I have created a function in SML that traverses the starting elements of a list and if first,second,third.. element are the same deletes these elements and updates a variable's value.What i have written:
let
val min=7
in
fun seen2 (set:int list) =
if hd set=hd(tl set) then
min=min-1
seen2(tl set)
else
tl set
end
The output of this function is meant to be a list with the elements i mentioned deleted.For example if it gets this list as input->[1,1,1,1,2,3,4] and min is set as 7 from before i excpect it to give [2,3,4] as a result and min to updated to 4.The min variable should be stored because this function will probably be called again and min may get further updated.This code gives me syntax errors.At the end the final min must be printed so i think this has to be something liek a global value(?).How could i accomplish this?
traverses the starting elements of a list and if first,second,third.. element are the same deletes these elements
If by "first,second,third.." you mean arbitrarily many, then this is what you want to be doing:
fun removeDuplicatesBeginning [] = []
| removeDuplicatesBeginning (x::y::zs) =
if (* are the first two elements the same? *)
then (* apply removeDuplicatesBeginning recursively
to sub-list with one of them removed *)
else (* no repeats, we're done recursing, only remove first element *)
Express your desired behavior using tests, e.g.
val test_1 = removeDuplicatesBeginning [1,1,1,1,2,3,4] = [2,3,4]
And don't forget corner cases, e.g.
val test_2 = removeDuplicatesBeginning [2,3,4] = [3,4]
val test_3 = removeDuplicatesBeginning [1,2,1] = [2,1]
Avoid...
setting elements at the beginning, like let val min = 7 in ... end. There's no point.
You can't write let ... in fun ... end, because fun ... is a declaration, and you can only have expressions between in and end for let. (The other thing is possible with local ... in ... end, but you still don't want to do this. There's no point.)
using hd and tl. Use pattern matching on the input list's elements (x::xs, or x::y::zs).
Related
I have the following question "Given a list of integer pairs, write a function to return a list of even numbers in that list in sml".
this is what I've achieved so far
val x = [(6, 2), (3, 4), (5, 6), (7, 8), (9, 10)];
fun isEven(num : int) =
if num mod 2 = 0 then num else 0;
fun evenNumbers(list : (int * int) list) =
if null list then [] else
if isEven(#1 (hd list)) <> 0
then if isEven(#2 (hd list)) <> 0
then #1 (hd list) :: #1 (hd list) :: evenNumbers(tl list)
else []
else if isEven(#2 (hd list)) <> 0
then #1 (hd list) :: evenNumbers(tl list)
else [];
evenNumbers(x);
the result should be like this [6,2,4,6,8,10]
any help would be appreciated.
I see two obvious problems.
If both the first and second number are even, you do
#1 (hd list) :: #1 (hd list) :: evenNumbers(tl list)
which adds the first number twice and ignores the second.
If the first number is odd and the second even, you do
#1 (hd list) :: evenNumbers(tl list)
which adds the number that you know is odd and ignores the one you know is even.
Programming with selectors and conditionals gets complicated very quickly (as you've noticed).
With pattern matching, you could write
fun evenNumbers [] = []
| evenNumber ((x,y)::xys) = ...
and reduce the risk of using the wrong selector.
However, this still makes for complicated logic, and there is a better way.
Consider the simpler problem of filtering the odd numbers out of a list of numbers, not pairs.
If you transform the input into such a list, you only need to solve that simpler problem (and there's a fair chance that you've already solved something very similar in a previous exercise).
Exercise: implement this transformation. Its type will be ('a * 'a) list -> 'a list.
Also, your isEven is more useful if it produces a truth value (if you ask someone, "is 36 even?", "36" is a very strange answer).
fun isEven x = x mod 2 = 0
Now, evenNumbers can be implemented as "just" a combination of other, more general, functions.
So running your current code,
- evenNumbers [(6, 2), (3, 4), (5, 6), (7, 8), (9, 10)];
val it = [6,6,3,5,7,9] : int list
suggests that you're not catching all even numbers, and that you're catching some odd numbers.
The function isEven sounds very much like you want to have the type int -> bool like so:
fun isEven n =
n mod 2 = 0
Instead of addressing the logic error of your current solution, I would like to propose a syntactically much simpler approach which is to use pattern matching and fewer explicit type annotations. One basis for such a solution could look like:
fun evenNumbers [] = ...
| evenNumbers ((x,y)::pairs) = ...
Using pattern matching is an alternative to if-then-else: the [] pattern is equivalent to if null list ... and the (x,y)::pairs pattern matches when the input list is non-empty (holds at least one element, being (x,y). At the same time, it deconstructs this one element into its parts, x and y. So in the second function body you can express isEven x and isEven y.
As there is a total of four combinations of whether x and y are even or not, this could easily end up with a similarly complicated nest of if-then-else's. For this I might do either one of two things:
Use case-of (and call evenNumbers recursively on pairs):
fun evenNumbers [] = ...
| evenNumbers ((x,y)::pairs) =
case (isEven x, isEven y) of
... => ...
| ... => ...
Flatten the list of pairs into a list of integers and filter it:
fun flatten [] = ...
| flatten ((x,y)::pairs) = ...
val evenNumbers pairs = ...
I am fairly new to functional programming and I do not understand my error here. I am trying to make a function that takes an integer list and returns both the sum of the even elements and the sum of the odd elements. The error I am getting is in line 1, and it states: "Error: right-hand-side of clause doesn't agree with function result type [overload conflict] ...". I don't understand the error, and I would appreciate any help in understanding my error.
fun add(nil) = 0
| add([x]) = x
| add(x :: xs) =
let
val evenList = xs;
val oddList = x :: xs
in
(hd evenList + add(tl(tl(evenList))), hd oddList + add(tl(tl(oddList))))
end;
The reason for the type error is that the function should return a pair, but your base cases don't.
I suspect you got to that code by thinking about skipping every other element, dividing the list by skipping.
There's a different way to approach this.
Consider the list [a,b,c,d].
Counting from 1, the elements are numbered
1 2 3 4
a b c d
Now consider the positions in the tail of the list.
They are
1 2 3
b c d
That is, odd positions in the tail are even positions in the entire list, and even positions in the tail are odd in the entire list.
This means that if we recursively compute "odds and evens" in the tail, we will get the sums from the tail, where its "odds" is our "evens", and if we add our head to the tail's "evens", we will get the "odds" we want.
All we need now is a good base case – and the sums of an empty list must be (0, 0).
Something like this:
fun add [] = (0,0)
| add (x::xs) = case add xs of
(odds, evens) => (x + evens, odds)
or, you can deconstruct the recursive result with a let-binding instead of case:
fun add [] = (0,0)
| add (x::xs) = let val (odds, evens) = add xs
in
(x + evens, odds)
end
I am trying to write a function in SML which when given a list of general elements, reorders its elements into equivalent classes and returns a list of these classes (type "a list list).
Leave the elements in the classes in the same order as in the original list.
A given function defines the equivalence of the elements and it returns true if the elements are equivalent or false otherwise.
I cannot seem to get a grip on the solution.
fun sample x y = x = y
Required type: fn : (''a -> ''a -> bool) -> ''a list -> ''a list list
Thank you very much for the help.
The helper function does not work correctly, all I want to do with it is see if a given element belongs to any of the classes and put it accordingly inside or create a new sublist which contains it.
fun srt listoflists func new =
case listoflists of [] => [[]]
| a::b => if func (new, hd a) = true then (new::a)::b
else if func (new, hd a) = false then a::(srt b func new) else [new]::a::b
The sample functions checks equivalence of two elements when divided by 11.
Tests are not all working, it is not adding 17 into a new class.
srt [[7,7,7],[5,5,5],[11,11,11],[13,13,13]] eq 7;
val it = [[7,7,7,7],[5,5,5],[11,11,11],[13,13,13]] : int list list
- srt [[7,7,7],[5,5,5],[11,11,11],[13,13,13]] eq 5;
val it = [[7,7,7],[5,5,5,5],[11,11,11],[13,13,13]] : int list list
- srt [[7,7,7],[5,5,5],[11,11,11],[13,13,13]] eq 11;
val it = [[7,7,7],[5,5,5],[11,11,11,11],[13,13,13]] : int list list
- srt [[7,7,7],[5,5,5],[11,11,11],[13,13,13]] eq 13;
val it = [[7,7,7],[5,5,5],[11,11,11],[13,13,13,13]] : int list list
- srt [[7,7,7],[5,5,5],[11,11,11],[13,13,13]] eq 17;
val it = [[7,7,7],[5,5,5],[11,11,11],[13,13,13],[]] : int list list
- srt [[7,7,7],[5,5,5],[11,11,11],[13,13,13],[111,111,111]] eq 111;
val it = [[7,7,7],[5,5,5],[11,11,11],[13,13,13],[111,111,111,111]]
How to correct this and also once this helper function works, how to encorporate it exactly into the main function that is required.
Thank you very much.
Your example code seems like you are getting close, but has several issues
1) The basis cases is where new should be added, so in that case you should return the value [[new]] rather than [[]]
2) Your problem description suggests that func be of type ''a -> ''a -> bool but your code for srt seems to be assuming it is of type (''a * ''a) -> bool. Rather than subexpressions like func (new, hd a) you need func new (hd a) (note the parentheses location).
3) if func returns a bool then comparing the output to true is needlessly verbose, instead of if func new (hd a) = true then ... simply have if func new (hd a) then ...
4) Since you are adding [new] in the basis cases, your second clause is needlessly verbose. I see no reason to have any nested if expressions.
Since this seems to be homework, I don't want to say much more. Once you get the helper working correctly it should be fairly straightforward to use it (in the recursive case) of the overall function. Note that you could use (a # [new])::b rather than (new::a)::b if you want to avoid the need for a final mapping of rev across the final return value. # is more expensive than :: (it is O(n) rather than O(1)), but for small examples it really doesn't matter and could even be slightly better since it would avoid the final step of reversing the lists.
How does one get the first key,value pair from F# Map without knowing the key?
I know that the Map type is used to get a corresponding value given a key, e.g. find.
I also know that one can convert the map to a list and use List.Head, e.g.
List.head (Map.toList map)
I would like to do this
1. without a key
2. without knowing the types of the key and value
3. without using a mutable
4. without iterating through the entire map
5. without doing a conversion that iterates through the entire map behind the seen, e.g. Map.toList, etc.
I am also aware that if one gets the first key,value pair it might not be of use because the map documentation does not note if using map in two different calls guarantees the same order.
If the code can not be written then an existing reference from a site such as MSDN explaining and showing why not would be accepted.
TLDR;
How I arrived at this problem was converting this function:
let findmin l =
List.foldBack
(fun (_,pr1 as p1) (_,pr2 as p2) -> if pr1 <= pr2 then p1 else p2)
(List.tail l) (List.head l)
which is based on list and is used to find the minimum value in the associative list of string * int.
An example list:
["+",10; "-",10; "*",20; "/",20]
The list is used for parsing binary operator expressions that have precedence where the string is the binary operator and the int is the precedence. Other functions are preformed on the data such that using F# map might be an advantage over list. I have not decided on a final solution but wanted to explore this problem with map while it was still in the forefront.
Currently I am using:
let findmin m =
if Map.isEmpty m then
None
else
let result =
Map.foldBack
(fun key value (k,v) ->
if value <= v then (key,value)
else (k,v))
m ("",1000)
Some(result)
but here I had to hard code in the initial state ("",1000) when what would be better is just using the first value in the map as the initial state and then passing the remainder of the map as the starting map as was done with the list:
(List.tail l) (List.head l)
Yes this is partitioning the map but that did not work e.g.,
let infixes = ["+",10; "-",10; "*",20; "/",20]
let infixMap = infixes |> Map.ofList
let mutable test = true
let fx k v : bool =
if test then
printfn "first"
test <- false
true
else
printfn "rest"
false
let (first,rest) = Map.partition fx infixMap
which results in
val rest : Map<string,int> = map [("*", 20); ("+", 10); ("-", 10)]
val first : Map<string,int> = map [("/", 20)]
which are two maps and not a key,value pair for first
("/",20)
Notes about answers
For practical purposes with regards to the precedence parsing seeing the + operations before - in the final transformation is preferable so returning + before - is desirable. Thus this variation of the answer by marklam
let findmin (map : Map<_,_>) = map |> Seq.minBy (fun kvp -> kvp.Value)
achieves this and does this variation by Tomas
let findmin m =
Map.foldBack (fun k2 v2 st ->
match st with
| Some(k1, v1) when v1 < v2 -> st
| _ -> Some(k2, v2)) m None
The use of Seq.head does return the first item in the map but one must be aware that the map is constructed with the keys sorted so while for my practical example I would like to start with the lowest value being 10 and since the items are sorted by key the first one returned is ("*",20) with * being the first key because the keys are strings and sorted by such.
For me to practically use the answer by marklam I had to check for an empty list before calling and massage the output from a KeyValuePair into a tuple using let (a,b) = kvp.Key,kvp.Value
I don't think there is an answer that fully satisfies all your requirements, but:
You can just access the first key-value pair using m |> Seq.head. This is lazy unlike converting the map to list. This does not guarantee that you always get the same first element, but realistically, the implementation will guarantee that (it might change in the next version though).
For finding the minimum, you do not actually need the guarantee that Seq.head returns the same element always. It just needs to give you some element.
You can use other Seq-based functons as #marklam mentioned in his answer.
You can also use fold with state of type option<'K * 'V>, which you can initialize with None and then you do not have to worry about finding the first element:
m |> Map.fold (fun st k2 v2 ->
match st with
| Some(k1, v1) when v1 < v2 -> st
| _ -> Some(k2, v2)) None
Map implements IEnumerable<KeyValuePair<_,_>> so you can treat it as a Seq, like:
let findmin (map : Map<_,_>) = map |> Seq.minBy (fun kvp -> kvp.Key)
It's even simpler than the other answers. Map internally uses an AVL balanced tree so the entries are already ordered by key. As mentioned by #marklam Map implements IEnumerable<KeyValuePair<_,_>> so:
let m = Map.empty.Add("Y", 2).Add("X", 1)
let (key, value) = m |> Seq.head
// will return ("X", 1)
It doesn't matter what order the elements were added to the map, Seq.head can operate on the map directly and return the key/value mapping for the min key.
Sometimes it's required to explicitly convert Map to Seq:
let m = Map.empty.Add("Y", 2).Add("X", 1)
let (key, value) = m |> Map.toSeq |> Seq.head
The error message I've seen for this case says "the type 'a * 'b does not match the type Collections.Generic.KeyValuePair<string, int>". It may also be possible add type annotations rather than Map.toSeq.
I am trying to produce the solution for an intersection of two sets using tail recursion and an empty list [] as an accu:
let rec setintersect list list =
let rec setintersect2 a b c =
match a with
| [] -> (match b with [] -> (setsimplify c) | h::t -> (setsimplify c))
| h1::t1 -> (match b with [] -> (setsimplify c) |h2::t2 -> (if (elementof h1 b) then (setintersect2 t1 b (c#[h1])) else (setintersect2 t1 b c))) in
setintersect2 list list [];;
Elementof takes takes "an int and a list" and is correctly working to give true if x is an element of the list, false otherwise..
Here is the problem:
# setintersect [5;2;1] [2;6;9];;
- : int list = [2; 6; 9]
and it should give [2].
What am I doing wrong?
I feel like there's something really simple that I am misunderstanding!
Edit:
Thanks for the responses so far.
setsimplify just removes the duplicates.
so [2,2,3,5,6,6] becomes [2,3,5,6]. Tested and made sure it is working properly.
I am not supposed to use anything from the List library either. Also, I must use "tail recursion" with the accumulator being a list that I build as I go.
Here is the thought:
Check the head element in list1, IF it exists in list2, THEN recurse with the "tail of list1, list2, and list c with that element added to it". ELSE, then recurse with "tail of list1, list2 and list c(as it is)".
end conditions are either list1 or list2 are empty or both together are empty, return list c (as it is).
let rec setintersect list list = is wrong: the two arguments should be named differently (you should of course update the call to setintersect2 accordingly), otherwise the second will shadow the first. I would have thought that OCaml would have at least warned you about this fact, but it appears that it is not the case.
Apart from that, the code seems to do the trick. There are a couple of things that could be improved though:
setintersect itself is not recursive (only setintersect2 is), you thus don't need the rec
you should find a different name for the argument of setintersect2. In particular, it is not obvious which is the accumulator (acc or accu will be understood by most OCaml programmers in these circumstances).
c#[h1] is inefficient: you will traverse c completely each time you append an element. It's better to do h1::c and reverse the result at the end
As a bonus point, if you append element at the beginning of c, and assume that a is ordered, you don't have to call setsimplify at the end of the call: just check whether c is empty, and if this is not the case, append h1 only if it is not equal to the head of c.
First, You didn't list out your setsimplify function.
To write an ocaml function, try to split it first, and then combine if possible.
To solve this task, you just go through all elements in l1, and for every element, you check whether it is in l2 or not, right?
So definitely you need a function to check whether an element is in a list or not, right?
let make one:
let rec mem x = function
| [] -> false
| hd::tl -> hd = x || mem x tl
Then you can do your intersection:
let rec inter l1 l2 =
match l1 with
| [] -> []
| hd::tl -> if mem hd l2 then hd::(inter tl l2) else inter tl l2
Note that the above function is not tail-recursive, I guess you can change it to tail-recursive as an excise.
If you use std library, then it is simple:
let intersection l1 l2 = List.filter (fun x -> List.mem x l2) l1