So my team and I have created a function, wordMarkovChain, which generates a random string with nWords words, whose word-pairs are distributed according to the cooccurrence histogram wCooc.
Now we wanted to create a function which will test the newly created wordMarkovChain, so I can confirm that it's working as it should.
The function, diffw2, which compares two cooccurrence histograms as the average sum of squared differences, needs to take in two parameters, c1 and c2, of type wordCooccurrences and return a double, where c1 and c2 are two cooccurrence histograms of M elements such that c1(i, j) is the number
of times word number i is found following word number j.
The math behind the function should look like this: 1/M^2 * Summation from i=0 to M-1 Summation from i=0 to M-1 (c1(i, j) - c2(i, j))^2.
Sorry, I can't post a picture:(
Our two types which we have created seems to be the problem. c1 and c2 can have a different length but the type wordHistogram inside the type wordCoocurence can also have a different length.
The question is, how can we create such a function?
We tried with for loops, but we think it needs to be a recursive function. We are quite new to the whole concept of programming and are looking for some guidance. Please bear in mind that we do not possess vast knowledge of F#, especially not their in build functions.
CODE
// Types
type wordHistogram = (string * int) list
type wordCooccurrences = (string * wordHistogram) list
let diffw2 (c1 : wordCooccurrences) (c2 : wordCooccurrences) : double =
let mutable res = 0
let z1, z2 = c1 |> List.unzip
let z3, z4 = c2 |> List.unzip
let m1 = c1 |> List.length
let m2 = c2 |> List.length
let m = m1 + m2
for i in 0 .. (m - 1) do
for j in 0 .. (m - 1) do
for k in 0 .. ((z2.[j] |> List.length) - 1) do
res <- res + (snd z2.[j].[k] - snd z4.[j].[k]) * (snd z2.[j].[k] - snd z4.[j].[k])
(1.0 / (float(m * m))) * float(res)
This might get you closer to what you need. This could instead be done using recursive functions, but I usually prefer to use built-in higher-order functions like fold or fold2 when they can do the job. See https://msdn.microsoft.com/en-us/visualfsharpdocs/conceptual/list.fold2%5B't1%2C't2%2C'state%5D-function-%5Bfsharp%5D
Each call of List.fold2 receives an "accumulator" with an initial value (or "state") of 0.0. This accumulator is passed to the lambda function in the parameter named "acc". As the lambda is applied to each element of the two input lists, it adds the intermediate result to that acc value, which is then returned as the result of the List.fold2 call.
type wordHistogram = (string * int) list
type wordCooccurrences = (string * wordHistogram) list
let diffw2 (c1 : wordCooccurrences) (c2 : wordCooccurrences) =
let m1 = c1 |> List.length
let m2 = c2 |> List.length
let m = m1 + m2
let res =
(0.0, c1, c2) |||>
List.fold2 (fun acc (_str1, hist1) (_str2, hist2) ->
let intermedRes =
(0.0, hist1, hist2) |||>
List.fold2 (fun acc (_str1, freq1) (_str2, freq2) ->
let diff = float (freq1 - freq2)
acc + diff * diff
)
acc + intermedRes
)
(1.0 / float(m * m)) * res
I think you are trying to do the following
let diffw2_ (c1 : wordCooccurrences) (c2 : wordCooccurrences) =
List.zip c1 c2
|> List.sumBy (fun ((_, w1), (_, w2)) ->
List.zip w1 w2
|> List.sumBy (fun ((_,i1), (_,i2)) ->
(float (i1 - i2)) ** 2.0
)
) |> (fun totalSumOfSquares -> totalSumOfSquares / (c1 |> List.length |> float))
This just returns the average square differences between corresponding elements of c1 and c2. It assumes that c1 and c2 have the same structure.
With F# (and functional programming in general), you typically avoid using mutable variables and for loops, in favor of pure functions.
I'm trying to write Taylor series in F#.
Have a look at my code
let rec iter a b f i =
if a > b then i;
else f a (iter (a+1) b f i)
let sum a b = iter a b (+) 0 // from 0
// e^x = 1 + x + (x^2)/2 + ... (x^n)/n! + ...
let fact n = iter 1 n (*) 1 // factorial
let pow x n = iter 1 n (fun n acc -> acc * x) 1
let exp x =
iter 0 x
(fun n acc ->
acc + (pow x n) / float (fact n)) 0
In the last row I am trying cast int fact n to float, but seems like I'm wrong because this code isn't compileable :(
Am I doing the right algorithm?
Can I call my code functional-first?
The code doesn't compile, because:
You're trying to divide an integer pow x n by a float. Division has to have operands of the same type.
You're specifying the terminal case of the wrong type. Literal 0 is integer. If you want float zero, use 0.0 or abbreviated 0.
Try this:
let exp x =
iter 0 x
(fun n acc ->
acc + float (pow x n) / float (fact n)) 0.
P.S. In the future, please provide the exact error messages and/or unexpected results that you're getting. Simply saying "doesn't work" is not a good description of a problem.
I am interested to implement fold3, fold4 etc., similar to List.fold and List.fold2. e.g.
// TESTCASE
let polynomial (x:double) a b c = a*x + b*x*x + c*x*x*x
let A = [2.0; 3.0; 4.0; 5.0]
let B = [1.5; 1.0; 0.5; 0.2]
let C = [0.8; 0.01; 0.001; 0.0001]
let result = fold3 polynomial 0.7 A B C
// 2.0 * (0.7 ) + 1.5 * (0.7 )^2 + 0.8 * (0.7 )^3 -> 2.4094
// 3.0 * (2.4094) + 1.0 * (2.4094)^2 + 0.01 * (2.4094)^3 -> 13.173
// 4.0 * (13.173) + 0.5 * (13.173)^2 + 0.001 * (13.173)^3 -> 141.75
// 5.0 * (141.75) + 0.2 * (141.75)^2 + 0.0001 * (141.75)^3 -> 5011.964
//
// Output: result = 5011.964
My first method is grouping the 3 lists A, B, C, into a list of tuples, and then apply list.fold
let fold3 f x A B C =
List.map3 (fun a b c -> (a,b,c)) A B C
|> List.fold (fun acc (a,b,c) -> f acc a b c) x
// e.g. creates [(2.0,1.5,0.8); (3.0,1.0,0.01); ......]
My second method is to declare a mutable data, and use List.map3
let mutable result = 0.7
List.map3 (fun a b c ->
result <- polynomial result a b c // Change mutable data
// Output intermediate data
result) A B C
// Output from List.map3: [2.4094; 13.17327905; 141.7467853; 5011.963942]
// result mutable: 5011.963942
I would like to know if there are other ways to solve this problem. Thank you.
For fold3, you could just do zip3 and then fold:
let polynomial (x:double) (a, b, c) = a*x + b*x*x + c*x*x*x
List.zip3 A B C |> List.fold polynomial 0.7
But if you want this for the general case, then you need what we call "applicative functors".
First, imagine you have a list of functions and a list of values. Let's assume for now they're of the same size:
let fs = [ (fun x -> x+1); (fun x -> x+2); (fun x -> x+3) ]
let xs = [3;5;7]
And what you'd like to do (only natural) is to apply each function to each value. This is easily done with List.map2:
let apply fs xs = List.map2 (fun f x -> f x) fs xs
apply fs xs // Result = [4;7;10]
This operation "apply" is why these are called "applicative functors". Not just any ol' functors, but applicative ones. (the reason for why they're "functors" is a tad more complicated)
So far so good. But wait! What if each function in my list of functions returned another function?
let f1s = [ (fun x -> fun y -> x+y); (fun x -> fun y -> x-y); (fun x -> fun y -> x*y) ]
Or, if I remember that fun x -> fun y -> ... can be written in the short form of fun x y -> ...
let f1s = [ (fun x y -> x+y); (fun x y -> x-y); (fun x y -> x*y) ]
What if I apply such list of functions to my values? Well, naturally, I'll get another list of functions:
let f2s = apply f1s xs
// f2s = [ (fun y -> 3+y); (fun y -> 5+y); (fun y -> 7+y) ]
Hey, here's an idea! Since f2s is also a list of functions, can I apply it again? Well of course I can!
let ys = [1;2;3]
apply f2s ys // Result: [4;7;10]
Wait, what? What just happened?
I first applied the first list of functions to xs, and got another list of functions as a result. And then I applied that result to ys, and got a list of numbers.
We could rewrite that without intermediate variable f2s:
let f1s = [ (fun x y -> x+y); (fun x y -> x-y); (fun x y -> x*y) ]
let xs = [3;5;7]
let ys = [1;2;3]
apply (apply f1s xs) ys // Result: [4;7;10]
For extra convenience, this operation apply is usually expressed as an operator:
let (<*>) = apply
f1s <*> xs <*> ys
See what I did there? With this operator, it now looks very similar to just calling the function with two arguments. Neat.
But wait. What about our original task? In the original requirements we don't have a list of functions, we only have one single function.
Well, that can be easily fixed with another operation, let's call it "apply first". This operation will take a single function (not a list) plus a list of values, and apply this function to each value in the list:
let applyFirst f xs = List.map f xs
Oh, wait. That's just map. Silly me :-)
For extra convenience, this operation is usually also given an operator name:
let (<|>) = List.map
And now, I can do things like this:
let f x y = x + y
let xs = [3;5;7]
let ys = [1;2;3]
f <|> xs <*> ys // Result: [4;7;10]
Or this:
let f x y z = (x + y)*z
let xs = [3;5;7]
let ys = [1;2;3]
let zs = [1;-1;100]
f <|> xs <*> ys <*> zs // Result: [4;-7;1000]
Neat! I made it so I can apply arbitrary functions to lists of arguments at once!
Now, finally, you can apply this to your original problem:
let polynomial a b c (x:double) = a*x + b*x*x + c*x*x*x
let A = [2.0; 3.0; 4.0; 5.0]
let B = [1.5; 1.0; 0.5; 0.2]
let C = [0.8; 0.01; 0.001; 0.0001]
let ps = polynomial <|> A <*> B <*> C
let result = ps |> List.fold (fun x f -> f x) 0.7
The list ps consists of polynomial instances that are partially applied to corresponding elements of A, B, and C, and still expecting the final argument x. And on the next line, I simply fold over this list of functions, applying each of them to the result of the previous.
You could check the implementation for ideas:
https://github.com/fsharp/fsharp/blob/master/src/fsharp/FSharp.Core/array.fs
let fold<'T,'State> (f : 'State -> 'T -> 'State) (acc: 'State) (array:'T[]) =
checkNonNull "array" array
let f = OptimizedClosures.FSharpFunc<_,_,_>.Adapt(f)
let mutable state = acc
for i = 0 to array.Length-1 do
state <- f.Invoke(state,array.[i])
state
here's a few implementations for you:
let fold2<'a,'b,'State> (f : 'State -> 'a -> 'b -> 'State) (acc: 'State) (a:'a array) (b:'b array) =
let mutable state = acc
Array.iter2 (fun x y->state<-f state x y) a b
state
let iter3 f (a: 'a[]) (b: 'b[]) (c: 'c[]) =
let f = OptimizedClosures.FSharpFunc<_,_,_,_>.Adapt(f)
if a.Length <> b.Length || a.Length <> c.Length then failwithf "length"
for i = 0 to a.Length-1 do
f.Invoke(a.[i], b.[i], c.[i])
let altIter3 f (a: 'a[]) (b: 'b[]) (c: 'c[]) =
if a.Length <> b.Length || a.Length <> c.Length then failwithf "length"
for i = 0 to a.Length-1 do
f (a.[i]) (b.[i]) (c.[i])
let fold3<'a,'b,'State> (f : 'State -> 'a -> 'b -> 'c -> 'State) (acc: 'State) (a:'a array) (b:'b array) (c:'c array) =
let mutable state = acc
iter3 (fun x y z->state<-f state x y z) a b c
state
NB. we don't have an iter3, so, implement that. OptimizedClosures.FSharpFunc only allow up to 5 (or is it 7?) params. There are a finite number of type slots available. It makes sense. You can go higher than this, of course, without using the OptimizedClosures stuff.
... anyway, generally, you don't want to be iterating too many lists / arrays / sequences at once. So I'd caution against going too high.
... the better way forward in such cases may be to construct a record or tuple from said lists / arrays, first. Then, you can just use map and iter, which are already baked in. This is what zip / zip3 are all about (see: "(array1.[i],array2.[i],array3.[i])")
let zip3 (array1: _[]) (array2: _[]) (array3: _[]) =
checkNonNull "array1" array1
checkNonNull "array2" array2
checkNonNull "array3" array3
let len1 = array1.Length
if len1 <> array2.Length || len1 <> array3.Length then invalidArg3ArraysDifferent "array1" "array2" "array3" len1 array2.Length array3.Length
let res = Microsoft.FSharp.Primitives.Basics.Array.zeroCreateUnchecked len1
for i = 0 to res.Length-1 do
res.[i] <- (array1.[i],array2.[i],array3.[i])
res
I'm working with arrays at the moment, so my solution pertained to those. Sorry about that. Here's a recursive version for lists.
let fold3 f acc a b c =
let mutable state = acc
let rec fold3 f a b c =
match a,b,c with
| [],[],[] -> ()
| [],_,_
| _,[],_
| _,_,[] -> failwith "length"
| ahead::atail, bhead::btail, chead::ctail ->
state <- f state ahead bhead chead
fold3 f atail btail ctail
fold3 f a b c
i.e. we define a recursive function within a function which acts upon/mutates/changes the outer scoped mutable acc variable (a closure in functional speak). Finally, this gets returned.
It's pretty cool how much type information gets inferred about these functions. In the array examples above, mostly I was explicit with 'a 'b 'c. This time, we let type inference kick in. It knows we're dealing with lists from the :: operator. That's kind of neat.
NB. the compiler will probably unwind this tail-recursive approach so that it is just a loop behind-the-scenes. Generally, get a correct answer before optimising. Just mentioning this, though, as food for later thought.
I think the existing answers provide great options if you want to generalize folding, which was your original question. However, if I simply wanted to call the polynomial function on inputs specified in A, B and C, then I would probably do not want to introduce fairly complex constructs like applicative functors with fancy operators to my code base.
The problem becomes a lot easier if you transpose the input data, so that rather than having a list [A; B; C] with lists for individual variables, you have a transposed list with inputs for calculating each polynomial. To do this, we'll need the transpose function:
let rec transpose = function
| (_::_)::_ as M -> List.map List.head M :: transpose (List.map List.tail M)
| _ -> []
Now you can create a list with inputs, transpose it and calculate all polynomials simply using List.map:
transpose [A; B; C]
|> List.map (function
| [a; b; c] -> polynomial 0.7 a b c
| _ -> failwith "wrong number of arguments")
There are many ways to solve this problem. Few are mentioned like first zip3 all three list, then run over it. Using Applicate Functors like Fyodor Soikin describes means you can turn any function with any amount of arguments into a function that expects list instead of single arguments. This is a good general solution that works with any numbers of lists.
While this is a general good idea, i'm sometimes shocked that so few use more low-level tools. In this case it is a good idea to use recursion and learn more about recursion.
Recursion here is the right-tool because we have immutable data-types. But you could consider how you would implement it with mutable lists and looping first, if that helps. The steps would be:
You loop over an index from 0 to the amount of elements in the lists.
You check if every list has an element for the index
If every list has an element then you pass this to your "folder" function
If at least one list don't have an element, then you abort the loop
The recursive version works exactly the same. Only that you don't use an index to access the elements. You would chop of the first element from every list and then recurse on the remaining list.
Otherwise List.isEmpty is the function to check if a List is empty. You can chop off the first element with List.head and you get the remaining list with the first element removed by List.tail. This way you can just write:
let rec fold3 f acc l1 l2 l3 =
let h = List.head
let t = List.tail
let empty = List.isEmpty
if (empty l1) || (empty l2) && (empty l3)
then acc
else fold3 f (f acc (h l1) (h l2) (h l3)) (t l1) (t l2) (t l3)
The if line checks if every list has at least one element. If that is true
it executes: f acc (h l1) (h l2) (h l3). So it executes f and passes it the first element of every list as an argument. The result is the new accumulator of
the next fold3 call.
Now that you worked on the first element of every list, you must chop off the first element of every list, and continue with the remaining lists. You achieve that with List.tail or in the above example (t l1) (t l2) (t l3). Those are the next remaining lists for the next fold3 call.
Creating a fold4, fold5, fold6 and so on isn't really hard, and I think it is self-explanatory. My general advice is to learn a little bit more about recursion and try to write recursive List functions without Pattern Matching. Pattern Matching is not always easier.
Some code examples:
fold3 (fun acc x y z -> x + y + z :: acc) [] [1;2;3] [10;20;30] [100;200;300] // [333;222;111]
fold3 (fun acc x y z -> x :: y :: z :: acc) [] [1;2;3] [10;20;30] [100;200;300] // [3; 30; 300; 2; 20; 200; 1; 10; 100]
I'm trying to create an infinite Stream in F# that contains armstrong numbers. An armstrong number is one whose cubes of its digits add up to the number. For example, 153 is an armstrong number because 1^3 + 5^3 + 3^3 = 153. so far, I have created several functions to help me do so. They are:
type 'a stream = Cons of 'a * (unit -> 'a stream);;
let rec upfrom n = Cons (n, fun() -> upfrom (n+1));;
let rec toIntArray = function
| 0 -> []
| n -> n % 10 :: toIntArray (n / 10);;
let rec makearmstrong = function
| [] -> 0
| y::ys -> (y * y * y) + makearmstrong ys;;
let checkarmstrong n = n = makearmstrong(toIntArray n);;
let rec take n (Cons(x,xsf)) =
match n with
| 0 -> []
| _ -> x :: take (n-1)(xsf());;
let rec filter p (Cons (x, xsf)) =
if p x then Cons (x, fun() -> filter p (xsf()))
else filter p (xsf());;
And finally:
let armstrongs = filter (fun n -> checkarmstrong n)(upfrom 1);;
Now, when I do take 4 armstrongs;;, (or any number less than 4) this works perfectly and gives me [1;153;370;371] but if I do take 5 armstrongs;;nothing happens, it seems like the program freezes.
I believe the issue is that there are no numbers after 407 that are the sums of their cubes (see http://oeis.org/A046197), but when your code evaluates the equivalent of take 1 (Cons(407, filter checkarmstrong (upfrom 408))) it's going to force the evaluation of the tail and filter will recurse forever, never finding a matching next element. Also note that your definition of Armstrong numbers differs from, say, Wikipedia's, which states that the power the digits are raised to should be the number of digits in the number.
I thought these two function were the same, but it seems that I was wrong.
I define two function f and g in this way:
let rec f n k =
match k with
|_ when (k < 0) || (k > n) -> 0
|_ when k = n -> 100
|_ -> (f n (k+1)) + 1
let rec g n k =
match k with
|_ when (k < 0) || (k > n) -> 0
| n -> 100
|_ -> (g n (k+1)) + 1
let x = f 10 5
let y = g 10 5
The results are:
val x : int = 105
val y : int = 100
Could anyone tell me what's the difference between these two functions?
EDIT
Why does it work here?
let f x =
match x with
| 1 -> 100
| 2 -> 200
|_ -> -1
List.map f [-1..3]
and we get
val f : x:int -> int
val it : int list = [-1; -1; 100; 200; -1]
The difference is that
match k with
...
when k = n -> 100
is a case that matches when some particular condition is true (k = n). The n used in the condition refers to the n that is bound as the function parameter. On the other hand
match k with
...
n -> 100
is a case that only needs to match k against a pattern variable n, which can always succeed. The n in the pattern isn't the same n as the n passed into the function.
For comparison, try the code
let rec g n k =
match k with
|_ when (k < 0) || (k > n) -> 0
| n -> n
|_ -> (g n (k+1)) + 1
and you should see that when you get to the second case, the value returned is the value of the pattern variable n, which has been bound to the value of k.
This behavior is described in the Variable Patterns section of the MSDN F# Language Reference, Pattern Matching:
Variable Patterns
The variable pattern assigns the value being matched to a variable
name, which is then available for use in the execution expression to
the right of the -> symbol. A variable pattern alone matches any
input, but variable patterns often appear within other patterns,
therefore enabling more complex structures such as tuples and arrays
to be decomposed into variables. The following example demonstrates a
variable pattern within a tuple pattern.
let function1 x =
match x with
| (var1, var2) when var1 > var2 -> printfn "%d is greater than %d" var1 var2
| (var1, var2) when var1 < var2 -> printfn "%d is less than %d" var1 var2
| (var1, var2) -> printfn "%d equals %d" var1 var2
function1 (1,2)
function1 (2, 1)
function1 (0, 0)
The use of when is described in more depth in Match Expressions.
The first function is ok, it calls recursively itself n-k times and returns 100 when matches with the conditional where k = n. So, it returns all the calls adding 1 n-k times. with your example, with n=10 and k=5 it is ok the result had been 105.
The problem is the second function. I tested here. See I changed the pattern n->100 to z->100 and it still matches there and never calls itself recursively. So, it always returns 100 if it does not fail in the first conditional. I think F# don't allow that kind of match so it is better to put a conditional to get what you want.