A taxi cab number, in case your wondering, is the least number that satisfied the sum of 2 unique cubes in for n unique sets of 2 unique cubes. Ta(2) is 1729.
[<EntryPoint>]
let main argv =
let Attempt (start : int) =
let stop = start+20
let integerList = [start..stop]
let list = List.init 3 (fun x -> integerList.[x])
//Is there a simple way to make initialize the list with random indices of integerList?
let Cube x = x*x*x
let newlist = list |> List.map (fun x -> Cube x)
let partitionList (x : List<int>) (y : int) = List.sum [x.[y];x.[y+1]]
let intLIST = [0..2]
let partitionList' = [for i in intLIST do yield partitionList newlist i]
let x = Set.ofList partitionList'
let y = Set.ofList partitionList'
//I was going to try to use some kind of equality operator to determine whether the two sets were equal, which could tell me whether we had actually found a Taxicab number by the weakened definition.
System.Console.Write(list)
System.Console.Write(newlist)
let rnd = System.Random()
//My primary question is how can I convert a random to an integer to use in start for the function Attempt?
System.Console.ReadKey() |> ignore
printfn("%A") argv
0
Dirty way to initialize list with random indexes of another list:
let randomIndexes count myList =
let rand = System.Random()
seq {
for n = 1 to count do
yield rand.Next(List.length myList) }
|> Seq.distinct
//|> Seq.sort // if you need them sorted
|> List.ofSeq
let result = randomIndexes 5 [3;2;4;5]
printfn "%A" result
I have the following demand: getting the array of tuples from the first array according to the elements’ appearance in the second array:
let totals = [| ("old1", "new1"); ("old2", "new2"); ("old3", "new3"); ("old4", "new4") |]
let changes = [| "new1"; "new4" |]
I want to have this:
let updates = [| ("old1", "new1"); ("old4", "new4") |]
If the both arrays totals and changes have the same length, then I think it is easy:
let updates = Array.zip changes totals
|> Array.choose(fun (a, B) -> if a = fst(B) then Some (B) else None)
Unfortunately, totals and changes have different number of elements; therefore, I can not find an easy way to get the elements I need.
The solution posted by pad is correct and will work fine for small number of elements in changes. However, it iterates over the array changes for every element in total, so it may be inefficient for large arrays.
As an alternative, you can turn changes into an F# set type which allows more efficient membership test:
// Create set containing 'changes'
let changesSet = Set.ofArray changes
// Filter totals where the second element is in 'changesSet'
totals |> Array.filter (fun (_, v) -> changesSet.Contains(v))
// Same thing using function composition
totals |> Array.filter (snd >> changesSet.Contains)
You should select pairs in totals which have second elements occur in changes:
let updates =
totals
|> Array.filter (fun (_, x) -> changes |> Array.exists (fun y -> y = x))
I have a sequence of integers representing dice in F#.
In the game in question, the player has a pool of dice and can choose to play one (governed by certain rules) and keep the rest.
If, for example, a player rolls a 6, 6 and a 4 and decides to play one the sixes, is there a simple way to return a sequence with only one 6 removed?
Seq.filter (fun x -> x != 6) dice
removes all of the sixes, not just one.
Non-trivial operations on sequences are painful to work with, since they don't support pattern matching. I think the simplest solution is as follows:
let filterFirst f s =
seq {
let filtered = ref false
for a in s do
if filtered.Value = false && f a then
filtered := true
else yield a
}
So long as the mutable implementation is hidden from the client, it's still functional style ;)
If you're going to store data I would use ResizeArray instead of a Sequence. It has a wealth of functions built in such as the function you asked about. It's simply called Remove. Note: ResizeArray is an abbreviation for the CLI type List.
let test = seq [1; 2; 6; 6; 1; 0]
let a = new ResizeArray<int>(test)
a.Remove 6 |> ignore
Seq.toList a |> printf "%A"
// output
> [1; 2; 6; 1; 0]
Other data type options could be Array
let removeOneFromArray v a =
let i = Array.findIndex ((=)v) a
Array.append a.[..(i-1)] a.[(i+1)..]
or List
let removeOneFromList v l =
let rec remove acc = function
| x::xs when x = v -> List.rev acc # xs
| x::xs -> remove (x::acc) xs
| [] -> acc
remove [] l
the below code will work for a list (so not any seq but it sounds like the sequence your using could be a List)
let rec removeOne value list =
match list with
| head::tail when head = value -> tail
| head::tail -> head::(removeOne value tail)
| _ -> [] //you might wanna fail here since it didn't find value in
//the list
EDIT: code updated based on correct comment below. Thanks P
EDIT: After reading a different answer I thought that a warning would be in order. Don't use the above code for infite sequences but since I guess your players don't have infite dice that should not be a problem but for but for completeness here's an implementation that would work for (almost) any
finite sequence
let rec removeOne value seq acc =
match seq.Any() with
| true when s.First() = value -> seq.Skip(1)
| true -> seq.First()::(removeOne value seq.Skip(1))
| _ -> List.rev acc //you might wanna fail here since it didn't find value in
//the list
However I recommend using the first solution which Im confident will perform better than the latter even if you have to turn a sequence into a list first (at least for small sequences or large sequences with the soughtfor value in the end)
I don't think there is any function that would allow you to directly represent the idea that you want to remove just the first element matching the specified criteria from the list (e.g. something like Seq.removeOne).
You can implement the function in a relatively readable way using Seq.fold (if the sequence of numbers is finite):
let removeOne f l =
Seq.fold (fun (removed, res) v ->
if removed then true, v::res
elif f v then true, res
else false, v::res) (false, []) l
|> snd |> List.rev
> removeOne (fun x -> x = 6) [ 1; 2; 6; 6; 1 ];
val it : int list = [1; 2; 6; 1]
The fold function keeps some state - in this case of type bool * list<'a>. The Boolean flag represents whether we already removed some element and the list is used to accumulate the result (which has to be reversed at the end of processing).
If you need to do this for (possibly) infinite seq<int>, then you'll need to use GetEnumerator directly and implement the code as a recursive sequence expression. This is a bit uglier and it would look like this:
let removeOne f (s:seq<_>) =
// Get enumerator of the input sequence
let en = s.GetEnumerator()
let rec loop() = seq {
// Move to the next element
if en.MoveNext() then
// Is this the element to skip?
if f en.Current then
// Yes - return all remaining elements without filtering
while en.MoveNext() do
yield en.Current
else
// No - return this element and continue looping
yield en.Current
yield! loop() }
loop()
You can try this:
let rec removeFirstOccurrence item screened items =
items |> function
| h::tail -> if h = item
then screened # tail
else tail |> removeFirstOccurrence item (screened # [h])
| _ -> []
Usage:
let updated = products |> removeFirstOccurrence product []
I've trying to learn F#. I'm a complete beginner, so this might be a walkover for you guys :)
I have the following function:
let removeEven l =
let n = List.length l;
let list_ = [];
let seq_ = seq { for x in 1..n do if x % 2 <> 0 then yield List.nth l (x-1)}
for x in seq_ do
let list_ = list_ # [x];
list_;
It takes a list, and return a new list containing all the numbers, which is placed at an odd index in the original list, so removeEven [x1;x2;x3] = [x1;x3]
However, I get my already favourite error-message: Incomplete construct at or before this point in expression...
If I add a print to the end of the line, instead of list_:
...
print_any list_;
the problem is fixed. But I do not want to print the list, I want to return it!
What causes this? Why can't I return my list?
To answer your question first, the compiler complains because there is a problem inside the for loop. In F#, let serves to declare values (that are immutable and cannot be changed later in the program). It isn't a statement as in C# - let can be only used as part of another expression. For example:
let n = 10
n + n
Actually means that you want the n symbol to refer to the value 10 in the expression n + n. The problem with your code is that you're using let without any expression (probably because you want to use mutable variables):
for x in seq_ do
let list_ = list_ # [x] // This isn't assignment!
list_
The problematic line is an incomplete expression - using let in this way isn't allowed, because it doesn't contain any expression (the list_ value will not be accessed from any code). You can use mutable variable to correct your code:
let mutable list_ = [] // declared as 'mutable'
let seq_ = seq { for x in 1..n do if x % 2 <> 0 then yield List.nth l (x-1)}
for x in seq_ do
list_ <- list_ # [x] // assignment using '<-'
Now, this should work, but it isn't really functional, because you're using imperative mutation. Moreover, appending elements using # is really inefficient thing to do in functional languages. So, if you want to make your code functional, you'll probably need to use different approach. Both of the other answers show a great approach, although I prefer the example by Joel, because indexing into a list (in the solution by Chaos) also isn't very functional (there is no pointer arithmetic, so it will be also slower).
Probably the most classical functional solution would be to use the List.fold function, which aggregates all elements of the list into a single result, walking from the left to the right:
[1;2;3;4;5]
|> List.fold (fun (flag, res) el ->
if flag then (not flag, el::res) else (not flag, res)) (true, [])
|> snd |> List.rev
Here, the state used during the aggregation is a Boolean flag specifying whether to include the next element (during each step, we flip the flag by returning not flag). The second element is the list aggregated so far (we add element by el::res only when the flag is set. After fold returns, we use snd to get the second element of the tuple (the aggregated list) and reverse it using List.rev, because it was collected in the reversed order (this is more efficient than appending to the end using res#[el]).
Edit: If I understand your requirements correctly, here's a version of your function done functional rather than imperative style, that removes elements with odd indexes.
let removeEven list =
list
|> Seq.mapi (fun i x -> (i, x))
|> Seq.filter (fun (i, x) -> i % 2 = 0)
|> Seq.map snd
|> List.ofSeq
> removeEven ['a'; 'b'; 'c'; 'd'];;
val it : char list = ['a'; 'c']
I think this is what you are looking for.
let removeEven list =
let maxIndex = (List.length list) - 1;
seq { for i in 0..2..maxIndex -> list.[i] }
|> Seq.toList
Tests
val removeEven : 'a list -> 'a list
> removeEven [1;2;3;4;5;6];;
val it : int list = [1; 3; 5]
> removeEven [1;2;3;4;5];;
val it : int list = [1; 3; 5]
> removeEven [1;2;3;4];;
val it : int list = [1; 3]
> removeEven [1;2;3];;
val it : int list = [1; 3]
> removeEven [1;2];;
val it : int list = [1]
> removeEven [1];;
val it : int list = [1]
You can try a pattern-matching approach. I haven't used F# in a while and I can't test things right now, but it would be something like this:
let rec curse sofar ls =
match ls with
| even :: odd :: tl -> curse (even :: sofar) tl
| even :: [] -> curse (even :: sofar) []
| [] -> List.rev sofar
curse [] [ 1; 2; 3; 4; 5 ]
This recursively picks off the even elements. I think. I would probably use Joel Mueller's approach though. I don't remember if there is an index-based filter function, but that would probably be the ideal to use, or to make if it doesn't exist in the libraries.
But in general lists aren't really meant as index-type things. That's what arrays are for. If you consider what kind of algorithm would require a list having its even elements removed, maybe it's possible that in the steps prior to this requirement, the elements can be paired up in tuples, like this:
[ (1,2); (3,4) ]
That would make it trivial to get the even-"indexed" elements out:
thelist |> List.map fst // take first element from each tuple
There's a variety of options if the input list isn't guaranteed to have an even number of elements.
Yet another alternative, which (by my reckoning) is slightly slower than Joel's, but it's shorter :)
let removeEven list =
list
|> Seq.mapi (fun i x -> (i, x))
|> Seq.choose (fun (i,x) -> if i % 2 = 0 then Some(x) else None)
|> List.ofSeq
I am trying to write an efficient algorithm that will effectively let me merge data sets (like a sql join). I think I need to use Array.tryFindIndex, but the syntax has me lost.
Based on the data below, I am calling arrX my "host" array, and want to return an int array that has its length, and tells me the positions of each of its elements in arrY (returning -1 if its not in there). (Once I know these indices I can then use them on arrays of data that of length arrY.length)
let arrX= [|"A";"B";"C";"D";"E";"F"|]
let arrY = [|"E";"A";"C"|];
let desiredIndices = [|1; -1; 2; -1; 0; -1|]
It looks like I need to use an option type somehow, and I think a mapi2 in there as well.
Does anyone know how to get this done? (I think it could be a very useful code snippet for people who are merging data sets from different sources)
Thanks!
//This code does not compile, can't figure out what to do here
let d = Array.tryFindIndex (fun x y -> x = y) arrX
The tryFindIndex function searches for a single element in the array specified as the second argument. The lambda function gets only a single parameter and it should return true if the parameter is the element you are looking for. The type signature of the tryFindIndex function shows this:
('a -> bool) -> 'a [] -> int option
(In your example, you're giving it a function that takes two parameters of type 'a -> 'a -> bool, which is incompatible with the expected type). The tryFindIndex function returns an option type, which means that it gives you None if no element matches the predicate, otherwise it gives you Some(idx) containing the index of the found element.
To get the desired array of indices, you need to run tryFindIndex for every element of the input array (arrX). This can be done using the Array.map function. If you want to get -1 if the element wasn't found, you can use pattern matching to convert None to -1 and Some(idx) to idx:
let desired =
arrX |> Array.map (fun x ->
let res = Array.tryFindIndex (fun y -> x = y) arrY
match res with
| None -> -1
| Some idx -> idx)
The same thing can be written using sequence expression (instead of map), which may be more readable:
let desired =
[| for x in arrX do
let res = Array.tryFindIndex (fun y -> x = y) arrY
match res with
| None -> yield -1
| Some idx -> yield idx |]
Anyway, if you need to implement a join-like operation, you can do it more simply using sequence expressions. In the following example, I also added some values (in addition to the string keys), so that you can better see how it works:
let arrX= [|"A",1; "B",2; "C",3; "D",4; "E",5; "F",6|]
let arrY = [|"E",10; "A",20; "C",30|]
[| for x, i in arrX do
for y, j in arrY do
if x = y then
yield x, i, j |]
// Result: [|("A", 1, 20); ("C", 3, 30); ("E", 5, 10)|]
The sequence expression simply loops over all arrX elements and for each of them, it loops over all arrY element. Then it tests whether the keys are the same and if they are, it produces a single element. This isn't particularly efficient, but in most of the cases, it should work fine.
Write a custom function that returns -1 if nothing is found, or returns the index if it's found. Next, use Array.map to map a new array using this function:
let arrX= [|"A";"B";"C";"D";"E";"F"|]
let arrY = [|"E";"A";"C"|];
let indexOrNegativeOne x =
match Array.tryFindIndex (fun y -> y = x) arrY with
| Some(y) -> y
| None -> -1
let desired = arrX |> Array.map indexOrNegativeOne
printfn "%A" desired