I have finished the project I have been working on but I am wanting to go back and cleanup my code. In this one instance I used a mutable variable however I want my code to contain no mutable variables. How would I rewrite this code section to return a bool but have it not mutable?
let mutable duplicates = false
for el in (combo|>Seq.head) do
let exists = Seq.exists (fun x -> x = el) (combo|>Seq.item 1)
duplicates <- exists
Any help would be appreciated, cheers!
let t = Seq.item 1 combo
let duplicates = Seq.head combo |> Seq.exists (fun el -> Seq.contains el t)
The usual caveats about handling seqs in this manner apply.
let s1 = combo |> Seq.head
let s2 = combo |> Seq.item 1
let duplicates = System.Linq.Enumerable.Intersect(s1, s2) |> Seq.isEmpty |> not
Related
I have two lists listA and listB where I want to return true if listB contains any element also in listA.
let listA = ["A";"B";"C"]
let listB = ["D";"E";"A"]
Should return true in this case. I feel like this should be easy to solve and I'm missing something fundamental somewhere.
For example, why can't I do like this?
let testIntersect = for elem in listA do List.exists (fun x -> x = elem) listB
You can't write something like your example code because a plain for doesn't return a result, it just evaluates an expression for its side-effects. You could write the code in a for comprehension:
let testIntersect listA listB =
[for elem in listA do yield List.exists (fun x -> x = elem) listB]
Of course, this then returns a bool list rather than a single bool.
val testIntersect :
listA:seq<'a> -> listB:'a list -> bool list when 'a : equality
let listA = ["A";"B";"C"]
let listB = ["D";"E";"A"]
testIntersect listA listB
val it : bool list = [true; false; false]
So, we can use the List.exists function to ensure that a true occurs at least once:
let testIntersect listA listB =
[for elem in listA do yield List.exists (fun x -> x = elem) listB]
|> List.exists id
val testIntersect :
listA:seq<'a> -> listB:'a list -> bool list when 'a : equality
val listA : string list = ["A"; "B"; "C"]
val listB : string list = ["D"; "E"; "A"]
val it : bool = false
It's pretty inefficient to solve this problem using List though, it's better to use Set. With Set, you can calculate intersection in O(log N * log M) time rather than O(N*M).
let testSetIntersect listA listB =
Set.intersect (Set.ofList listA) (Set.ofList listB)
|> Set.isEmpty
|> not
One function that you could use is List.except, which is not yet documented (!) but can be seen in this pull request that was merged a couple of years ago. You'd probably use it like this:
let testIntersect a b =
let b' = b |> List.except a
// If b' is shorter than b, then b contained at least one element of a
List.length b' < List.length b
However, this runs through list B about three times, once to do the except algorithm and once each to do both the length calls. So another approach might be to do what you did, but turn list A into a set so that the exists call won't be O(N):
let testIntersect a b =
let setA = a |> Set.ofList
match b |> List.tryFind (fun x -> setA |> Set.contains x) with
| Some _ -> true
| None -> false
The reason I used tryFind is because List.find would throw an exception if the predicate didn't match any items of the list.
Edit: An even better approach is to use List.exists, which I temporarily forgot about (thanks to Honza Brestan for reminding me about it):
let testIntersect a b =
let setA = a |> Set.ofList
b |> List.exists (fun x -> setA |> Set.contains x)
Which, of course, is pretty much what you were originally wanting to do in your testIntersect code sample. The only difference is that you were using the for ... in syntax in your code sample, which wouldn't work. In F#, the for loop is exclusively for expressions that return unit (and thus, probably have side effects). If you want to return a value, the for loop won't do that. So using the functions that do return value, like List.exists, is the approach you want to take.
let testIntersect listA listB =
(Set.ofList listA) - (Set.ofList listB) |> Set.isEmpty |> not
I'm a beginner in F#, and this is my first attempt at programming something serious. I'm sorry the code is a bit long, but there are some issues with mutability that I don't understand.
This is an implementation of the Karger MinCut Algorithm to calculate the mincut on a non-directed graph component. I won't discuss here how the algo works,
for more info https://en.wikipedia.org/wiki/Karger%27s_algorithm
What is important is it's a randomized algorithm, which is running a determined number of trial runs, and taking the "best" run.
I realize now that I could avoid a lot of the problems below if I did construct a specific function for each random trial, but I'd like to understand EXACTLY what is wrong in the implementation below.
I'm running the code on this simple graph (the mincut is 2 when we cut the graph
into 2 components (1,2,3,4) and (5,6,7,8) with only 2 edges between those 2 components)
3--4-----5--6
|\/| |\/|
|/\| |/\|
2--1-----7--8
the file simplegraph.txt should encode this graph as follow
(1st column = node number, other columns = links)
1 2 3 4 7
2 1 3 4
3 1 2 4
4 1 2 3 5
5 4 6 7 8
6 5 7 8
7 1 5 6 8
8 5 6 7
This code may look too much as imperative programming yet, I'm sorry for that.
So There is a main for i loop calling each trial.
the first execution, (when i=1) looks smooth and perfect,
but I have runtime error execution when i=2, because it looks some variables,
like WG are not reinitialized correctly, causing out of bound errors.
WG, WG1 and WGmin are type wgraphobj, which are a record of Dictionary objects
WG1 is defined outside the main loop and i make no new assignments to WG1.
[but its type is mutable though, alas]
I defined first WG with the instruction
let mutable WG = WG1
then at the beginning of the for i loop,
i write
WG <- WG1
and then later, i modify the WG object in each trial to make some calculations.
when the trial is finished and we go to the next trial (i is increased) i want to reset WG to its initial state being like WG1.
but it seems its not working, and I don't get why...
Here is the full code
MyModule.fs [some functions not necessary for execution]
namespace MyModule
module Dict =
open System.Collections.Generic
let toSeq d = d |> Seq.map (fun (KeyValue(k,v)) -> (k,v))
let toArray (d:IDictionary<_,_>) = d |> toSeq |> Seq.toArray
let toList (d:IDictionary<_,_>) = d |> toSeq |> Seq.toList
let ofMap (m:Map<'k,'v>) = new Dictionary<'k,'v>(m) :> IDictionary<'k,'v>
let ofList (l:('k * 'v) list) = new Dictionary<'k,'v>(l |> Map.ofList) :> IDictionary<'k,'v>
let ofSeq (s:('k * 'v) seq) = new Dictionary<'k,'v>(s |> Map.ofSeq) :> IDictionary<'k,'v>
let ofArray (a:('k * 'v) []) = new Dictionary<'k,'v>(a |> Map.ofArray) :> IDictionary<'k,'v>
Karger.fs
open MyModule.Dict
open System.IO
let x = File.ReadAllLines "\..\simplegraph.txt";;
// val x : string [] =
let splitAtTab (text:string)=
text.Split [|'\t';' '|]
let splitIntoKeyValue (s:seq<'T>) =
(Seq.head s, Seq.tail s)
let parseLine (line:string)=
line
|> splitAtTab
|> Array.filter (fun s -> not(s=""))
|> Array.map (fun s-> (int s))
|> Array.toSeq
|> splitIntoKeyValue
let y =
x |> Array.map parseLine
open System.Collections.Generic
// let graph = new Map <int, int array>
let graphD = new Dictionary<int,int seq>()
y |> Array.iter graphD.Add
let graphM = y |> Map.ofArray //immutable
let N = y.Length // number of nodes
let Nruns = 2
let remove_table = new Dictionary<int,bool>()
[for i in 1..N do yield (i,false)] |> List.iter remove_table.Add
// let remove_table = seq [|for a in 1 ..N -> false|] // plus court
let label_head_table = new Dictionary<int,int>()
[for i in 1..N do yield (i,i)] |> List.iter label_head_table.Add
let label = new Dictionary<int,int seq>()
[for i in 1..N do yield (i,[i])] |> List.iter label.Add
let mutable min_cut = 1000000
type wgraphobj =
{ Graph : Dictionary<int,int seq>
RemoveTable : Dictionary<int,bool>
Label : Dictionary<int,int seq>
LabelHead : Dictionary<int,int> }
let WG1 = {Graph = graphD;
RemoveTable = remove_table;
Label = label;
LabelHead = label_head_table}
let mutable WGmin = WG1
let IsNotRemoved x = //
match x with
| (i,false) -> true
| (i,true) -> false
let IsNotRemoved1 WG i = //
(i,WG.RemoveTable.[i]) |>IsNotRemoved
let GetLiveNode d =
let myfun x =
match x with
| (i,b) -> i
d |> toList |> List.filter IsNotRemoved |> List.map myfun
let rand = System.Random()
// subsets a dictionary given a sub_list of keys
let D_Subset (dict:Dictionary<'T,'U>) (sub_list:list<'T>) =
let z = Dictionary<'T,'U>() // create new empty dictionary
sub_list |> List.filter (fun k -> dict.ContainsKey k)
|> List.map (fun k -> (k, dict.[k]))
|> List.iter (fun s -> z.Add s)
z
// subsets a dictionary given a sub_list of keys to remove
let D_SubsetC (dict:Dictionary<'T,'U>) (sub_list:list<'T>) =
let z = dict
sub_list |> List.filter (fun k -> dict.ContainsKey k)
|> List.map (fun k -> (dict.Remove k)) |>ignore
z
// subsets a sequence by values in a sequence
let S_Subset (S:seq<'T>)(sub_list:seq<'T>) =
S |> Seq.filter (fun s-> Seq.exists (fun elem -> elem = s) sub_list)
let S_SubsetC (S:seq<'T>)(sub_list:seq<'T>) =
S |> Seq.filter (fun s-> not(Seq.exists (fun elem -> elem = s) sub_list))
[<EntryPoint>]
let main argv =
let mutable u = 0
let mutable v = 0
let mutable r = 0
let mutable N_cut = 1000000
let mutable cluster_A_min = seq [0]
let mutable cluster_B_min = seq [0]
let mutable WG = WG1
let mutable LiveNodeList = [0]
// when i = 2, i encounter problems with mutability
for i in 1 .. Nruns do
WG <- WG1
printfn "%d" i
for k in 1..(N-2) do
LiveNodeList <- GetLiveNode WG.RemoveTable
r <- rand.Next(0,N-k)
u <- LiveNodeList.[r] //selecting a live node
let uuu = WG.Graph.[u] |> Seq.map (fun s -> WG.LabelHead.[s] )
|> Seq.filter (IsNotRemoved1 WG)
|> Seq.distinct
let n_edge = uuu |> Seq.length
let x = rand.Next(1,n_edge)
let mutable ok = false //maybe we can take this out
while not(ok) do
// selecting the edge from node u
v <- WG.LabelHead.[Array.get (uuu |> Seq.toArray) (x-1)]
let vvv = WG.Graph.[v] |> Seq.map (fun s -> WG.LabelHead.[s] )
|> Seq.filter (IsNotRemoved1 WG)
|> Seq.distinct
let zzz = S_SubsetC (Seq.concat [uuu;vvv] |> Seq.distinct) [u;v]
WG.Graph.[u] <- zzz
let lab_u = WG.Label.[u]
let lab_v = WG.Label.[v]
WG.Label.[u] <- Seq.concat [lab_u;lab_v] |> Seq.distinct
if (k<N-1) then
WG.RemoveTable.[v]<-true
//updating Label_head for all members of Label.[v]
WG.LabelHead.[v]<- u
for j in WG.Label.[v] do
WG.LabelHead.[j]<- u
ok <- true
printfn "u= %d v=%d" u v
// end of for k in 1..(N-2)
// counting cuts
// u,v contain the 2 indexes of groupings
let cluster_A = WG.Label.[u]
let cluster_B = S_SubsetC (seq[for i in 1..N do yield i]) cluster_A // defined as complementary of A
// let WG2 = {Graph = D_Subset WG1.Graph (cluster_A |> Seq.toList)
// RemoveTable = remove_table
// Label = D_Subset WG1.Graph (cluster_A |> Seq.toList)
// LabelHead = label_head_table}
let cross_edge = // returns keyvalue pair (k,S')
let IsInCluster cluster (k,S) =
(k,S_Subset S cluster)
graphM |> toSeq |> Seq.map (IsInCluster cluster_B)
N_cut <-
cross_edge |> Seq.map (fun (k:int,v:int seq)-> Seq.length v)
|> Seq.sum
if (N_cut<min_cut) then
min_cut <- N_cut
WGmin <- WG
cluster_A_min <- cluster_A
cluster_B_min <- cluster_B
// end of for i in 1..Nruns
0 // return an integer exit code
Description of the algo: (i don't think its too essential to solve my problem)
at each trial, there are several steps. at each step, we merge 2 nodes into 1, (removing effectively 1) updating the graph. we do that 6 times until there are only 2 nodes left, which we define as 2 clusters, and we look at the number of cross edges between those 2 clusters. if we are "lucky" those 2 clusters would be (1,2,3,4) and (5,6,7,8) and find the right number of cuts.
at each step, the object WG is updated with the effects of merging 2 nodes
with only LiveNodes (the ones which are not eliminated as a result of merging 2 nodes) being perfectly kept up to date.
WG.Graph is the updated graph
WG.Label contains the labels of the nodes which have been merged into the current node
WG.LabelHead contains the label of the node into which that node has been merged
WG.RemoveTable says if the node has been removed or not.
Thanks in advance for anyone willing to take a look at it !
"It seems not working", because wgraphobj is a reference type, which is allocated on the stack, which means that when you're mutating the innards of WG, you're also mutating the innards of WG1, because they're the same innards.
This is precisely the kind of mess you get yourself into if you use mutable state. This is why people recommend to not use it. In particular, your use of mutable dictionaries undermines the robustness of your algorithm. I recommend using the F#'s own efficient immutable dictionary (called Map) instead.
Now, in response to your comment about WG.Graph <- GraphD giving compile error.
WG is mutable, but WG.Graph is not (but the contents of WG.Graph are again mutable). There is a difference, let me try to explain it.
WG is mutable in the sense that it points to some object of type wgraphobj, but you can make it, in the course of your program, to point at another object of the same type.
WG.Graph, on the other hand, is a field packed inside WG. It points to some object of type Dictionary<_,_>. And you cannot make it point to another object. You can create a different wgraphobj, in which the field Graph point to a different dictionary, but you cannot change where the field Graph of the original wgraphobj points.
In order to make the field Graph itself mutable, you can declare it as such:
type wgraphobj = {
mutable Graph: Dictionary<int, int seq>
...
Then you will be able to mutate that field:
WG.Graph <- GraphD
Note that in this case you do not need to declare the value WG itself as mutable.
However, it seems to me that for your purposes you can actually go the way of creating a new instance wgraphobj with the field Graph changed, and assigning it to the mutable reference WG:
WG.Graph <- { WG with Graph = GraphD }
I wrote this function which merges two lists together but as I'm fairly new to functional programming I was wondering whether there is a better (simpler) way to do it?
let a = ["a"; "b"; "c"]
let b = ["d"; "b"; "a"]
let merge a b =
// take all a and add b
List.fold (fun acc elem ->
let alreadyContains = acc |> List.exists (fun item -> item = elem)
if alreadyContains = true then
acc
else
elem :: acc |> List.rev
) b a
let test = merge a b
Expected result is: ["a"; "b"; "c"; "d"], I'm reverting the list in order to keep the original order. I thought I would be able to achieve the same using List.foldBack (and dropping List.rev) but it results in an error:
Type mismatch. Expecting a
'a
but given a
'a list
The resulting type would be infinite when unifying ''a' and ''a list'
Why is there a difference when using foldBack?
You could use something like the following
let merge a b =
a # b
|> Seq.distinct
|> List.ofSeq
Note that this will preserve order and remove any duplicates.
In F# 4.0 this will be simplified to
let merge a b = a # b |> List.distinct
If I wanted to write this in a way that is similar to your original version (using fold), then the main change I would do is to move List.rev outside of the function (you are calling List.rev every time you add a new element, which is wrong if you're adding even number of elements!)
So, a solution very similar to yours would be:
let merge a b =
(b, a)
||> List.fold (fun acc elem ->
let alreadyContains = acc |> List.exists (fun item -> item = elem)
if alreadyContains = true then acc
else elem :: acc)
|> List.rev
This uses the double-pipe operator ||> to pass two parameters to the fold function (this is not necessary, but I find it a bit nicer) and then passes the result to List.rev.
Do you know the nicest way to make this work :
let toTableau2D (seqinit:seq<'a*'b*'c>) =
let myfst = fun (a,b,c) -> a
let myscd = fun (a,b,c) -> b
let mytrd = fun (a,b,c) -> c
let inputd = seqinit |> groupBy2 myfst myscd
there must be a better way than rewriting fst..
UPDATE
After pad advice, I rewrote packing the previous 'a*'b into a single structure
My code now looks like
let toTableau (seqinit:seq<'a*'b>) =
let inputd = seqinit |> Seq.groupBy fst |> toMap
let keys = seqinit |> Seq.map fst |> Set.ofSeq |> List.ofSeq
...
Why don't you just write it explicitly:
let toTableau2D (a, b, c) =
let toto = a
// ...
If you want to refer to seqinit later on, you always can reconstruct the triple or use the named pattern:
let toTableau2D ((a, b, c) as seqinit) =
let toto = a
// Do something with seqinit
// ...
EDIT:
Unless you use reflection, you cannot have fst function for any kind of tuples. In your example, writing some utility functions and reusing them doesn't hurt:
let fst3 (a, _, _) = a
let snd3 (_, b, _) = b
let thd3 (_, _, c) = c
let toTableau2D (seqinit: seq<'a*'b*'c>) =
let inputd = seqinit |> groupBy2 fst3 snd3
// ...
If you want to make this work for arbitrary number of tuple elements, consider changing tuples to lists and employing pattern matching on lists.
+1 to what #pad said. Otherwise (if you just simplified what you're trying to do and are stuck with seqinit defined that way) I guess you can always do:
let toTableau2D (seqinit:'a*'b*'c) =
let toto, _, _ = seqinit
//...
I am prototyping how I am going to handle Double.NaN values in an F# array, and the first step, trying to simply count how many there are, has me stumped. The value "howMany" comes back as zero in my code, but I know there are 2, because I set 2 value to be Double.NaN. Can anyone point out what I am missing? Thanks!
let rnd = new System.Random()
let fakeAlphas = Array.init 10 (fun _ -> rnd.NextDouble());;
fakeAlphas.[0] <- Double.NaN;
fakeAlphas.[1] <- Double.NaN;
let countNA arr = arr |> Array.filter (fun x -> x = Double.NaN) |> Array.length;;
let howMany = countNA fakeAlphas;;
To answer the general question in the title:
let HowManySatisfy pred = Seq.filter pred >> Seq.length
for example
let nums = [1;2;3;4;5]
let countEvens = nums |> HowManySatisfy (fun n -> n%2=0)
printfn "%d" countEvens
Double.NaN = n is false for all n. See the MSDN page for Double.NaN.
Instead use Double.IsNaN. See the MSDN page for more information.
I think you need to use the Double.IsNan method. So your filter function would be:
(fun x -> Double.IsNan x)
I believe the issue is that NaN never equals anything -- even another NaN!