I believe I have some basic misunderstanding about catching exceptions in SML.
I wrote the following code:
fun my_g acc p =
let
val r = my_g acc
in
case p of
Wildcard => acc
| Variable x => if List.exists (fn y => y = x) acc then raise NoAnswer else x::acc
| TupleP ps => List.foldl (fn (p,i) => (my_g i p)) acc ps
| ConstructorP(_,p) => r p
| _ => acc
end
(* val check_pat = fn : pattern -> bool *)
fun check_pat p =
if my_g [] p <> [] then
true
else
true
handle NoAnswer => false
I'm happy to explain the code in gory detail, but basically I'm looking to see if strings are repeated in a list. If I find a string repeated, I raise an exception. Notice I'm handling the exception in function check_pat, which calls function my_g. However, when I run the code with some test methods, I get uncaught exception NoAnswer
Can I catch an exception thrown in one function in another (calling) function?
What am I doing wrong?
Thanks, Dave
Additional details for Andreas and future viewers. The original hints were to first unfold the structure and get a list of strings and only then go through and look for duplicates. I felt that was inefficient and it would be best to look for duplicates as you unfolded. Unfortunately, my SML knowledge is not good enough to come up with a super clean solution. Really, I don't care about the return value of my_g. If it doesn't throw an exception, then there are no duplicates. As simple as that. But it seemed the syntax rules were forcing me to check the return value. Now that you've solved the "handle" issue for me, I might revisit the problem. I was hoping to just write:
(my_g [] p
true)
handle NoAnswer => false
but that didn't seem to work. More broadly, although I think my solution is more efficient than first unfolding the entire list just to then look for duplicates, I suspect the idea of using an exception like I did is not good style. In languages I'm familiar with (C++, C#), an exception means some exceptional or unexpected occurred. Finding a duplicate string is certainly not exceptional. Again, I'm sure there is another way to stop at the first duplicate without using exceptions. I'm just not proficient enough in SML to know it. Thanks!
This is simply a matter of parenthesization: handle binds tighter than if, so you have effectively written
if ... then ... else (... handle ...)
Instead you want
(if ... then ... else ...) handle ...
so you need to put in parentheses.
BTW, I can't make sense of your use of if -- why a conditional when both branches produce the same result? Also, if A then true else B is a verbose way of saying A orelse B.
Edit in reply to edit in the question: If you want to ignore the result of an expression and return something else instead then you can use the semicolon operator:
(my_g [] p; true)
However, in general, using exceptions for non-exceptional control flow is not recommended. There is a cleaner way to write this function:
fun ids (Variable x) = [x]
| ids (Tuple ps) = List.concat (List.map ids ps)
| ids (Constructor(_,p)) = ids p
| ids _ = []
fun hasDups [] = false
| hasDups (x::xs) = List.exists (fn y => y = x) xs orelse hasDups xs
fun checkPat p = not (hasDups (ids p))
Edit 2: In the normal case (where there are no duplicates), this solution isn't slower than the other. So it's not necessarily worth taking the shortcut. If you insist, though, there are various options that don't require exceptions. For example:
fun checkPat'(_, NONE) = NONE
| checkPat'(Variable x, SOME xs) = if List.exists (fn y => y = x) xs then NONE else SOME (x::xs)
| checkPat'(Tuple ps, xso) = List.foldl checkPat' xso ps
| checkPat'(Constructor(_,p), xso) = checkPat'(p, xso)
| checkPat'(_, xso) = xso
fun checkPat p = isSome (checkPat'(p, SOME []))
Or, if you are willing to use a bit of mutable state:
fun checkPat' xs (Variable x) = List.exists (fn y => y = x) (!xs) before (xs := x :: !xs)
| checkPat' xs (Tuple ps) = List.all (checkPat' xs) ps
| checkPat' xs (Constructor(_,p)) = checkPat' xs p
| checkPat' xs _ = true
fun checkPat p = checkPat' (ref []) p
Related
I'm trying to learn haskell by solving some online problems and training exercises.
Right now I'm trying to make a function that'd remove adjacent duplicates from a list.
Sample Input
"acvvca"
"1456776541"
"abbac"
"aabaabckllm"
Expected Output
""
""
"c"
"ckm"
My first though was to make a function that'd simply remove first instance of adjacent duplicates and restore the list.
module Test where
removeAdjDups :: (Eq a) => [a] -> [a]
removeAdjDups [] = []
removeAdjDups [x] = [x]
removeAdjDups (x : y : ys)
| x == y = removeAdjDups ys
| otherwise = x : removeAdjDups (y : ys)
*Test> removeAdjDups "1233213443"
"122133"
This func works for first found pairs.
So now I need to apply same function over the result of the function.
Something I think foldl can help with but I don't know how I'd go about implementing it.
Something along the line of
removeAdjDups' xs = foldl (\acc x -> removeAdjDups x acc) xs
Also is this approach the best way to implement the solution or is there a better way I should be thinking of?
Start in last-first order: first remove duplicates from the tail, then check if head of the input equals to head of the tail result (which, by this moment, won't have any duplicates, so the only possible pair is head of the input vs. head of the tail result):
main = mapM_ (print . squeeze) ["acvvca", "1456776541", "abbac", "aabaabckllm"]
squeeze :: Eq a => [a] -> [a]
squeeze (x:xs) = let ys = squeeze xs in case ys of
(y:ys') | x == y -> ys'
_ -> x:ys
squeeze _ = []
Outputs
""
""
"c"
"ckm"
I don't see how foldl could be used for this. (Generally, foldl pretty much combines the disadvantages of foldr and foldl'... those, or foldMap, are the folds you should normally be using, not foldl.)
What you seem to intend is: repeating the removeAdjDups, until no duplicates are found anymore. The repetition is a job for
iterate :: (a -> a) -> a -> [a]
like
Prelude> iterate removeAdjDups "1233213443"
["1233213443","122133","11","","","","","","","","","","","","","","","","","","","","","","","","","","",""...
This is an infinite list of ever reduced lists. Generally, it will not converge to the empty list; you'll want to add some termination condition. If you want to remove as many dups as necessary, that's the fixpoint; it can be found in a very similar way to how you implemented removeAdjDups: compare neighbor elements, just this time in the list of reductions.
bipll's suggestion to handle recursive duplicates is much better though, it avoids unnecessary comparisons and traversing the start of the list over and over.
List comprehensions are often overlooked. They are, of course syntactic sugar but some, like me are addicted. First off, strings are lists as they are. This functions could handle any list, too as well as singletons and empty lists. You can us map to process many lists in a list.
(\l -> [ x | (x,y) <- zip l $ (tail l) ++ " ", x /= y]) "abcddeeffa"
"abcdefa"
I don't see either how to use foldl. It's maybe because, if you want to fold something here, you have to use foldr.
main = mapM_ (print . squeeze) ["acvvca", "1456776541", "abbac", "aabaabckllm"]
-- I like the name in #bipll answer
squeeze = foldr (\ x xs -> if xs /= "" && x == head(xs) then tail(xs) else x:xs) ""
Let's analyze this. The idea is taken from #bipll answer: go from right to left. If f is the lambda function, then by definition of foldr:
squeeze "abbac" = f('a' f('b' f('b' f('a' f('c' "")))
By definition of f, f('c' "") = 'c':"" = "c" since xs == "". Next char from the right: f('a' "c") = 'a':"c" = "ac" since 'a' != head("c") = 'c'. f('b' "ac") = "bac" for the same reason. But f('b' "bac") = tail("bac") = "ac" because 'b' == head("bac"). And so forth...
Bonus: by replacing foldr with scanr, you can see the whole process:
Prelude> squeeze' = scanr (\ x xs -> if xs /= "" && x == head(xs) then tail(xs) else x:xs) ""
Prelude> zip "abbac" (squeeze' "abbac")
[('a',"c"),('b',"ac"),('b',"bac"),('a',"ac"),('c',"c")]
I'm working on a program that appends either a '+' or '-' to an element of a list, depending on whether the index of that element is odd or even (i.e an alternating sums list).
However, I'm having trouble identifying what the index of each element is. I have code that I believe should append the correct symbol, using if statements and mod
fun alternating([]) = 0
| alternating(l) =
if List.nth(l,hd(l)) mod 2 == 0 then '+'#hd(l)#alternating(tl(l))
else '-'#hd(l)#alternating(tl(l))
However, List.nth(l,hd(l)) always returns the element at the second index, not the first.
On the off chance that you really just want to negate integers them so you can pass them into some kind of summation, I would just negate the argument if it's odd. Using mutual recursion one can do it without any explicit index bookkeeping:
fun alternate l =
let
fun alternate1 [] = []
| alternate1 (x::xs) = (~x) :: alternate2 xs
and alternate2 [] = []
| alternate2 (x::xs) = x :: alternate1 xs
in
alternate1 l
end
It works like so:
- alternate [1,2,3,4];
val it = [~1,2,~3,4] : int list
I would strongly encourage you to use pattern matching instead of hd.
Edit discussing hd
As a rule of thumb, if you need hd you probably need tl as well. hd is a partial function--it's going to throw Empty if your list is empty. If you pattern match, you conveniently get variables for the head and tail of the list right there, and you get a visual reminder that you need to handle the empty list. It's more aesthetically pleasing, IMO, to see:
fun foo [] = ...
| foo (x::xs) = ...
than the equivalent
fun foo l =
if null l
then ...
else (hd l) ... (tl l)
In other words, you get shorter, cleaner code with an automatic reminder to make it correct. Win/win. To my knowledge there's no significant advantage to doing it the other way. Of course, you may find yourself in a situation where you know the list will have at least one element and you don't need to do anything else. You still have to consider the cases you're given, but it's a good rule of thumb.
If you want to decorate your list with an index you could try something like the following
fun add_index l =
let
fun add_index_helper (nil, _) = nil
| add_index_helper (h::tl,i) = (h,i) :: add_index_helper (tl,1+i)
in
add_index_helper (l,0)
end
val x = add_index [0,1,4,9,16,25]
but you can also just directly compute parity with the same method
fun add_sign l =
let
fun add_sign_helper (nil, _) = nil
| add_sign_helper (h::tl,i) = (h,i) :: add_sign_helper (tl,1-i)
in
add_sign_helper (l,0)
end
val y = add_sign [0,1,4,9,16,25]
then you can map the parity to a string
fun sign_to_char (x,0) = (x,"+")
| sign_to_char (x,_) = (x,"-")
val z = List.map sign_to_char y
or you can just add the sign directly
fun add_char l =
let
fun add_char_helper (nil, _) = nil
| add_char_helper (h::tl,0) = (h,"+") :: add_char_helper (tl,1)
| add_char_helper (h::tl,_) = (h,"-") :: add_char_helper (tl,0)
in
add_char_helper (l,0)
end
val zz = add_char [0,1,4,9,16,25]
Alternatively if you had a string list and you wanted to add chars you could try something like this
fun signs L =
let
datatype parity = even | odd
fun signs_helper ( nil ,_) = nil
| signs_helper (x::xs,even) = ("+" ^ x) :: signs_helper(xs,odd)
| signs_helper (x::xs,odd) = ("-" ^ x) :: signs_helper(xs,even)
in
signs_helper (L,even)
end
val z = signs ["x","2y","3z","4"]
(* this gives you val z = ["+x","-2y","+3z","-4"] : string list *)
So i'm new to sml and am trying to understand the ins/out out of it. Recently i tried creating a filter which takes two parameters: a function (that returns a boolean), and a list of values to run against the function. What the filter does is it returns the list of values which return true against the function.
Code:
fun filter f [] = [] |
filter f (x::xs) =
if (f x)
then x::(filter f xs)
else (filter f xs);
So that works. But what i'm trying to do now is just a return a tuple that contains the list of true values, and false. I'm stuck on my conditional and I can't really see another way. Any thoughts on how to solve this?
Code:
fun filter2 f [] = ([],[]) |
filter2 f (x::xs) =
if (f x)
then (x::(filter2 f xs), []) (* error *)
else ([], x::(filter2 f xs)); (* error *)
I think there are several ways to do this.
Reusing Filter
For instance, we could use a inductive approach based on the fact that your tuple would be formed by two elements, the first is the list of elements that satisfy the predicate and the second the list of elements that don't. So, you could reuse your filter function as:
fun partition f xs = (filter f xs, filter (not o f) xs)
This is not the best approach, though, because it evaluates the lists twice, but if the lists are small, this is quite evident and very readable.
Folding
Another way to think about this is in terms of fold. You could think that you are reducing your list to a tuple list, and as you go, you split your items depending on a predicate. Somwewhat like this:
fun parition f xs =
let
fun split x (xs,ys) =
if f x
then (x::xs,ys)
else (xs, x::ys)
val (trueList, falseList) = List.foldl (fn (x,y) => split x y)
([],[]) xs
in
(List.rev trueList, List.rev falseList)
end
Parition
You could also implement your own folding algorithm in the same way as the List.parition method of SML does:
fun partition f xs =
let
fun iter(xs, (trueList,falseList)) =
case xs of
[] => (List.rev trueList, List.rev falseList)
| (x::xs') => if f x
then iter(xs', (x::trueList,falseList))
else iter(xs', (trueList,x::falseList))
in
iter(xs,([],[]))
end
Use SML Basis Method
And ultimately, you can avoid all this and use SML method List.partition whose documentation says:
partition f l
applies f to each element x of l, from left to right, and returns a
pair (pos, neg) where pos is the list of those x for which f x
evaluated to true, and neg is the list of those for which f x
evaluated to false. The elements of pos and neg retain the same
relative order they possessed in l.
This method is implemented as the previous example.
So I will show a good way to do it, and a better way to do it (IMO). But the 'better way' is just for future reference when you learn:
fun filter2 f [] = ([], [])
| filter2 f (x::xs) = let
fun ftuple f (x::xs) trueList falseList =
if (f x)
then ftuple f xs (x::trueList) falseList
else ftuple f xs trueList (x::falseList)
| ftuple _ [] trueList falseList = (trueList, falseList)
in
ftuple f (x::xs) [] []
end;
The reason why yours does not work is because when you call x::(filter2 f xs), the compiler is naively assuming that you are building a single list, it doesn't assume that it is a tuple, it is stepping into the scope of your function call. So while you think to yourself result type is tuple of lists, the compiler gets tunnel vision and thinks result type is list. Here is the better version in my opinion, you should look up the function foldr if you are curious, it is much better to employ this technique since it is more readable, less verbose, and much more importantly ... more predictable and robust:
fun filter2 f l = foldr (fn(x,xs) => if (f x) then (x::(#1(xs)), #2(xs)) else (#1(xs), x::(#2(xs)))) ([],[]) l;
The reason why the first example works is because you are storing default empty lists that accumulate copies of the variables that either fit the condition, or do not fit the condition. However, you have to explicitly tell SML compiler to make sure that the type rules agree. You have to make absolutely sure that SML knows that your return type is a tuple of lists. Any mistake in this chain of command, and this will result in failure to execute. Hence, when working with SML, always study your type inferences. As for the second one, you can see that it is a one-liner, but I will leave you to research that one on your own, just google foldr and foldl.
What would be the syntax (if possible at all) for returning the list of lists ([[a]]) but without the use of empty list ([]:[a])?
(similar as the second commented guard (2) below, which is incorrect)
This is a function that works correctly:
-- Split string on every (shouldSplit == true)
splitWith :: (Char -> Bool) -> [Char] -> [[Char]]
splitWith shouldSplit list = filter (not.null) -- would like to get rid of filter
(imp' shouldSplit list)
where
imp' _ [] = [[]]
imp' shouldSplit (x:xs)
| shouldSplit x = []:imp' shouldSplit xs -- (1) this line is adding empty lists
-- | shouldSplit x = [imp' shouldSplit xs] -- (2) if this would be correct, no filter needed
| otherwise = let (z:zs) = imp' shouldSplit xs in (x:z):zs
This is the correct result
Prelude> splitWith (== 'a') "miraaaakojajeja234"
["mir","koj","jej","234"]
However, it must use "filter" to clean up its result, so I would like to get rid of function "filter".
This is the result without the use of filter:
["mir","","","","koj","jej","234"]
If "| shouldSplit x = imp' shouldSplit xs" is used instead the first guard, the result is incorrect:
["mirkojjej234"]
The first guard (1) adds empty list so (I assume) compiler can treat the result as a list of lists ([[a]]).
(I'm not interested in another/different solutions of the function, just the syntax clarification.)
.
.
.
ANSWER:
Answer from Dave4420 led me to the answer, but it was a comment, not an answer so I can't accept it as answer. The solution of the problem was that I'm asking the wrong question. It is not the problem of syntax, but of my algorithm.
There are several answers with another/different solutions that solve the empty list problem, but they are not the answer to my question. However, they expanded my view of ways on how things can be done with basic Haskell syntax, and I thank them for it.
Edit:
splitWith :: (Char -> Bool) -> String -> [String]
splitWith p = go False
where
go _ [] = [[]]
go lastEmpty (x:xs)
| p x = if lastEmpty then go True xs else []:go True xs
| otherwise = let (z:zs) = go False xs in (x:z):zs
This one utilizes pattern matching to complete the task of not producing empty interleaving lists in a single traversal:
splitWith :: Eq a => (a -> Bool) -> [a] -> [[a]]
splitWith f list = case splitWith' f list of
[]:result -> result
result -> result
where
splitWith' _ [] = []
splitWith' f (a:[]) = if f a then [] else [[a]]
splitWith' f (a:b:tail) =
let next = splitWith' f (b : tail)
in if f a
then if a == b
then next
else [] : next
else case next of
[] -> [[a]]
nextHead:nextTail -> (a : nextHead) : nextTail
Running it:
main = do
print $ splitWith (== 'a') "miraaaakojajeja234"
print $ splitWith (== 'a') "mirrraaaakkkojjjajeja234"
print $ splitWith (== 'a') "aaabbbaaa"
Produces:
["mir","koj","jej","234"]
["mirrr","kkkojjj","jej","234"]
["bbb"]
The problem is quite naturally expressed as a fold over the list you're splitting. You need to keep track of two pieces of state - the result list, and the current word that is being built up to append to the result list.
I'd probably write a naive version something like this:
splitWith p xs = word:result
where
(result, word) = foldr func ([], []) xs
func x (result, word) = if p x
then (word:result,[])
else (result, x:word)
Note that this also leaves in the empty lists, because it appends the current word to the result whenever it detects a new element that satisfies the predicate p.
To fix that, just replace the list cons operator (:) with a new operator
(~:) :: [a] -> [[a]] -> [[a]]
that only conses one list to another if the original list is non-empty. The rest of the algorithm is unchanged.
splitWith p xs = word ~: result
where
(result, word) = foldr func ([], []) xs
func x (result, word) = if p x
then (word ~: result, [])
else (result, x:word)
x ~: xs = if null x then xs else x:xs
which does what you want.
I guess I had a similar idea to Chris, I think, even if not as elegant:
splitWith shouldSplit list = imp' list [] []
where
imp' [] accum result = result ++ if null accum then [] else [accum]
imp' (x:xs) accum result
| shouldSplit x =
imp' xs [] (result ++ if null accum
then []
else [accum])
| otherwise = imp' xs (accum ++ [x]) result
This is basically just an alternating application of dropWhile and break, isn't it:
splitWith p xs = g xs
where
g xs = let (a,b) = break p (dropWhile p xs)
in if null a then [] else a : g b
You say you aren't interested in other solutions than yours, but other readers might be. It sure is short and seems clear. As you learn, using basic Prelude functions becomes second nature. :)
As to your code, a little bit reworked in non-essential ways (using short suggestive function names, like p for "predicate" and g for a main worker function), it is
splitWith :: (Char -> Bool) -> [Char] -> [[Char]]
splitWith p list = filter (not.null) (g list)
where
g [] = [[]]
g (x:xs)
| p x = [] : g xs
| otherwise = let (z:zs) = g xs
in (x:z):zs
Also, there's no need to pass the predicate as an argument to the worker (as was also mentioned in the comments). Now it is arguably a bit more readable.
Next, with a minimal change it becomes
splitWith :: (Char -> Bool) -> [Char] -> [[Char]]
splitWith p list = case g list of ([]:r)-> r; x->x
where
g [] = [[]]
g (x:xs)
| p x = case z of []-> r; -- start a new word IF not already
_ -> []:r
| otherwise = (x:z):zs
where -- now z,zs are accessible
r#(z:zs) = g xs -- in both cases
which works as you wanted. The top-level case is removing at most one empty word here, which serves as a separator marker at some point during the inner function's work. Your filter (not.null) is essentially fused into the worker function g here, with the conditional opening1 of a new word (i.e. addition1 of an empty list).
Replacing your let with where allowed for the variables (z etc.) to became accessible in both branches of the second clause of the g definition.
In the end, your algorithm was close enough, and the code could be fixed after all.
1 when thinking "right-to-left". In reality the list is constructed left-to-right, in guarded recursion ⁄ tail recursion modulo cons fashion.
Anyone able to offer any advice for a function in SML that will take 2 lists and return the XOR of them, so that if you have the lists [a,b,c,d], [c,d,e,f] the function returns [a,b,e,f] ?
I have tried to do it with 2 functions, but even that does not work properly.
fun del(nil,L2) = nil
|del(x::xs,L2)=
if (List.find (fn y => y = x) L2) <> (SOME x) then
del(xs, L2) # [x]
else
del(xs, L2);
fun xor(L3,L4) =
rev(del(L3,L4)) # rev(del(L4,L3));
Your attempt seems almost correct, except that fn x => x = x does not make sense, since it always returns true. I think you want fn y => y = x instead.
A couple of other remarks:
You can replace your use of List.find with List.filter which is closer to what you want.
Don't do del(xs,L) # [x] for the recursive step. Appending to the end of the list has a cost linear to the length of the first list, so if you do it in every step, your function will have quadratic runtime. Do x :: del(xs,L) instead, which also allows you to drop the list reversals in the end.
What you call "XOR" here is usually called the symmetric difference, at least for set-like structures.
The simplest way would be to filter out duplicates from each list and then concatenate the two resulting lists. Using List.filter you can remove any element that is a member (List.exists) of the other list.
However that is quite inefficient, and the below code is more an example of how not to do it in real life, though it is "functionally" nice to look at :)
fun symDiff a b =
let
fun diff xs ys =
List.filter (fn x => not (List.exists ( fn y => x = y) ys)) xs
val a' = diff a b
val b' = diff b a
in
a' # b'
end
This should be a better solution, that is still kept simple. It uses the SML/NJ specific ListMergeSort module for sorting the combined list a # b.
fun symDiff1 a b =
let
val ab' = ListMergeSort.sort op> (a # b)
(* Remove elements if they occur more than once. Flag indicates whether x
should be removed when no further matches are found *)
fun symDif' (x :: y :: xs) flag =
(case (x = y, flag) of
(* Element is not flagged for removal, so keep it *)
(false, false) => x :: symDif' (y :: xs) false
(* Reset the flag and remove x as it was marked for removal *)
| (false, true) => symDif' (y::xs) false
(* Remove y and flag x for removal if it wasn't already *)
| (true, _) => symDif' (x::xs) true)
| symDif' xs _ = xs
in
symDif' ab' false
end
However this is still kind of stupid. As the sorting function goes through all elements in the combined list, and thus it also ought to be the one that is "responsible" for removing duplicates.