How do I concatenate a list of string option?
let m = [ ""; "12"; "a"; "b"]
// I can join these with
m |> List.toSeq |> String.concat "\n"
// now I got a list of string option list
let l = [Some ""; None; Some "a"; Some "b"]
l |> List.toSeq |> ????
you first use List.choose to "extract" the Some values of the list :
l |> List.choose id |> String.concat "\n"
Note you don't need List.toSeq as seq is an alias for IEnumerable and 't list implement it already.
Related
I need to extract the sequence of equal chars in a text.
For example:
The string "aaaBbbcccccccDaBBBzcc11211" should be converted to a list of strings like
["aaa";"B";"bb";"ccccccc";"D";"a";"BBB";"z";"cc";"11";"2";"11"].
That's my solution until now:
let groupSequences (text:string) =
let toString chars =
System.String(chars |> Array.ofList)
let rec groupSequencesRecursive acc chars = seq {
match (acc, chars) with
| [], c :: rest ->
yield! groupSequencesRecursive [c] rest
| _, c :: rest when acc.[0] <> c ->
yield (toString acc)
yield! groupSequencesRecursive [c] rest
| _, c :: rest when acc.[0] = c ->
yield! groupSequencesRecursive (c :: acc) rest
| _, [] ->
yield (toString acc)
| _ ->
yield ""
}
text
|> List.ofSeq
|> groupSequencesRecursive []
groupSequences "aaaBbbcccccccDaBBBzcc11211"
|> Seq.iter (fun x -> printfn "%s" x)
|> ignore
I'm a F# newbie.
This solution can be better?
Here a completely generic implementation:
let group xs =
let folder x = function
| [] -> [[x]]
| (h::t)::ta when h = x -> (x::h::t)::ta
| acc -> [x]::acc
Seq.foldBack folder xs []
This function has the type seq<'a> -> 'a list list when 'a : equality, so works not only on strings, but on any (finite) sequence of elements, as long as the element type supports equality comparison.
Used with the input string in the OP, the return value isn't quite in the expected shape:
> group "aaaBbbcccccccDaBBBzcc11211";;
val it : char list list =
[['a'; 'a'; 'a']; ['B']; ['b'; 'b']; ['c'; 'c'; 'c'; 'c'; 'c'; 'c'; 'c'];
['D']; ['a']; ['B'; 'B'; 'B']; ['z']; ['c'; 'c']; ['1'; '1']; ['2'];
['1'; '1']]
Instead of a string list, the return value is a char list list. You can easily convert it to a list of strings using a map:
> group "aaaBbbcccccccDaBBBzcc11211" |> List.map (List.toArray >> System.String);;
val it : System.String list =
["aaa"; "B"; "bb"; "ccccccc"; "D"; "a"; "BBB"; "z"; "cc"; "11"; "2"; "11"]
This takes advantage of the String constructor overload that takes a char[] as input.
As initially stated, this implementation is generic, so can also be used with other types of lists; e.g. integers:
> group [1;1;2;2;2;3;4;4;3;3;3;0];;
val it : int list list = [[1; 1]; [2; 2; 2]; [3]; [4; 4]; [3; 3; 3]; [0]]
How about with groupby
"aaaBbbcccccccD"
|> Seq.groupBy id
|> Seq.map (snd >> Seq.toArray)
|> Seq.map (fun t -> new string (t))
If you input order matters, here is a method that works
"aaaBbbcccccccDaBBBzcc11211"
|> Seq.pairwise
|> Seq.toArray
|> Array.rev
|> Array.fold (fun (accum::tail) (ca,cb) -> if ca=cb then System.String.Concat(accum,string ca)::tail else string(ca)::accum::tail) (""::[])
This one is also based on recursion though the matching gets away with smaller number of checks.
let chop (txt:string) =
let rec chopInner txtArr (word: char[]) (res: List<string>) =
match txtArr with
| h::t when word.[0] = h -> chopInner t (Array.append word [|h|]) res
| h::t when word.[0] <> h ->
let newWord = word |> (fun s -> System.String s)
chopInner t [|h|] (List.append res [newWord])
| [] ->
let newWord = word |> (fun s -> System.String s)
(List.append res [newWord])
let lst = txt.ToCharArray() |> Array.toList
chopInner lst.Tail [|lst.Head|] []
And the result is as expected:
val text : string = "aaaBbbcccccccDaBBBzcc11211"
> chop text;;
val it : string list =
["aaa"; "B"; "bb"; "ccccccc"; "D"; "a"; "BBB"; "z"; "cc"; "11"; "2"; "11"]
When you're folding, you'll need to carry along both the previous value and the accumulator holding the temporary results. The previous value is wrapped as option to account for the first iteration. Afterwards, the final result is extracted and reversed.
"aaaBbbcccccccDaBBBzcc11211"
|> Seq.map string
|> Seq.fold (fun state ca ->
Some ca,
match state with
| Some cb, x::xs when ca = cb -> x + ca::xs
| _, xss -> ca::xss )
(None, [])
|> snd
|> List.rev
// val it : string list =
// ["aaa"; "B"; "bb"; "ccccccc"; "D"; "a"; "BBB"; "z"; "cc"; "11"; "2"; "11"]
Just interesting why everyone publishing solutions based on match-with? Why not go plain recursion?
let rec groups i (s:string) =
let rec next j = if j = s.Length || s.[i] <> s.[j] then j else next(j+1)
if i = s.Length then []
else let j = next i in s.Substring(i, j - i) :: (groups j s)
"aaaBbbcccccccDaBBBzcc11211" |> groups 0
val it : string list = ["aaa"; "B"; "bb"; "ccccccc"; "D"; "a"; "BBB"; "z"; "cc"; "11"; "2"; "11"]
As someone other here:
Know thy fold ;-)
let someString = "aaaBbbcccccccDaBBBzcc11211"
let addLists state elem =
let (p, ls) = state
elem,
match p = elem, ls with
| _, [] -> [ elem.ToString() ]
| true, h :: t -> (elem.ToString() + h) :: t
| false, h :: t -> elem.ToString() :: ls
someString
|> Seq.fold addLists ((char)0, [])
|> snd
|> List.rev
I have C# code like this
// Class definition
Class Letter {
char c;
int id;
}
// A C# function to obtain an array of letters
Array[Letter] getLetters();
A function call from F#
let L = getLetters()
I want to break L into a list of strings grouped by id, assuming id starts from 0 to N. How can I do that in F#? I'm a F# novice.
let rslt letters =
letters
|> Seq.groupBy (fun l -> l.id) // Groups Letters into sequence of pair (id * seq<Letter>)
|> Seq.map (fun (_, str) -> str |> Seq.map (fun l -> l.c) ) // Maps the sequence to seq<seq<<char>>
|> Seq.map (fun ls -> System.String.Join("", ls)) // Maps the sequence to seq<string>
|> List.ofSeq // Creates list of strings from seq<string>
I have the following code.
let s1 = [(12, "abcde12345"); (23, "bcdef2345"); (12, "xyzafg3838")]
let s2 = ["bcd"; "345"]
What's the best way to find all items in s1 which second item has sub-string of any one in s2?
(12, "abcde12345"); (23, "bcdef2345")
In my real code s1 is a Seq.
Seq.filter (fun (_, x) -> List.exists (x.Contains) s2) s1
I figured out one.
s1 |> Seq.filter (fun i -> List.exists (fun e -> (snd i).Contains(e)) s2)
Concat all of the items from the second set into a regular expression, then apply it on each item in the first set.
open System
open System.Text.RegularExpressions
let setA = [ "One"; "Two"; "Three" ]
let setB = [ "o"; "n" ];
let pattern = String.Join("|", setB);
let regex = new Regex(pattern);
let results = setA |> List.filter (fun str -> regex.Match(str).Success)
results |> List.iter (fun result -> Console.WriteLine(result))
a little rusty from my Scheme days, I'd like to take 2 lists: one of numbers and one of strings, and fold them together into a single string where each pair is written like "{(ushort)5, "bla bla bla"},\n". I have most of it, i'm just not sure how to write the Fold properly:
let splitter = [|","|]
let indexes =
indexStr.Split(splitter, System.StringSplitOptions.None) |> Seq.toList
let values =
valueStr.Split(splitter, System.StringSplitOptions.None) |> Seq.toList
let pairs = List.zip indexes values
printfn "%A" pairs
let result = pairs |> Seq.fold
(fun acc a -> String.Format("{0}, \{(ushort){1}, \"{2}\"\}\n",
acc, (List.nth a 0), (List.nth a 1)))
Your missing two things. The initial state of the fold which is an empty string and you can't use list comprehension on tuples in F#.
let splitter = [|","|]
let indexes =
indexStr.Split(splitter, System.StringSplitOptions.None) |> Seq.toList
let values =
valueStr.Split(splitter, System.StringSplitOptions.None) |> Seq.toList
let pairs = List.zip indexes values
printfn "%A" pairs
let result =
pairs
|> Seq.fold (fun acc (index, value) ->
String.Format("{0}{{(ushort){1}, \"{2}\"}},\n", acc, index, value)) ""
fold2 version
let result =
List.fold2
(fun acc index value ->
String.Format("{0}{{(ushort){1}, \"{2}\"}},\n", acc, index, value))
""
indexes
values
If you are concerned with speed you may want to use string builder since it doesn't create a new string every time you append.
let result =
List.fold2
(fun (sb:StringBuilder) index value ->
sb.AppendFormat("{{(ushort){0}, \"{1}\"}},\n", index, value))
(StringBuilder())
indexes
values
|> string
Fold probably isn't the best method for this task. Its a lot easier to map and concat like this:
let l1 = "a,b,c,d,e".Split([|','|])
let l2 = "1,2,3,4,5".Split([|','|])
let pairs =
Seq.zip l1 l2
|> Seq.map (fun (x, y) -> sprintf "(ushort)%s, \"%s\"" x y)
|> String.concat "\n"
I think you want List.fold2. For some reason the List module has a fold2 member but Seq doesn't. Then you can dispense with the zip entirely.
The types of your named variables and the type of the result you hope for are all implicit, so it's difficult to help, but if you are trying to accumulate a list of strings you might consider something along the lines of
let result = pairs |> Seq.fold
(fun prev (l, r) ->
String.Format("{0}, \{(ushort){1}, \"{2}\"\}\n", prev, l, r)
"" pairs
My F#/Caml is very rusty so I may have the order of arguments wrong. Also note your string formation is quadratic; in my own code I would go with something more along these lines:
let strings =
List.fold2 (fun ss l r ->
String.format ("\{(ushort){0}, \"{1}\"\}\n", l, r) :: ss)
[] indexes values
let result = String.concat ", " strings
This won't cost you quadratic time and it's a little easier to follow. I've checked MSDN and believe I have the correct order of arguments on fold2.
Keep in mind I know Caml not F# and so I may have details or order of arguments wrong.
Perhaps this:
let strBuilder = new StringBuilder()
for (i,v) in Seq.zip indexes values do
strBuilder.Append(String.Format("{{(ushort){0}, \"{1}\"}},\n", i,v))
|> ignore
with F# sometimes is better go imperative...
map2 or fold2 is the right way to go. Here's my take, using the (||>) operator:
let l1 = [| "a"; "b"; "c"; "d"; "e" |]
let l2 = [| "1"; "2"; "3"; "4"; "5" |]
let pairs = (l1, l2) ||> Seq.map2 (sprintf ("(ushort)%s, \"%s\""))
|> String.concat "\n"
I'm trying this at the moment, but I haven't quite got the method signature worked out... anyone? messages is a field of seq[string]
let messageString = List.reduce(messages, fun (m1, m2) -> m1 + m2 + Environment.NewLine)
> String.concat " " ["Juliet"; "is"; "awesome!"];;
val it : string = "Juliet is awesome!"
Not exactly what you're looking for, but
let strings = [| "one"; "two"; "three" |]
let r = System.String.Concat(strings)
printfn "%s" r
You can do
let strings = [ "one"; "two"; "three" ]
let r = strings |> List.fold (+) ""
printfn "%s" r
or
let strings = [ "one"; "two"; "three" ]
let r = strings |> List.fold (fun r s -> r + s + "\n") ""
printfn "%s" r
I'd use String.concat unless you need to do fancier formatting and then I'd use StringBuilder.
(StringBuilder(), [ "one"; "two"; "three" ])
||> Seq.fold (fun sb str -> sb.AppendFormat("{0}\n", str))
just one more comment,
when you are doing with string, you'd better use standard string functions.
The following code is for EulerProject problem 40.
let problem40 =
let str = {1..1000000} |> Seq.map string |> String.concat ""
let l = [str.[0];str.[9];str.[99];str.[999];str.[9999];str.[99999];str.[999999];]
l |> List.map (fun x-> (int x) - (int '0')) |> List.fold (*) 1
if the second line of above program uses fold instead of concat, it would be extremely slow because each iteration of fold creates a new long string.
System.String.Join(Environment.NewLine, List.to_array messages)
or using your fold (note that it's much more inefficient)
List.reduce (fun a b -> a ^ Environment.NewLine ^ b) messages