How to modify below code to Return "string" so that returned output displayed on my MVC page and also would like to accept enteredChar from user.
Is there better way to do create this pyramid?
Current code:
let enteredChar = 'F' // As Interactive window doesn't support to Read Input
let mylist = ['A'..enteredChar]
let mylistlength = mylist |> List.length
let myfunc i x tlist1 =
(for j = 0 to mylistlength-i-2 do printf "%c" ' ')
let a1 = [for p in tlist1 do if p < x then yield p]
for p in a1 do printf "%c" p
printf "%c" x
let a2 = List.rev a1
for p in a2 do printf "%c" p
printfn "%s" " "
mylist |> List.iteri(fun i x -> myfunc i x mylist)
Output:
A
ABA
ABCBA
ABCDCBA
ABCDEDCBA
ABCDEFEDCBA
A few small optimizations could be:
Use StringBuilder instead of printf which is quite slow with long strings.
Use Array instead of List since Array works better with String.
Here is a version producing a pyramid string, which is kept closely to your code:
open System
open System.Text
let generateString c =
let sb = StringBuilder()
let generate i x arr =
String.replicate (Array.length arr-i-1) " " |> sb.Append |> ignore
let a1 = Array.filter (fun p -> p < x) arr
String(a1) |> sb.Append |> ignore
sb.Append x |> ignore
String(Array.rev a1) |> sb.Append |> ignore
sb.AppendLine " " |> ignore
let arr = [|'A'..c|]
arr |> Array.iteri(fun i x -> generate i x arr)
sb.ToString()
generateString 'F' |> printfn "%s"
As an alternative to Daniel's solution, you can achieve what you want with minimal changes to the code logic. Instead of using printf that writes the output to the console, you can use Printf.bprintf which writes the output to a specified StringBuilder. Then you can simply get the resulting string from the StringBuilder.
The modified function will look like this. I added parameter str and replaced printf with Printf.bprintf str (and printfn with bprintf together with additional \n char):
let myfunc i x tlist1 str =
(for j = 0 to mylistlength-i-2 do Printf.bprintf str "%c" ' ')
let a1 = [for p in tlist1 do if p < x then yield p]
for p in a1 do Printf.bprintf str "%c" p
Printf.bprintf str "%c" x
let a2 = List.rev a1
for p in a2 do Printf.bprintf str "%c" p
Printf.bprintf str "%s\n" " "
To call the function, you first create StringBuilder and then pass it to myfunc in every call. At the end, you can get the result using ToString method:
let str = StringBuilder()
mylist |> List.iteri(fun i x -> myfunc i x mylist str)
str.ToString()
I think Daniel's solution looks nicer, but this is the most direct way to tunr your printing code into a string-building code (and it can be done, pretty much, using Search & Replace).
If I understand your question (this likely belongs on Code Review) here's one way to rewrite your function:
let showPyramid (output: TextWriter) lastChar =
let chars = [|'A' .. lastChar|]
let getRowChars n =
let rec loop acc i =
[|
if i < n then let c = chars.[i] in yield c; yield! loop (c::acc) (i+1)
else yield! List.tail acc
|]
loop [] 0
let n = chars.Length
for r = 1 to n do
output.WriteLine("{0}{1}{0}", String(' ', n - r), String(getRowChars r))
Example
showPyramid Console.Out 'F'
or, to output to a string
use output = new StringWriter()
showPyramid output 'F'
let pyramid = output.ToString()
EDIT
After seeing Tomas' answer I realized I skipped over "return a string" in your question. I updated the code and added examples to show how you could do that.
let pyramid (ch:char) =
let ar = [| 'A'..ch |]
let len = ar.Length
Array.mapi
(fun i ch ->
let ar = ar.[0..i]
String.replicate (len - i - 1) " " + new string(ar) + new string((Array.rev ar).[1..]))
ar
|> String.concat "\n"
pyramid 'F' |> printfn "%s"
Here's another approach that seems to be a good demonstration of functional composition. I bet it's the shortest solution among the answers here. :)
let charsToString = Seq.map string >> String.concat String.Empty
let pyramid lastChar =
let src = '-'::['A'..lastChar] |> List.toArray
let len = Array.length src - 1
fun row col -> row-abs(col-len+1)+1 |> max 0 |> Array.get src // (1)
>> Seq.init (len*2-1) >> charsToString // (2)
|> Seq.init len // (3)
pyramid 'X' |> Seq.iter (printfn "%s")
First, we generate an unusual array of initial data. Its element [0] contains a space or whatever separator you want to have; I preferred dash (-) for debugging purposes.
The (1) line makes a function that calculates what character to be placed. The result of row-abs(col-len+1)+1 can be either positive (and there is a char to be placed) or zeronegative, and there should be a space. Note that there is no if statement: it is hidden within the max function;
The (2) line composes a function int -> string for generating an individual row;
The (3) line passes the function above as argument for sequence initializer.
The three lines can be written in a more verbose way:
let genCell row col = row-abs(col-len+1)+1 |> max 0 |> Array.get src
let genRow = genCell >> Seq.init (len*2-1) >> charsToString
Seq.init len genRow
Note genRow needs no formal argument due to functional composition: the argument is being bound into genCell, returning a function of a single argument, exactly what Seq.init needs.
Related
Let's says I have a string of a length N that contains only 0 or 1. I want to split that string in multiples strings and each string should contains only one digit.
Example:
00011010111
Should be split into:
000
11
0
1
0
111
The only solution I can think of if using a for loop with a string builder (Written in pseudo code below, more c# like sorry):
result = new list<string>
tmpChar = ""
tmpString = ""
for each character c in MyString
if tmpchar != c
if tmpsString != ""
result.add tmpString
tmpString = ""
endIf
tmpchar = c
endIf
tmpString += tmpChar
endFor
Do you have any other solution and maybe a clever solution that use a more functional approach?
I think Seq.scan would be a good fit for this, this is a very procedural problem in nature, preserving the order like that. But here is code that I believe does what you are asking.
"00011010111"
|> Seq.scan (fun (s, i) x ->
match s with
| Some p when p = x -> Some x, i
| _ -> Some x, i + 1 ) (None, 0)
|> Seq.countBy id
|> Seq.choose (function
| (Some t, _), n -> Some(t, n)
| _ -> None )
|> Seq.toList
Perhaps something along the lines of:
let result =
let rec groupWhileSame xs result =
match xs with
| a when a |> Seq.isEmpty -> result
| _ ->
let head = xs |> Seq.head
let str = xs |> Seq.takeWhile ((=) head)
let rest = xs |> Seq.skipWhile ((=) head)
groupWhileSame rest (Seq.append result [str])
groupWhileSame (myStr) []
Seq.fold (fun (acc:(string list)) x ->
match acc with
| y::rst when y.StartsWith(string x) -> (string x) + y::rst
| _ -> (string x)::acc)
[]
"00011010111"
Consider this function (which is generic):
let chunk s =
if Seq.isEmpty s then []
else
let rec chunk items chunks =
if Seq.isEmpty items then chunks
else
let chunks' =
match chunks with
| [] -> [(Seq.head items, 1)]
| x::xs ->
let c,n = x in let c' = Seq.head items in
if c = c' then (c, n + 1) :: xs else (c', 1) :: x :: xs
chunk (Seq.tail items) chunks'
chunk s [] |> List.rev
It returns a list of tuples, where each tuple represents an item and its repetitions.
So
"00011010111" |> Seq.toList |> chunk
actually returns
[('0', 3); ('1', 2); ('0', 1); ('1', 1); ('0', 1); ('1', 3)]
Basically, we're doing run length encoding (which is admittedly a bit wasteful in the case of your example string).
To get the list of strings that you want, we use code like following:
"00011010111"
|> Seq.toList
|> chunk
|> List.map (fun x -> let c,n = x in new string(c, n))
Here's a working version of OP's proposal with light syntax:
let chunk (s: string) =
let result = System.Collections.Generic.List<string>()
let mutable tmpChar = ""
let mutable tmpString = ""
for c in s do
if tmpChar <> string c then
if tmpString <> "" then
result.Add tmpString
tmpString <- ""
tmpChar <- string c
tmpString <- tmpString + tmpChar
result.Add tmpString
result
No attempt was made to follow a functional style.
I have a F# list and I'm taking two elements of that list.
If the list has 10 elements in it :
let rnd = new Random()
let elem1 = list.Item(rnd.Next(0,9))
let elem2 = list.Item(rnd.Next(0,9))
There is a chance elem1 and elem2 are equal.
I have checked some workarounds and most of them work using a do while, but I don't want to implement a function that may never end in F#.
Is there a way to create a restriction in the random function?
First random : 0 <= x <= 9
Second random : 0 <= y <= 9 <> x
A simple solution:
let rnd = new Random()
let ndx1 = rnd.Next(9)
let ndx2 =
let x = rnd.Next(8)
if x < ndx1 then x else x + 1
let elem1, elem2 = list.[ndx1], list.[ndx2]
Another way, using maths and calling the random function once:
let r = Random().Next(9 * 8)
let x = 1 + r + r / 9
let elem1, elem2 = list.[x / 9], list.[x % 9]
which may be generalised to:
let getTwoElements lst =
let c = List.length lst
let x, y = Math.DivRem(Random().Next(c * (c-1)) * (c+1) / c + 1, c)
lst.[x], lst.[y]
A more declarative approach, taking into account your comment about points in the image:
let rnd = System.Random()
/// this will build you a list of 10 pairs of indices where a <> b.
let indices =
Seq.initInfinite (fun _ -> rnd.Next(0,10), rnd.Next(0,10))
|> Seq.filter (fun (a,b) -> a <> b)
|> Seq.take 10
|> List.ofSeq
/// map indices into actual points.
let elems =
let points = list |> Array.ofList
List.map (fun (a, b) -> points.[a], points.[b]) indices
As a side note, do not use random access on lists. They're not made for that and performance of that is poor. Convert them to an array first.
There are lots of way to achieve this. A simple one would be something like this:
open System
open System.Linq
let rnd = new Random()
let elem1 = list.Item(rnd.Next(0,9))
let elem2 = list.Where(fun x->x <> elem1).ElementAt(rnd.Next(0,8))
Anyone can tell how to print a Stack data structure in OCaml?
The build-in Stack type definition is like :
type 'a t = { mutable c : 'a list }
exception Empty
let create () = { c = [] }
let clear s = s.c <- []
let push x s = s.c <- x :: s.c
let pop s = match s.c with hd::tl -> s.c <- tl; hd | [] -> raise Empty
let length s = List.length s.c
let iter f s = List.iter f s.c
Want to print and keep its elements in place, which means do not use pop and push.
Better to use the pattern matching to complete the problem.
Code should be like:
let print_stack stack =???
This looks like it could be homework. You should show something you've tried that didn't work, and explain why you think it didn't work. This is a lot more valuable that just having somebody give you the answer.
If it's not homework: if you think about it, you can find a good implementation at another place in the standard library. The implementation of Stack.iter tells you where to look.
It seems like the function Stack.iter does exactly what you want:
let print_stack print_elem stack = Stack.iter print_elem
Where. print_elem prints one element of the stack.
e.g. let print_elem_int n = (print_int n; print_newline ())
Finally get the answer:
let rec print_s {c=l}=
match l with
| [] -> raise Empty
| [x] -> print_int x; print_string " "
| h :: ts -> print_int h; print_string " "; print_s {c=ts}
;;
Improved version:
let print_s2 {c=l}=
let rec print_l list =
match list with
| [] -> raise Empty
| [x] -> print_int x; print_string " "
| h :: ts -> print_int h; print_string " "; print_l ts
in
print_l l
;;
I'm trying to learn F# by rewriting some C# algorithms I have into idiomatic F#.
One of the first functions I'm trying to rewrite is a batchesOf where:
[1..17] |> batchesOf 5
Which would split the sequence into batches with a max of five in each, i.e:
[[1; 2; 3; 4; 5]; [6; 7; 8; 9; 10]; [11; 12; 13; 14; 15]; [16; 17]]
My first attempt at doing this is kind of ugly where I've resorted to using a mutable ref object after running into errors trying to use mutable type inside the closure. Using ref is particularly unpleasant since to dereference it you have to use the ! operator which when inside a condition expression can be counter intuitive to some devs who will read it as logical not. Another problem I ran into is where Seq.skip and Seq.take are not like their Linq aliases in that they will throw an error if size exceeds the size of the sequence.
let batchesOf size (sequence: _ seq) : _ list seq =
seq {
let s = ref sequence
while not (!s |> Seq.isEmpty) do
yield !s |> Seq.truncate size |> List.ofSeq
s := System.Linq.Enumerable.Skip(!s, size)
}
Anyway what would be the most elegant/idiomatic way to rewrite this in F#? Keeping the original behaviour but preferably without the ref mutable variable.
Implementing this function using the seq<_> type idiomatically is difficult - the type is inherently mutable, so there is no simple nice functional way. Your version is quite inefficient, because it uses Skip repeatedly on the sequence. A better imperative option would be to use GetEnumerator and just iterate over elements using IEnumerator. You can find various imperative options in this snippet: http://fssnip.net/1o
If you're learning F#, then it is better to try writing the function using F# list type. This way, you can use idiomatic functional style. Then you can write batchesOf using pattern matching with recursion and accumulator argument like this:
let batchesOf size input =
// Inner function that does the actual work.
// 'input' is the remaining part of the list, 'num' is the number of elements
// in a current batch, which is stored in 'batch'. Finally, 'acc' is a list of
// batches (in a reverse order)
let rec loop input num batch acc =
match input with
| [] ->
// We've reached the end - add current batch to the list of all
// batches if it is not empty and return batch (in the right order)
if batch <> [] then (List.rev batch)::acc else acc
|> List.rev
| x::xs when num = size - 1 ->
// We've reached the end of the batch - add the last element
// and add batch to the list of batches.
loop xs 0 [] ((List.rev (x::batch))::acc)
| x::xs ->
// Take one element from the input and add it to the current batch
loop xs (num + 1) (x::batch) acc
loop input 0 [] []
As a footnote, the imperative version can be made a bit nicer using computation expression for working with IEnumerator, but that's not standard and it is quite advanced trick (for example, see http://fssnip.net/37).
A friend asked me this a while back. Here's a recycled answer. This works and is pure:
let batchesOf n =
Seq.mapi (fun i v -> i / n, v) >>
Seq.groupBy fst >>
Seq.map snd >>
Seq.map (Seq.map snd)
Or an impure version:
let batchesOf n =
let i = ref -1
Seq.groupBy (fun _ -> i := !i + 1; !i / n) >> Seq.map snd
These produce a seq<seq<'a>>. If you really must have an 'a list list as in your sample then just add ... |> Seq.map (List.ofSeq) |> List.ofSeq as in:
> [1..17] |> batchesOf 5 |> Seq.map (List.ofSeq) |> List.ofSeq;;
val it : int list list = [[1; 2; 3; 4; 5]; [6; 7; 8; 9; 10]; [11; 12; 13; 14; 15]; [16; 17]]
Hope that helps!
This can be done without recursion if you want
[0..20]
|> Seq.mapi (fun i elem -> (i/size),elem)
|> Seq.groupBy (fun (a,_) -> a)
|> Seq.map (fun (_,se) -> se |> Seq.map (snd));;
val it : seq<seq<int>> =
seq
[seq [0; 1; 2; 3; ...]; seq [5; 6; 7; 8; ...]; seq [10; 11; 12; 13; ...];
seq [15; 16; 17; 18; ...]; ...]
Depending on how you think this may be easier to understand. Tomas' solution is probably more idiomatic F# though
Hurray, we can use List.chunkBySize, Seq.chunkBySize and Array.chunkBySize in F# 4, as mentioned by Brad Collins and Scott Wlaschin.
This isn't perhaps idiomatic but it works:
let batchesOf n l =
let _, _, temp', res' = List.fold (fun (i, n, temp, res) hd ->
if i < n then
(i + 1, n, hd :: temp, res)
else
(1, i, [hd], (List.rev temp) :: res))
(0, n, [], []) l
(List.rev temp') :: res' |> List.rev
Here's a simple implementation for sequences:
let chunks size (items:seq<_>) =
use e = items.GetEnumerator()
let rec loop i acc =
seq {
if i = size then
yield (List.rev acc)
yield! loop 0 []
elif e.MoveNext() then
yield! loop (i+1) (e.Current::acc)
else
yield (List.rev acc)
}
if size = 0 then invalidArg "size" "must be greater than zero"
if Seq.isEmpty items then Seq.empty else loop 0 []
let s = Seq.init 10 id
chunks 3 s
//output: seq [[0; 1; 2]; [3; 4; 5]; [6; 7; 8]; [9]]
My method involves converting the list to an array and recursively chunking the array:
let batchesOf (sz:int) lt =
let arr = List.toArray lt
let rec bite curr =
if (curr + sz - 1 ) >= arr.Length then
[Array.toList arr.[ curr .. (arr.Length - 1)]]
else
let curr1 = curr + sz
(Array.toList (arr.[curr .. (curr + sz - 1)])) :: (bite curr1)
bite 0
batchesOf 5 [1 .. 17]
[[1; 2; 3; 4; 5]; [6; 7; 8; 9; 10]; [11; 12; 13; 14; 15]; [16; 17]]
I found this to be a quite terse solution:
let partition n (stream:seq<_>) = seq {
let enum = stream.GetEnumerator()
let rec collect n partition =
if n = 1 || not (enum.MoveNext()) then
partition
else
collect (n-1) (partition # [enum.Current])
while enum.MoveNext() do
yield collect n [enum.Current]
}
It works on a sequence and produces a sequence. The output sequence consists of lists of n elements from the input sequence.
You can solve your task with analog of Clojure partition library function below:
let partition n step coll =
let rec split ss =
seq {
yield(ss |> Seq.truncate n)
if Seq.length(ss |> Seq.truncate (step+1)) > step then
yield! split <| (ss |> Seq.skip step)
}
split coll
Being used as partition 5 5 it will provide you with sought batchesOf 5 functionality:
[1..17] |> partition 5 5;;
val it : seq<seq<int>> =
seq
[seq [1; 2; 3; 4; ...]; seq [6; 7; 8; 9; ...]; seq [11; 12; 13; 14; ...];
seq [16; 17]]
As a premium by playing with n and step you can use it for slicing overlapping batches aka sliding windows, and even apply to infinite sequences, like below:
Seq.initInfinite(fun x -> x) |> partition 4 1;;
val it : seq<seq<int>> =
seq
[seq [0; 1; 2; 3]; seq [1; 2; 3; 4]; seq [2; 3; 4; 5]; seq [3; 4; 5; 6];
...]
Consider it as a prototype only as it does many redundant evaluations on the source sequence and not likely fit for production purposes.
This version passes all my tests I could think of including ones for lazy evaluation and single sequence evaluation:
let batchIn batchLength sequence =
let padding = seq { for i in 1 .. batchLength -> None }
let wrapped = sequence |> Seq.map Some
Seq.concat [wrapped; padding]
|> Seq.windowed batchLength
|> Seq.mapi (fun i el -> (i, el))
|> Seq.filter (fun t -> fst t % batchLength = 0)
|> Seq.map snd
|> Seq.map (Seq.choose id)
|> Seq.filter (fun el -> not (Seq.isEmpty el))
I am still quite new to F# so if I'm missing anything - please do correct me, it will be greatly appreciated.
This is probably trivial, and I do have a solution but I'm not happy with it. Somehow, (much) simpler forms don't seem to work and it gets messy around the corner cases (either first, or last matching pairs in a row).
To keep it simple, let's define the matching rule as any two or more numbers that have a difference of two. Example:
> filterTwins [1; 2; 4; 6; 8; 10; 15; 17]
val it : int list = [2; 4; 6; 8; 10; 15; 17]
The code I currently use is this, which just feels sloppy and overweight:
let filterTwins list =
let func item acc =
let prevItem, resultList = acc
match prevItem, resultList with
| 0, []
-> item, []
| var, [] when var - 2 = item
-> item, item::var::resultList
| var, hd::tl when var - 2 = item && hd <> var
-> item, item::var::resultList
| var, _ when var - 2 = item
-> item, item::resultList
| _
-> item, resultList
List.foldBack func list (0, [])
|> snd
I intended my own original exercise to experiment with List.foldBack, large lists and parallel programming (which went well) but ended up messing with the "easy" part...
Guide through the answers
Daniel's last, 113 characters*, easy to follow, slow
Kvb's 2nd, 106 characters* (if I include the function), easy, but return value requires work
Stephen's 2nd, 397 characters*, long winded and comparably complex, but fastest
Abel's, 155 characters*, based on Daniel's, allows duplicates (this wasn't a necessity, btw) and is relatively fast.
There were more answers, but the above were the most distinct, I believe. Hope I didn't hurt anybody's feelings by accepting Daniel's answer as solution: each and every one solution deserves to be the selected answer(!).
* counting done with function names as one character
Would this do what you want?
let filterTwins l =
let rec filter l acc flag =
match l with
| [] -> List.rev acc
| a :: b :: rest when b - 2 = a ->
filter (b::rest) (if flag then b::acc else b::a::acc) true
| _ :: t -> filter t acc false
filter l [] false
This is terribly inefficient, but here's another approach using more built-in functions:
let filterTwinsSimple l =
l
|> Seq.pairwise
|> Seq.filter (fun (a, b) -> b - 2 = a)
|> Seq.collect (fun (a, b) -> [a; b])
|> Seq.distinct
|> Seq.toList
Maybe slightly better:
let filterTwinsSimple l =
seq {
for (a, b) in Seq.pairwise l do
if b - 2 = a then
yield a
yield b
}
|> Seq.distinct
|> Seq.toList
How about this?
let filterPairs f =
let rec filter keepHead = function
| x::(y::_ as xs) when f x y -> x::(filter true xs)
| x::xs ->
let rest = filter false xs
if keepHead then x::rest else rest
| _ -> []
filter false
let test = filterPairs (fun x y -> y - x = 2) [1; 2; 4; 6; 8; 10; 15; 17]
Or if all of your list's items are unique, you could do this:
let rec filterPairs f s =
s
|> Seq.windowed 2
|> Seq.filter (fun [|a;b|] -> f a b)
|> Seq.concat
|> Seq.distinct
let test = filterPairs (fun x y -> y - x = 2) [1; 2; 4; 6; 8; 10; 15; 17]
EDIT
Or here's another alternative which I find elegant. First define a function for breaking a list into a list of groups of consecutive items satisfying a predicate:
let rec groupConsec f = function
| [] -> []
| x::(y::_ as xs) when f x y ->
let (gp::gps) = groupConsec f xs
(x::gp)::gps
| x::xs -> [x]::(groupConsec f xs)
Then, build your function by collecting all results back together, discarding any singletons:
let filterPairs f =
groupConsec f
>> List.collect (function | [_] -> [] | l -> l)
let test = filterPairs (fun x y -> y - x = 2) [1; 2; 4; 6; 8; 10; 15; 17]
The following solution is in the spirit of your own, but I use a discriminate union to encapsulate aspects of the algorithm and reign in the madness a bit:
type status =
| Keep of int
| Skip of int
| Tail
let filterTwins xl =
(Tail, [])
|> List.foldBack
(fun cur (prev, acc) ->
match prev with
| Skip(prev) when prev - cur = 2 -> (Keep(cur), cur::prev::acc)
| Keep(prev) when prev - cur = 2 -> (Keep(cur), cur::acc)
| _ -> (Skip(cur), acc))
xl
|> snd
Here's another solution which uses a similar discriminate union strategy as my other answer but it works on sequences lazily so you can watch those twin (primes?) roll in as they come:
type status =
| KeepTwo of int * int
| KeepOne of int
| SkipOne of int
| Head
let filterTwins xl =
let xl' =
Seq.scan
(fun prev cur ->
match prev with
| KeepTwo(_,prev) | KeepOne prev when cur - prev = 2 ->
KeepOne cur
| SkipOne prev when cur - prev = 2 ->
KeepTwo(prev,cur)
| _ ->
SkipOne cur)
Head
xl
seq {
for x in xl' do
match x with
| KeepTwo(a,b) -> yield a; yield b
| KeepOne b -> yield b
| _ -> ()
}
for completeness sake, I'll answer this with what I eventually came up with, based on the friendly suggestions in this thread.
The benefits of this approach are that it doesn't need Seq.distinct, which I believe is an improvement as it allows for duplicates. However, it still needs List.rev which doesn't make it the fastest. Nor is it the most succinct code (see comparison of solution in question itself).
let filterTwins l =
l
|> Seq.pairwise
|> Seq.fold (fun a (x, y) ->
if y - x = 2 then (if List.head a = x then y::a else y::x::a)
else a) [0]
|> List.rev
|> List.tail