How can i make `printfn "%A"` print full complex structures? - f#

When I print a complex stuctured value via printfn "%A" the output is truncated as in the example below.
for fsi there are some options (ie fsi.PrintLength <- System.Int32.MaxValue) to change this behavior.
How do i get the full structure in a simple console App?
seq
[[adspath, {name = "adspath";
typ = "System.String";
count = 1;
value =
"GC://...";}];
[c, {name = "c";
typ = "System.String";
count = 1;
value = "DE";}];
[cn, {name = "cn";
typ = "System.String";
count = 1;
value = "I....";}];
[co, {name = "co";
typ = "System.String";
count = 1;
value = "DE";}]; ...]

As i cannot delete the question here is the answer. The Problem was that the sequence was not materalised. In my case I had a value of type System.Collections.Generic.IDictionary<string,<T>>
v
|> Seq.toList
|> printfn "%A"
solves it.

Related

Second Taxicab Number Generator

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

Reduce an array of array into a single array with flattening in f#

If I have the following
type Merchant = {
Id:int;
OtherItems:int[] }
let (merchant1:Merchant) = {
Id = 1;
OtherItems = [| 1; 2 |]}
let (merchant2:Merchant) = {
Id = 2;
OtherItems = [| 1; 2 |]}
let merchants = [| merchant1;merchant2|]
And I want to flatten to the following, how do I do it?
Id = 1 OtherItems 1
Id = 1 OtherItems 2
Id = 2 OtherItems 1
Id = 2 OtherItems 2
This is what I have come up with but cant seem to get any further
let x =
merchants
|> Array.map(fun merchant -> merchant, merchant.OtherItems)
Note: I could do the long way in c# oo style but want to use functional way
Here's a way using Array.collect:
let x =
merchants
|> Array.collect (fun m ->
m.OtherItems
|> Array.map (fun x -> { Id = m.Id; OtherItems = [|x|] }))
You could make this easier to understand my first introducing a function that flattens a single merchant:
let flatten merchant = [|
for x in merchant.OtherItems do
yield { Id = merchant.Id; OtherItems = [|x|] } |]
This function has the type Merchant -> Merchant [], so it turns a single merchant into an array of merchants, one for each OtherItems.
With that flatten function, you can use the standard, built-in collect function to flatten an array of merchants:
let x' = merchants |> Array.collect flatten
Both options produce this result:
[|
{Id = 1; OtherItems = [|1|];};
{Id = 1; OtherItems = [|2|];};
{Id = 2; OtherItems = [|1|];};
{Id = 2; OtherItems = [|2|];}
|]
Use sequence generation:
let mySequence =
seq {
for merchant in merchants do
for otherItem in merchant.OtherItems do
yield {Id=merchant.Id; OtherItems=[|otherItem|]}
}
or array generation
let myArray =
[|
for merchant in merchants do
for otherItem in merchant.OtherItems do
yield {Id=merchant.Id; OtherItems= [|otherItem|]}
|]

f# sequence of running total

Ok, this looks like it should be easy, but I'm just not getting it. If I have a sequence of numbers, how do I generate a new sequence made up of the running totals? eg for a sequence [1;2;3;4], I want to map it to [1;3;6;10]. In a suitably functional way.
Use List.scan:
let runningTotal = List.scan (+) 0 >> List.tail
[1; 2; 3; 4]
|> runningTotal
|> printfn "%A"
Seq.scan-based implementation:
let runningTotal seq' = (Seq.head seq', Seq.skip 1 seq') ||> Seq.scan (+)
{ 1..4 }
|> runningTotal
|> printfn "%A"
Another variation using Seq.scan (Seq.skip 1 gets rid of the leading zero):
> {1..4} |> Seq.scan (+) 0 |> Seq.skip 1;;
val it : seq<int> = seq [1; 3; 6; 10]
> Seq.scan (fun acc n -> acc + n) 0 [1;2;3;4];;
val it : seq<int> = seq [0; 1; 3; 6; ...]
With lists:
> [1;2;3;4] |> List.scan (fun acc n -> acc + n) 0 |> List.tail;;
val it : int list = [1; 3; 6; 10]
Edit: Another way with sequences:
let sum s = seq {
let x = ref 0
for i in s do
x := !x + i
yield !x
}
Yes, there's a mutable variable, but I find it more readable (if you want to get rid of the leading 0).
Figured it was worthwhile to share how to do this with Record Types in case that's also what you came here looking for.
Below is a fictitious example demonstrating the concept using runner laps around a track.
type Split = double
type Lap = { Num : int; Split : Split }
type RunnerLap = { Lap : Lap; TotalTime : double }
let lap1 = { Num = 1; Split = 1.23 }
let lap2 = { Num = 2; Split = 1.13 }
let lap3 = { Num = 3; Split = 1.03 }
let laps = [lap1;lap2;lap3]
let runnerLapsAccumulator =
Seq.scan
(fun rl l -> { rl with Lap = l; TotalTime = rl.TotalTime + l.Split }) // acumulator
{ Lap = { Num = 0; Split = 0.0 }; TotalTime = 0.0 } // initial state
let runnerLaps = laps |> runnerLapsAccumulator
printfn "%A" runnerLaps
Not sure this is the best way but it should do the trick
let input = [1; 2; 3; 4]
let runningTotal =
(input, 0)
|> Seq.unfold (fun (list, total) ->
match list with
| [] ->
None
| h::t ->
let total = total + h
total, (t, total) |> Some)
|> List.ofSeq

Check if an element is within a sequence

how to check if an element is contained within a sequence? I expected some Seq.contains, but i could not find it. Thanks
EDIT:
Or, for an easier task, how to make the diff between two sequences? Like, getting all the elements within a list that doesn not belong to another (or that do)?
Little bit simpler:
let contains x = Seq.exists ((=) x)
Seq.exists
let testseq = seq [ 1; 2; 3; 4 ]
let equalsTwo n = (n = 2)
let containsTwo = Seq.exists equalsTwo testseq
Set is your friend here:
let a = set [0;1;2;3]
let b = set [2;3;4;5]
let c = a - b
let d = b - a
let e = Set.intersect a b
let f = a + b
>
val c : Set<int> = seq [0; 1]
val d : Set<int> = seq [4; 5]
val e : Set<int> = seq [2; 3]
val f : Set<int> = seq [0; 1; 2; 3; ...]
Danny
Seq.exists again, but with slightly different syntax -
let testseq = seq [ 1; 2; 3; 4 ]
let testn = 2
testseq |> Seq.exists (fun x -> x = testn)
See MSDN F#: Seq.exists function: https://msdn.microsoft.com/en-us/library/ee353562.aspx
Lots of other good ones there too!
(Another question, another answer.)
This works, but I don't think that it's the most idomatic way to do it - (you'll need to wait until the US wakes up to find out):
let s1 = seq [ 1; 2; 3; 4 ]
let s2 = seq [ 3; 4; 5; 6 ]
seq {
for a in s1 do
if not (Seq.exists (fun n -> n = a) s2) then
yield a
}

Help Needed Creating a Binary Tree Given Truth Table

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)"

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