I'm new to F# and I'm struggling to figure out how I could combine the numbers in a float list to a float.
If I have the list
let floatList = [ 9.0; 8.0; 3.0 ]
I would like a function where the return value is a float of value 983.0. How would I go about this problem?
You go through the list, multiply the previously accumulated number by 10, then add the current element of the list:
((9 * 10) + 8) * 10 + 3 = 983
A handy way to go through the list while keeping an "accumulator" is List.fold:
floatList
|> List.fold (fun acc n -> acc*10 + n) 0
Chain rev, mapi(map with index) and sum functions
let digits = [| 9.0; 8.0; 3.0 |]
let result =
Array.rev digits
|> Array.mapi (fun i x -> x * (10.0 ** float i))
|> Array.sum
printfn "%f" result
Because array can return it's length without iterating over the collection, you can remove "reversing" and calculate offset for power based on the index and length
let digits = [| 9.0; 8.0; 3.0 |]
let result =
Array.mapi (fun i x -> x * (10.0 ** float (digits.Length - i))) digits
|> Array.sum
printfn "%f" result
This is a way to do it
let floatList = [ 9.0; 8.0; 3.0 ]
let rec loop mult list =
match list with
| head :: tail -> (mult * head) + (loop (mult / 10.) tail)
| [] -> 0.
let mult = 10. ** float ((List.length floatList) - 1) // len 1 = 1. len 2 = 10. len 3 = 100
let result = loop mult floatList
printfn "%f" result
Related
Let's say I have following equation . My goal is to create sequence which returns next elements of this. Here's my solution and it works:
let rec factorial(n:float) =
match n with
|0.0 -> 1.0
|n -> n * factorial(n-1.0)
let seq1 = Seq.initInfinite( fun i -> factorial(float(i)) / sqrt(float(i)+1.0) ))
Now, analogically, I would like to create sequence which return elements according to equation:
I've got some code, but it's wrong so how to make it work?
let seq2(x:float) = Seq.initInfinite(fun a -> let i = float(a)
(1.0/factorial(0.0)) + System.Math.Pow(x,i)/factorial(i) )
Can't you skip the (1.0/factorial(0.0)) part of the equation (or maybe I misunderstood the question).
edit: i.e
let seq2(x:float) =
Seq.initInfinite(fun a ->
let i = float(a) in
System.Math.Pow(x,i)/factorial(i))
edit: to truncate a seq you can use 'take' and to sum you can use 'sum'. As in
let seq2sum nbelems =
seq2 >> Seq.take nbelems >> Seq.sum
then you get seq2sum 12 3.0 equal to approx 20 :-)
The great thing about functional languages is that you can have your solution be as close an expression of the original definition as possible.
You can avoid explicit type declarations for most functions:
let rec factorial = function
| 0 -> 1
| n -> n * (factorial (n-1))
let e x n =
seq { 0 .. n }
|> Seq.map(fun i -> x ** (float i) / float (factorial i))
|> Seq.sum
In the infinite series, you will have to take the first n entries before you sum, as an infinite series will never finish evaluating:
let e' x n =
Seq.initInfinite(fun i -> x ** (float i) / float (factorial i))
|> Seq.take n
|> Seq.sum
e 1.0 10 //2.718281801
e' 1.0 10 //2.718281801
Trying to learn F#, by solving some programming puzzles. I don't want to add too many details about the problem as I don't want to spoil the fun for others.
Basically, the issue is to find all 4-uples { (i,j,k,l) | i ^ j ^ k ^ l != 0 } with no repetition (eg., (1,1,1,2) and (1,1,2,1) are the same and should be counted just once).
I have found a O(n^3) approach which works, please see countImperative(a,b,c,d) below. But I also tried to refactor the code as to get rid of the nested for loops. However, I could not do so without a significant performance penalty. It was my impression that F#'s syntactic sugar would allow a more concise style (using pipes and folds), letting the compiler do the heavy-lifting to produce comparably fast code (compared to my nested for loops). The big performance hit comes from the calculation of the partial2 sum.
Here's the code:
open System
open System.Diagnostics
open System.Collections
module quadruples =
[<EntryPoint>]
let main argv =
let input = "2000 2000 2000 2000"
let ordered = [ for x in input.Split([|' '|]) -> Convert.ToInt32(x) ] |> List.sort
let a,b,c,d = ordered.[0], ordered.[1], ordered.[2], ordered.[3]
let inner(a,b) = a * (a-1) / 2 + a * (b-a)
let sw = new Stopwatch()
sw.Start()
let partial1 = [ 1.. b ] |> List.fold (fun acc j -> acc + (int64 ((min a j) * inner(c-j+1, d-j+1)))) 0L
sw.Stop()
let elapsed1 = (sw.ElapsedMilliseconds |> double) / 1000.0
printfn "Partial1: %f s" elapsed1
sw.Restart()
let combinations = [ for i in 1..a do for j in i+1..b do yield (j,i^^^j) ]
let range = [ 1..c ]
let partial2 = combinations |> List.fold(fun acc (j,x) -> acc + (range |> List.skip(j-1) |> List.fold(fun acc k -> if k ^^^ x < k || k ^^^ x > d then acc + 1L else acc) 0L)) 0L
sw.Stop()
let elapsed2 = (sw.ElapsedMilliseconds |> double) / 1000.0
printfn "Partial2: %f s" elapsed2
printfn "Functional: %d, Elapsed: %f s" (partial1 + partial2) (elapsed1 + elapsed2)
// "imperative" approach
let countImperative(a,b,c,d) =
let mutable count = seq { 1..b } |> Seq.fold (fun acc j -> acc + (int64 ((min a j) * inner(c-j+1, d-j+1)))) 0L
for i in 1..a do
for j in i+1..b do
let x = i ^^^ j
for k in j..c do
let y = x ^^^ k
if y < k || y > d then
count <- count + 1L
count
sw.Restart();
let count = countImperative(a,b,c,d)
sw.Stop()
printfn "Imperative: %d, Elapsed: %f s" count ((sw.ElapsedMilliseconds |> double) / 1000.0)
0 // return an integer exit code
So my question was, if there is any way to speed up the code (specifically the calculation of partial2) while maintaining F#'s nice syntax.
I am working through an F# function that calculates variance. I'm trying to step through each iteration to get the correct answer but think im getitng off track somewhere because I keep getting the wrong answer. Could someone please walk me through one iteration of this function to get me back on track
let variance values
let average = Seq.average values
let length = Seq.length values
values
|> Seq.map (fun x -> 1.0 / float length * (x - average) ** 2.0)
|> Seq.sum
call is variance [1.0..6.0]
To me the first value passed is 1.0 so it would be (1.0 / 6 * (1.0-3.5) ** 2.0) and therefore .166 * -2.5 ** 2.0
I'm also unsure what the ** means in formula I'm assuming multiply.
Correct answer should be 2.9166666667
To make it easier to understand, you can rewrite the code as follows:
let variance values =
let average = Seq.average values
let length = Seq.length values
let sum = values
|> Seq.map (fun x -> (x - average) ** 2.0)
|> Seq.sum in sum / float length
variance [1.0..6.0] |> printfn "%A"
Print: 2.916666667
Link: https://dotnetfiddle.net/09PHXn
By iteration:
let variancetest values =
let average = Seq.average values
let length = Seq.length values
values
|> Seq.iteri(fun i x ->
printfn "%i [%f]: %f ^ 2 = %A" i x (x - average) ((x - average) ** 2.0))
let sum = values
|> Seq.map (fun x -> (x - average) ** 2.0)
|> Seq.sum
let flength = float length
printfn "Sum = %f" sum
printfn "1/length = %f" (1.0 / flength)
printfn "Result / length = %f" (sum / flength)
variancetest [1.0..6.0]
Print:
0 [1.000000]: -2.500000 ^ 2 = 6.25
1 [2.000000]: -1.500000 ^ 2 = 2.25
2 [3.000000]: -0.500000 ^ 2 = 0.25
3 [4.000000]: 0.500000 ^ 2 = 0.25
4 [5.000000]: 1.500000 ^ 2 = 2.25
5 [6.000000]: 2.500000 ^ 2 = 6.25
Sum = 17.500000
1/length = 0.166667
Result / length = 2.916667
https://dotnetfiddle.net/02r3qG
let highs = [| 2; 4; 6 |]
let lows = [| 1; 5; 10 |]
I want to get 2 arrays from the above: if the element in highs is smaller than the corresponding element in lows, then swap them. So, I can get the final 2 arrays:
let trueHighs = [| 2; 5; 10 |]
let trueLows = [| 1; 4; 6 |]
How do I do this?
Similar with JaredPar's answer but simpler:
let trueHighs, trueLows =
Array.zip highs lows
|> Array.map (fun (x, y) -> if x >= y then (x, y) else (y, x))
|> Array.unzip
Another more concise version:
let trueHighs, trueLows =
(highs, lows)
||> Array.map2 (fun x y -> if x >= y then (x, y) else (y, x))
|> Array.unzip
Here is the code you should use:
let n = highs.Length
let trueHighs = Array.init n (fun i -> max highs.[i] lows.[i])
let trueLows = Array.init n (fun i -> min highs.[i] lows.[i])
If performance is uber-critical, you're probably better off with an imperative approach.
let n = highs.Length
let trueHighs = Array.zeroCreate n
let trueLows = Array.zeroCreate n
for i = 0 to n-1 do
let hi = highs.[i]
let lo = lows.[i]
if hi > lo then
trueHighs.[i] <- hi
trueLows.[i] <- lo
else
trueHighs.[i] <- lo
trueLows.[i] <- hi
Try the following
let trueHighs, trueLows =
let zipped =
highs
|> Seq.ofArray
|> Seq.zip (lows |> Seq.ofArray)
|> Seq.map (fun (x, y) -> min x y, max x y)
let lows = zipped |> Seq.map fst |> Array.ofSeq
let highs = zipped |> Seq.map snd |> Array.ofSeq
highs, lows
My code (below) falls over with a stack overflow exception. Im assuming F# isnt like haskell and dosent play well with recursive lists. Whats the correct way of dealing with recursive lists like this in F# ? Should i pass it an int so it has a determined size?
let rec collatz num =
match num with
|x when x % 2 = 0 ->num :: collatz (x/2)
|x -> num :: collatz ((x * 3) + 1)
let smallList = collatz(4) |> Seq.take(4)
For an infinite list like this, you want to return a sequence. Sequences are lazy; lists are not.
let rec collatz num =
seq {
yield num
match num with
| x when x % 2 = 0 -> yield! collatz (x/2)
| x -> yield! collatz ((x * 3) + 1)
}
let smallList =
collatz 4
|> Seq.take 4
|> Seq.toList //[4; 2; 1; 4]
let collatz num =
let next x = if x % 2 = 0 then x / 2 else x * 3 + 1
(num, next num)
|>Seq.unfold (fun (n, x) -> Some (n, (x, next x)))