So I have a function SolveEquasion that returns a pair float*float[]. What is the best way to print the number and the array and continue working with the array? I made the following code but it seems there is a better way
...
|> SolveEquasion
|> (fun (det, solution) -> printfn "Determinant = %f\nSolution = %A" det (Array.toList solution), solution )
|> snd
I don't think your solution can improved if you want to do this in a pipeline. Another approach is to use a let binding, along with splitting up the pipelined operations, to avoid having a function that acts like the love child of map and iter.
let (det, solution) = SolveEquasion
printfn "Determinant = %f\nSolution = %A" det (Array.toList solution)
//do something else with solution
I think the original solution is fine, and we can improve its clarity by giving your anonymous function the name I've seen it given in some other libraries based around pipelining higher-order functions: tap.
let tap f x =
f x
x
(1.0, [| 2.0; 3.0 |])
|> tap (fun (s, a) -> printfn "%A %A" s a)
|> snd
Well, for one thing you can skip the use of snd by returning a single value rather than a tuple from the previous function:
...
|> SolveEquasion
|> (fun (det, solution) ->
printfn "Determinant = %f\nSolution = %A" det (Array.toList solution)
solution )
I'd probably use Daniel's approach and just assign the value you want to print to a symbol using let. Alternatively, you could define a variant of printf that takes some arguments and returns one of them. I'm not sure if there is a general scheme how this should be done - for your example it would take a two-element tuple:
let mprintf fmt (a, b) =
Printf.kprintf (fun s -> printf "%s" s; (a, b)) fmt a b
Then you can write:
...
|> SolveEquasion
|> mprintfn "Determinant = %f\nSolution = %A"
|> snd |> // ... more stuff with solution
Related
I'm learning F# (again) and I'm trying to sum some rows in excel. This is my attempt.
let sumRows (source: seq<double[]>) =
source
|> Seq.reduce (fun a b -> (a, b) ||> Seq.map2 (fun x y -> x + y) |> Seq.toArray)
Can it be done better? I already discovered double forward pipe operator, but now, whole part fun a b -> (a, b) ||> seems quite redundant...
You are right it is redundant, because the double pipe operator is convenient only when you need to convert a tuple into two separate parameters. In this case you already have them as 2 separate parameters so you could just pass them like this:
let sumRows (source: seq<double[]>) =
source
|> Seq.reduce (fun a b -> Seq.map2 (fun x y -> x + y) a b |> Seq.toArray)
We can get rid of the |> Seq.toArray by replacing Seq.map2 with Array.map2:
let sumRows (source: seq<double[]>) =
source
|> Seq.reduce (fun a b -> Array.map2 (fun x y -> x + y) a b)
now we can simplify further by removing the fun a b ->:
let sumRows (source: seq<double[]>) =
source
|> Seq.reduce (Array.map2 (fun x y -> x + y) )
Finally, did you know that an operator like + can be treated as a 2 parameters function
by putting in parenthesis (+)?
let sumRows2 (source: seq<double[]>) =
source
|> Seq.reduce (Array.map2 (+) )
All of this versions have the same signature and they are all correct. You choose whichever suits more your style.
BTW, you maybe tempted to go one further and do this:
let sumRows2 = Seq.reduce (Array.map2 (+) )
but it causes issues with the famous Value restriction error. There are workarounds like adding the type annotation or actually using it somewhere in the code, but the best workaround is to add the parameter, like we had before.
Let's say I have two lists:
let listOfValues = [100..105] //can be list of strings or whatever
let indexesToSortBy = [1;2;0;4;5;3]
Now I need listOfValues_sorted: 102;100;101;105;103;104
It can be done with zip and "conversion" to Tuple:
let listOfValues_sorted = listOfValues
|> Seq.zip indexesToSortBy
|> Seq.sortBy( fun x-> fst x)
|> Seq.iter(fun c -> printfn "%i" (snd c))
But I guess, there is better solution for that?
I think your solution is pretty close. I would do this
let listOfValues_sorted =
listOfValues
|> Seq.zip indexesToSortBy
|> Seq.sortBy fst
|> Seq.toList
|> List.unzip
|> List.head
you can collapse fun x -> fst x into simply fst. And then unzip and get what ever list you want
If indexesToSortBy is a complete set of indexes you could simply use:
indexesToSortBy |> List.map (fun x -> listOfValues |> List.item x )
Your example sounds precisely what the List.permute function is for:
let listOfValues = [100..105]
let indexesToSortBy = [|1;2;0;4;5;3|] // Note 0-based indexes
listOfValues |> List.permute (fun i -> indexesToSortBy.[i])
// Result: [102; 100; 101; 105; 103; 104]
Two things: First, I made indexesToSortBy an array since I'll be looking up a value inside it N times, and doing that in a list would lead to O(N^2) run time. Second, List.permute expects to be handed a 0-based index into the original list, so I subtracted 1 from all the indexes in your original indexToSortBy list. With these two changes, this produces exactly the same ordering as the let listOfValues_sorted = ... example in your question.
I know there's the back-pipe (<|) operator, referenced in several other SO answers. But that doesn't work well when combined with forward pipes (|>), which is common in chaining. However I'm looking for related options. Basically is there any built-in version of the below function definition? Or is this a bad/dangerous practice?
let inline (^%) f = f
let stuff =
[1;2;3]
|> Seq.filter ^% (>) 2
|> Seq.map ^% fun x -> x.ToString()
// compare to this, which doesn't compile (and would be hard to follow even if it did)
let stuff =
[1;2;3]
|> Seq.filter <| (>) 2
|> Seq.map <| fun x -> x.ToString()
There are some Haskell features, like optional infixing using backticks, and sections, which aren't available in F#. That makes certain constructs a bit more verbose.
Usually, I'd simply write a pipe of functions as the above like this:
let stuff =
[1;2;3]
|> Seq.filter (fun x -> x < 2)
|> Seq.map string
This is, in my opinion, much more readable. For example, using Seq.filter ^% (>) 2, I'd intuitively read that as meaning 'all values greater than 2', but that's not what it does:
> let inline (^%) f = f;;
val inline ( ^% ) : f:'a -> 'a
> let stuff =
[1;2;3]
|> Seq.filter ^% (>) 2
|> Seq.map ^% fun x -> x.ToString()
|> Seq.toList;;
val stuff : string list = ["1"]
If you leave the reader of the code in doubt of what the code does, you've just made everyone less productive. Using Seq.filter (fun x -> x < 2) may look more verbose, but is unambiguous to the reader.
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.
In APL one can use a bit vector to select out elements of another vector; this is called compression. For example 1 0 1/3 5 7 would yield 3 7.
Is there a accepted term for this in functional programming in general and F# in particular?
Here is my F# program:
let list1 = [|"Bob"; "Mary"; "Sue"|]
let list2 = [|1; 0; 1|]
[<EntryPoint>]
let main argv =
0 // return an integer exit code
What I would like to do is compute a new string[] which would be [|"Bob"; Sue"|]
How would one do this in F#?
Array.zip list1 list2 // [|("Bob",1); ("Mary",0); ("Sue",1)|]
|> Array.filter (fun (_,x) -> x = 1) // [|("Bob", 1); ("Sue", 1)|]
|> Array.map fst // [|"Bob"; "Sue"|]
The pipe operator |> does function application syntactically reversed, i.e., x |> f is equivalent to f x. As mentioned in another answer, replace Array with Seq to avoid the construction of intermediate arrays.
I expect you'll find many APL primitives missing from F#. For lists and sequences, many can be constructed by stringing together primitives from the Seq, Array, or List modules, like the above. For reference, here is an overview of the Seq module.
I think the easiest is to use an array sequence expression, something like this:
let compress bits values =
[|
for i = 0 to bits.Length - 1 do
if bits.[i] = 1 then
yield values.[i]
|]
If you only want to use combinators, this is what I would do:
Seq.zip bits values
|> Seq.choose (fun (bit, value) ->
if bit = 1 then Some value else None)
|> Array.ofSeq
I use Seq functions instead of Array in order to avoid building intermediary arrays, but it would be correct too.
One might say this is more idiomatic:
Seq.map2 (fun l1 l2 -> if l2 = 1 then Some(l1) else None) list1 list2
|> Seq.choose id
|> Seq.toArray
EDIT (for the pipe lovers)
(list1, list2)
||> Seq.map2 (fun l1 l2 -> if l2 = 1 then Some(l1) else None)
|> Seq.choose id
|> Seq.toArray
Søren Debois' solution is good but, as he pointed out, but we can do better. Let's define a function, based on Søren's code:
let compressArray vals idx =
Array.zip vals idx
|> Array.filter (fun (_, x) -> x = 1)
|> Array.map fst
compressArray ends up creating a new array in each of the 3 lines. This can take some time, if the input arrays are long (1.4 seconds for 10M values in my quick test).
We can save some time by working on sequences and creating an array at the end only:
let compressSeq vals idx =
Seq.zip vals idx
|> Seq.filter (fun (_, x) -> x = 1)
|> Seq.map fst
This function is generic and will work on arrays, lists, etc. To generate an array as output:
compressSeq sq idx |> Seq.toArray
The latter saves about 40% of computation time (0.8s in my test).
As ildjarn commented, the function argument to filter can be rewritten to snd >> (=) 1, although that causes a slight performance drop (< 10%), probably because of the extra function call that is generated.