F# sort by indexes - f#

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.

Related

F#.Data HTML Parser Extracting Strings From Nodes

I am trying to use FSharp.Data's HTML Parser to extract a string List of links from href attributes.
I can get the links printed out to console, however, i'm struggling to get them into a list.
Working snippet of a code which prints the wanted links:
let results = HtmlDocument.Load(myUrl)
let links =
results.Descendants("td")
|> Seq.filter (fun x -> x.HasClass("pagenav"))
|> Seq.map (fun x -> x.Elements("a"))
|> Seq.iter (fun x -> x |> Seq.iter (fun y -> y.AttributeValue("href") |> printf "%A"))
How do i store those strings into variable links instead of printing them out?
Cheers,
On the very last line, you end up with a sequence of sequences - for each td.pagenav you have a bunch of <a>, each of which has a href. That's why you have to have two nested Seq.iters - first you iterate over the outer sequence, and on each iteration you iterate over the inner sequence.
To flatten a sequence of sequences, use Seq.collect. Further, to convert a sequence to a list, use Seq.toList or List.ofSeq (they're equivalent):
let a = [ [1;2;3]; [4;5;6] ]
let b = a |> Seq.collect id |> Seq.toList
> val b : int list = [1; 2; 3; 4; 5; 6]
Applying this to your code:
let links =
results.Descendants("td")
|> Seq.filter (fun x -> x.HasClass("pagenav"))
|> Seq.map (fun x -> x.Elements("a"))
|> Seq.collect (fun x -> x |> Seq.map (fun y -> y.AttributeValue("href")))
|> Seq.toList
Or you could make it a bit cleaner by applying Seq.collect at the point where you first encounter a nested sequence:
let links =
results.Descendants("td")
|> Seq.filter (fun x -> x.HasClass("pagenav"))
|> Seq.collect (fun x -> x.Elements("a"))
|> Seq.map (fun y -> y.AttributeValue("href"))
|> Seq.toList
That said, I would rather rewrite this as a list comprehension. Looks even cleaner:
let links = [ for td in results.Descendants "td" do
if td.HasClass "pagenav" then
for a in td.Elements "a" ->
a.AttributeValue "href"
]

How do I identify the max length from a Map's value set?

How do I identify the max length from a Map's value set?
let numbers = [1;2;2;3;3;3;4;5;5]
let map = numbers |> Seq.groupBy id
|> Map.ofSeq
I want to do this:
map.Values |> List.max
or...
let longestSequence = Map.map (fun (k, v) -> List.max(List.ofSeq(v)));
you can get something similar to Dictionary.Values with Map.toSeq >> Seq.map snd so you can get the largest collected sequence in your map like this:
> map |> Map.toSeq |> Seq.map snd |> Seq.maxBy Seq.length;;
val it : seq<int> = seq [3; 3; 3]
of course when your list is already in a sorted stage it seems strange to take the detour over Map as
> numbers |> Seq.groupBy id |> Seq.map snd |> Seq.maxBy Seq.length;;
val it : seq<int> = seq [3; 3; 3]
will do the same ;)
also if you think about the problem here can write a List.fold (with a additional map of the result) doing this as well which will only require to traverse the (sorted) list once ... maybe you can try to do this yourself ^^

GroupBy Year then take Pairwise diffs except for the head value then Flatten Using Deedle and F#

I have the following variable:
data:seq<(DateTime*float)>
and I want to do something like the following F# code but using Deedle:
data
|> Seq.groupBy (fun (k,v) -> k.Year)
|> Seq.map (fun (k,v) ->
let vals = v |> Seq.pairwise
let first = seq { yield v |> Seq.head }
let diffs = vals |> Seq.map (fun ((t0,v0),(t1,v1)) -> (t1, v1 - v0))
(k, diffs |> Seq.append first))
|> Seq.collect snd
This works fine using F# sequences but I want to do it using Deedle series. I know I can do something like:
(data:Series<DateTime*float>) |> Series.groupBy (fun k v -> k.Year)...
But then I need to take the within group year diffs except for the head value which should just be the value itself and then flatten the results into on series...I am bit confused with the deedle syntax
Thanks!
I think the following might be doing what you need:
ts
|> Series.groupInto
(fun k _ -> k.Month)
(fun m s ->
let first = series [ fst s.KeyRange => s.[fst s.KeyRange]]
Series.merge first (Series.diff 1 s))
|> Series.values
|> Series.mergeAll
The groupInto function lets you specify a function that should be called on each of the groups
For each group, we create series with the differences using Series.diff and append a series with the first value at the beginning using Series.merge.
At the end, we get all the nested series & flatten them using Series.mergeAll.

Merging two lists in F#

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.

f# array.filter based on a bool array

if I have array A, and I have another bool array isChosen with the same length of A how can I build a new array from A where isChosen is true? something like A.[isChosen]? I cannot use Array.filter directly since isChosen is not a function of A elements and there is no Array.filteri like Array.mapi.
zip should help:
let l = [|1;2;3|]
let f = [|true; false; true|]
let r = [| for (v, f) in Seq.zip l f do if f then yield v|]
// or
let r = (l, f) ||> Seq.zip |> Seq.filter snd |> Seq.map fst |> Seq.toArray
Try the zip operator
seq.zip A isChosen
|> Seq.filter snd
|> Seq.map fst
|> Array.ofSeq
This will create a sequence of tuples where one value is from A and the other is from isChosen. This will pair the values together and make it very easy to filter them out in a Seq.filter expression
It's not as elegant or 'functional' as the other answers, but every once in a while I like a gentle reminder that you can use loops and array indices in F#:
let A = [|1;2;3|]
let isChosen = [|true; false; true|]
let r = [| for i in 0..A.Length-1 do
if isChosen.[i] then
yield A.[i] |]
printfn "%A" r
:)
And here are two more ways, just to demonstrate (even) more F# library functions:
let A = [|1;2;3|]
let isChosen = [|true;false;true|]
let B = Seq.map2 (fun x b -> if b then Some x else None) A isChosen
|> Seq.choose id
|> Seq.toArray
let C = Array.foldBack2 (fun x b acc -> if b then x::acc else acc) A isChosen []
|> List.toArray
My personal favorite for understandability (and therefore maintainability): desco's answer
let r = [| for (v, f) in Seq.zip l f do if f then yield v|]

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