So what I am trying to do is write a function that recursively checks if two ships hit each other. I have a solution but it doesn't do at all what i want. The outcome should be a list of all ships that collide with each other. so if x and y collide with each other the output list should be [x;y]
The code to build the ships is
type Ship =
{
Id : int
Name : string
X : int
Y : int
}
with
static member Create ( id , name , x , y ) =
{
Id = id
Name = name
X = x
Y = y
}
let ships =
[
Ship.Create(0 , " HMS Horizon " , 0 , 5)
Ship.Create(1 , " Titanic " , 2 , 5)
Ship.Create(2 , " The Iceberg " , 2 , 5)
Ship.Create(3 , " USS Mississipi " , 0 , 5)
Ship.Create(4 , " Battleship Yamato " , 3 , 5)
]
And here is my function I wrote to tackle the issue. But it doesn't work at all
let rec collisions(ships : List<Ship>) : List<Ship> =
match ships with
| [] -> []
| [x] -> [x]
| x :: y :: rest->
if (x.X = y.X) && (x.Y = y.Y) then
[x ; y] # collisions (rest)
elif (x.X <> y.X) || (x.Y <> y.Y) then
[y] # collisions (rest)
else
[]
I think I need to compare all elements of ships with each other and make a new list of the ships that overlap coordinate wise. Although I don't know how I would go about doing this. Also, i'd like to keep the structure of the recursive function if at all possible.
If I understood right, all you need to do is to group by coordinates. You'll get a list of Ship list with all possible collisions:
let collisions (ships : Ship list) =
ships |> List.groupBy (fun s -> s.X,s.Y) //group by coordinates
|> List.filter (fun (_,s) -> List.length s > 1) //keeps only collisions
|> List.map snd //drops coordinate ID created by groupBy
Related
I am wondering if there is a way to write this line without piping h to calcVol function twice?
| h :: t when (h |> calcVol) > maxVol -> maxLoop t (h |> calcVol)
Where h is a tuple containing three dimensions, and calcVol returns a float value.
I know that I could explicitly define a vol value as:
| h :: t ->
let vol = calcVol h
if vol > maxVol then...
I am wondering if there is a way to do this nicely in one line?
If all the uses of vol were before the arrow, you could do this:
| h :: t when let vol = (h |> calcVol) in vol > maxVol -> // Something
But let assignments in the when clause left of the arrow do not carry over to the right-hand side. Demonstration:
let f x = x + 5
let l = [1; 2]
match l with
| a :: b when let y = f a in y = 6 -> "Six"
| _ -> "Other"
This works, and returns "Six". But:
let f x = x + 5
let l = [1; 2]
match l with
| a :: b when let y = f a in y = 6 -> sprintf "Six = %d" y
| _ -> "Other"
This does not work, producing the error:
error FS0039: The value or constructor 'y' is not defined.
So unfortunately, you can't have the one-line version you want and you'll have to go with the longer approach (with a let followed by an if, as you demonstrate in the second half of your answer).
Using active patterns a solution could look like this:
let calcVol v = v
let (|MaxVol|) maxVol = function
| [] -> (maxVol, [])
| h :: t -> ((max (calcVol h) maxVol), t)
let rec maxLoop list m =
match list with
| [] -> m
| MaxVol m (c, t) -> maxLoop t c
let vs = [ -1; 42; 3 ]
maxLoop vs System.Int32.MinValue // 42
Another possibility with better readability might be to first calculate the volumes (e.g. by mapping) and then find the maximum. Difficult to tell without the complete code...
I'm new to F# and I want to find largest element form list of structures:
type Element = struct
val X: int
val Y: int
val RES: int
new (x, y, res) =
{X = x; Y = y; RES = res;}
override this.ToString() = sprintf "%i = %i * %i" this.RES this.X this.Y
end
X is larger than Y when X.RES > Y.RES. I wrote some code:
let max2 x y = if x.RES < y.RES then y else x //BAD LINE
let max_list list =
let rec loop hi list =
match list with
| h::t -> loop (max2 h hi) t
| [] -> hi
match list with
| h::t -> loop h t
| [] -> invalidArg "list" "Empty list"
and call:
let list = findPalindromes 1 1 List.empty //this call populates the "list"
printfn "%A" (max_list list)
This call generates 2 errors (pointing x.RES and y.RES) in line //BAD LINE:
error FS0072: Lookup on object of indeterminate type based on information prior to this program point. A type annotation may be needed prior to this program point to constrain the type of the object. This may allow the lookup to be resolved.
I know that I should cast x and y to Element, I've tried to do it but every time I failed.
How can I fix this code or implement this functionality other way?
The F# standard library has this built-in – List.maxBy:
findPalindromes 1 1 List.empty
|> List.maxBy (fun e -> e.RES)
|> printfn "%A"
As to the error you're getting with max2, type annotations solve it:
let max2 (x:Element) (y:Element) = if x.RES < y.RES then y else x
I'd like to take a List or Array, and given two elements in the collection, get all elements between them. But I want to do this in a circular fashion, such that given a list [1;2;3;4;5;6] and if I ask for the elements that lie between 4 then 2, I get back [5;6;1]
Being used to imperative programming I can easily do this with loops, but I imagine there may be a nicer idiomatic approach to it in F#.
Edit
Here is an approach I came up with, having found the Array.indexed function
let elementsBetween (first:int) (second:int) (elements: array<'T>) =
let diff = second - first
elements
|> Array.indexed
|> Array.filter (fun (index,element) -> if diff = 0 then false
else if diff > 0 then index > first && index < second
else if diff < 0 then index > first || index < second
else false
This approach will only work with arrays obviously but this seems pretty good. I have a feeling I could clean it up by replacing the if/then/else with pattern matching but am not sure how to do that cleanly.
You should take a look at MSDN, Collections.Seq Module for example.
Let's try to be clever:
let elementsBetween a e1 e2 =
let aa = a |> Seq.append a
let i1 = aa |> Seq.findIndex (fun e -> e = e1)
let i2 = aa |> Seq.skip i1 |> Seq.findIndex (fun e -> e = e2)
aa |> Seq.skip(i1+1) |> Seq.take(i2-1)
I am not on my normal computer with an f# compiler, so I haven't tested it yet. It should look something like this
[Edit] Thank you #FoggyFinder for showing me https://dotnetfiddle.net/. I have now tested the code below with it.
[Edit] This should find the circular range in a single pass.
let x = [1;2;3;4;5]
let findCircRange l first second =
let rec findUpTo (l':int list) f (s:int) : (int list * int list) =
match l' with
| i::tail ->
if i = s then tail, (f [])
else findUpTo tail (fun acc -> f (i::acc)) s
// In case we are passed an empty list.
| _ -> [], (f [])
let remainder, upToStart = findUpTo l id first
// concatenate the list after start with the list before start.
let newBuffer = remainder#upToStart
snd <| findUpTo newBuffer id second
let values = findCircRange x 4 2
printf "%A" values
findUpTo takes a list (l'), a function for creating a remainder list (f) and a value to look for (s). We recurse through it (tail recursion) to find the list up to the given value and the list after the given value. Wrap the buffer around by appending the end to the remainder. Pass it to the findUpTo again to find up to the end. Return the buffer up to the end.
We pass a function for accumulating found items. This technique allows us to append to the end of the list as the function calls unwind.
Of course, there is no error checking here. We are assuming that start and end do actually exist. That will be left to an exercise for the reader.
Here is a variation using your idea of diff with list and list slicing
<some list.[x .. y]
let between (first : int) (second : int) (l : 'a list) : 'a list =
if first < 0 then
failwith "first cannot be less than zero"
if second < 0 then
failwith "second cannot be less than zero"
if first > (l.Length * 2) then
failwith "first cannot be greater than length of list times 2"
if second > (l.Length * 2) then
failwith "second cannot be greater than length of list times 2"
let diff = second - first
match diff with
| 0 -> []
| _ when diff > 0 && (abs diff) < l.Length -> l.[(first + 1) .. (second - 1)]
| _ when diff > 0 -> (l#l).[(first + 1) .. (second - 1)]
| _ when diff < 0 && (abs diff) < l.Length -> l.[(second + 1) .. (second + first - 1)]
| _ when diff < 0 -> (l#l).[(second + 1) .. (second + first - 1)]
I have the first matrix which should account for each users (in lines) which products (in columns) they like.
Let's take 3 users and 5 products.
No user liked a product, so my matrix ILike equals a nul matrix :
let matrixILike = [[0.; 1.;2.;3.]
[1.;0.;0.;0.]
[2.;0.;0.;0.]
[3.;0.;0.;0.]
[4.;0.;0.;0.]
[5.;0.;0.;0.]]
Now user 1 likes product 2 and user 3 likes product 5 which can be summarized in the following matrix:
let matrixAction = [[1.;2.]
[3.;5.]]
So I would like to implement the matrix ILike thanks to the matrixAction to obtain a new updated matrixILike like this :
let matrixILike = [[0.; 1.;2.;3.]
[1.;0.;0.;0.]
[2.;1.;0.;0.]
[3.;0.;0.;0.]
[4.;0.;0.;0.]
[5.;0.;0.;1.]]
I try to do this with a "match with" code but it is not working.
for k = 0 to matrixAction.NumRows - 1 do
match (matrixAction.[k,0] , matrixAction.[k,1]) with
| (matrixILike.[x,0] , matrixILike.[0,y]) -> (matrixILike.[x,y] <- 1.)
| _ -> (matrixILike.[x,y] <- 0.)
matrixILike
If you have any suggestions I take it.
This is trivial if you change matrixILike to an array.
let matrixILike = [|
[|0.;1.;2.;3.|]
[|1.;0.;0.;0.|]
[|2.;0.;0.;0.|]
[|3.;0.;0.;0.|]
[|4.;0.;0.;0.|]
[|5.;0.;0.;0.|]
|]
let matrixAction = [
(1., 2.)
(3., 5.)
]
matrixAction
|> List.iter (fun (u, p) -> matrixILike.[int p].[int u] <- 1.)
Without changing your input parameters, this function will do the job.
let update actions =
let mapiTail f = function
| [] -> []
| h::t -> h :: List.mapi (f h) t
mapiTail (fun matHead _ ->
mapiTail (fun rowHead i x ->
if List.exists ((=) [matHead.[i+1];rowHead]) actions then 1. else x))
Usage:
update matrixAction matrixILike
It uses List.mapi which is the same as List.map but with additional parameter: the index.
First, in order to provide full disclosure, I want to point out that this is related to homework in a Machine Learning class. This question is not the homework question and instead is something I need to figure out in order to complete the bigger problem of creating an ID3 Decision Tree Algorithm.
I need to generate tree similar to the following when given a truth table
let learnedTree = Node(0,"A0", Node(2,"A2", Leaf(0), Leaf(1)), Node(1,"A1", Node(2,"A2", Leaf(0), Leaf(1)), Leaf(0)))
learnedTree is of type BinaryTree which I've defined as follows:
type BinaryTree =
| Leaf of int
| Node of int * string * BinaryTree * BinaryTree
ID3 algorithms take into account various equations to determine where to split the tree, and I've got all that figured out, I'm just having trouble creating the learned tree from my truth table. For example if I have the following table
A1 | A2 | A3 | Class
1 0 0 1
0 1 0 1
0 0 0 0
1 0 1 0
0 0 0 0
1 1 0 1
0 1 1 0
And I decide to split on attribute A1 I would end up with the following:
(A1 = 1) A1 (A1 = 0)
A2 | A3 | Class A2 | A3 | Class
0 0 1 1 0 1
0 1 0 0 0 0
1 0 1 0 0 0
0 1 1
Then I would split the left side and split the right side, and continue the recursive pattern until the leaf nodes are pure and I end up with a tree similar to the following based on the splitting.
let learnedTree = Node(0,"A0", Node(2,"A2", Leaf(0), Leaf(1)), Node(1,"A1", Node(2,"A2", Leaf(0), Leaf(1)), Leaf(0)))
Here is what I've kind of "hacked" together thus far, but I think I might be way off:
let rec createTree (listToSplit : list<list<float>>) index =
let leftSideSplit =
listToSplit |> List.choose (fun x -> if x.Item(index) = 1. then Some(x) else None)
let rightSideSplit =
listToSplit |> List.choose (fun x -> if x.Item(index) = 0. then Some(x) else None)
if leftSideSplit.Length > 0 then
let pureCheck = isListPure leftSideSplit
if pureCheck = 0 then
printfn "%s" "Pure left node class 0"
createTree leftSideSplit (index + 1)
else if pureCheck = 1 then
printfn "%s" "Pure left node class 1"
createTree leftSideSplit (index + 1)
else
printfn "%s - %A" "Recursing Left" leftSideSplit
createTree leftSideSplit (index + 1)
else printfn "%s" "Pure left node class 0"
Should I be using pattern matching instead? Any tips/ideas/help? Thanks a bunch!
Edit: I've since posted an implementation of ID3 on my blog at:
http://blogs.msdn.com/chrsmith
Hey Jim, I've been wanting to write a blog post implementing ID3 in F# for a while - thanks for giving me an execute. While this code doesn't implement the algorithm full (or correctly), it should be sufficient for getting you started.
In general you have the right approach - representing each branch as a discriminated union case is good. And like Brian said, List.partition is definitely a handy function. The trick to making this work correctly is all in determining the optimal attribute/value pair to split on - and to do that you'll need to calculate information gain via entropy, etc.
type Attribute = string
type Value = string
type Record =
{
Weather : string
Temperature : string
PlayTennis : bool
}
override this.ToString() =
sprintf
"{Weather = %s, Temp = %s, PlayTennis = %b}"
this.Weather
this.Temperature
this.PlayTennis
type Decision = Attribute * Value
type DecisionTreeNode =
| Branch of Decision * DecisionTreeNode * DecisionTreeNode
| Leaf of Record list
// ------------------------------------
// Splits a record list into an optimal split and the left / right branches.
// (This is where you use the entropy function to maxamize information gain.)
// Record list -> Decision * Record list * Record list
let bestSplit data =
// Just group by weather, then by temperature
let uniqueWeathers =
List.fold
(fun acc item -> Set.add item.Weather acc)
Set.empty
data
let uniqueTemperatures =
List.fold
(fun acc item -> Set.add item.Temperature acc)
Set.empty
data
if uniqueWeathers.Count = 1 then
let bestSplit = ("Temperature", uniqueTemperatures.MinimumElement)
let left, right =
List.partition
(fun item -> item.Temperature = uniqueTemperatures.MinimumElement)
data
(bestSplit, left, right)
else
let bestSplit = ("Weather", uniqueWeathers.MinimumElement)
let left, right =
List.partition
(fun item -> item.Weather = uniqueWeathers.MinimumElement)
data
(bestSplit, left, right)
let rec determineBranch data =
if List.length data < 4 then
Leaf(data)
else
// Use the entropy function to break the dataset on
// the category / value that best splits the data
let bestDecision, leftBranch, rightBranch = bestSplit data
Branch(
bestDecision,
determineBranch leftBranch,
determineBranch rightBranch)
// ------------------------------------
let rec printID3Result indent branch =
let padding = new System.String(' ', indent)
match branch with
| Leaf(data) ->
data |> List.iter (fun item -> printfn "%s%s" padding <| item.ToString())
| Branch(decision, lhs, rhs) ->
printfn "%sBranch predicate [%A]" padding decision
printfn "%sWhere predicate is true:" padding
printID3Result (indent + 4) lhs
printfn "%sWhere predicate is false:" padding
printID3Result (indent + 4) rhs
// ------------------------------------
let dataset =
[
{ Weather = "windy"; Temperature = "hot"; PlayTennis = false }
{ Weather = "windy"; Temperature = "cool"; PlayTennis = false }
{ Weather = "nice"; Temperature = "cool"; PlayTennis = true }
{ Weather = "nice"; Temperature = "cold"; PlayTennis = true }
{ Weather = "humid"; Temperature = "hot"; PlayTennis = false }
]
printfn "Given input list:"
dataset |> List.iter (printfn "%A")
printfn "ID3 split resulted in:"
let id3Result = determineBranch dataset
printID3Result 0 id3Result
You can use List.partition instead of your two List.choose calls.
http://research.microsoft.com/en-us/um/cambridge/projects/fsharp/manual/FSharp.Core/Microsoft.FSharp.Collections.List.html
(or now http://msdn.microsoft.com/en-us/library/ee353738(VS.100).aspx )
It isn't clear to me that pattern matching will buy you much here; the input type (list of lists) and processing (partitioning and 'pureness' check) doesn't really lend itself to that.
And of course when you finally get the 'end' (a pure list) you need to create a tree, and then presumably this function will create a Leaf when the input only has one 'side' and it's 'pure', but create a Node out of the left-side and right-side results for every other input. Maybe. I didn't quite grok the algorithm completely.
Hopefully that will help steer you a little bit. May be useful to draw up a few smaller sample inputs and outputs to help work out the various cases of the function body.
Thanks Brian & Chris! I was actually able to figure this out and I ended up with the following. This calculates the information gain for determining the best place to split. I'm sure there are probably better ways for me to arrive at this solution especially around the chosen data structures, but this is a start. I plan to refine things later.
#light
open System
let trainList =
[
[1.;0.;0.;1.;];
[0.;1.;0.;1.;];
[0.;0.;0.;0.;];
[1.;0.;1.;0.;];
[0.;0.;0.;0.;];
[1.;1.;0.;1.;];
[0.;1.;1.;0.;];
[1.;0.;0.;1.;];
[0.;0.;0.;0.;];
[1.;0.;0.;1.;];
]
type BinaryTree =
| Leaf of int
| Node of int * string * BinaryTree * BinaryTree
let entropyList nums =
let sumOfnums =
nums
|> Seq.sum
nums
|> Seq.map (fun x -> if x=0.00 then x else (-((x/sumOfnums) * Math.Log(x/sumOfnums, 2.))))
|> Seq.sum
let entropyBinaryList (dataListOfLists:list<list<float>>) =
let classList =
dataListOfLists
|> List.map (fun x -> x.Item(x.Length - 1))
let ListOfNo =
classList
|> List.choose (fun x -> if x = 0. then Some(x) else None)
let ListOfYes =
classList
|> List.choose (fun x -> if x = 1. then Some(x) else None)
let numberOfYes : float = float ListOfYes.Length
let numberOfNo : float = float ListOfNo.Length
let ListOfNumYesAndSumNo = [numberOfYes; numberOfNo]
entropyList ListOfNumYesAndSumNo
let conditionalEntropy (dataListOfLists:list<list<float>>) attributeNumber =
let NoAttributeList =
dataListOfLists
|> List.choose (fun x -> if x.Item(attributeNumber) = 0. then Some(x) else None)
let YesAttributeList =
dataListOfLists
|> List.choose (fun x -> if x.Item(attributeNumber) = 1. then Some(x) else None)
let numberOfYes : float = float YesAttributeList.Length
let numberOfNo : float = float NoAttributeList.Length
let noConditionalEntropy = (entropyBinaryList NoAttributeList) * (numberOfNo/(numberOfNo + numberOfYes))
let yesConditionalEntropy = (entropyBinaryList YesAttributeList) * (numberOfYes/(numberOfNo + numberOfYes))
[noConditionalEntropy; yesConditionalEntropy]
let findBestSplitIndex(listOfInstances : list<list<float>>) =
let IGList =
[0..(listOfInstances.Item(0).Length - 2)]
|> List.mapi (fun i x -> (i, (entropyBinaryList listOfInstances) - (List.sum (conditionalEntropy listOfInstances x))))
IGList
|> List.maxBy snd
|> fst
let isListPure (listToCheck : list<list<float>>) =
let splitList = listToCheck |> List.choose (fun x -> if x.Item(x.Length - 1) = 1. then Some(x) else None)
if splitList.Length = listToCheck.Length then 1
else if splitList.Length = 0 then 0
else -1
let rec createTree (listToSplit : list<list<float>>) =
let pureCheck = isListPure listToSplit
if pureCheck = 0 then
printfn "%s" "Pure - Leaf(0)"
else if pureCheck = 1 then
printfn "%s" "Pure - Leaf(1)"
else
printfn "%A - is not pure" listToSplit
if listToSplit.Length > 1 then // There are attributes we can split on
// Chose best place to split list
let splitIndex = findBestSplitIndex(listToSplit)
printfn "spliting at index %A" splitIndex
let leftSideSplit =
listToSplit |> List.choose (fun x -> if x.Item(splitIndex) = 1. then Some(x) else None)
let rightSideSplit =
listToSplit |> List.choose (fun x -> if x.Item(splitIndex) = 0. then Some(x) else None)
createTree leftSideSplit
createTree rightSideSplit
else
printfn "%s" "Not Pure, but can't split choose based on heuristics - Leaf(0 or 1)"