I have been trying to count elements in a list of integer 3-tuples, that equals a given integer using SML, but it's not working. Can anyone help me figure out what's wrong with the below code or straighten it up for me?
fun number_in_month(x : int*int*int list, m: int) =
if null x then 0
else
let fun inc x = x + 1;
in
val counter = 0;
if m = #2 (hd x) andalso m > 0 then inc counter
number_in_month((tl x), m)
` else
number_in_month((tl x), m)
end
This function is supposed to return the number of times m equals to the second element of each tuple in the list.
Clearly you have a hard time to let go of your imperative thinking.
Let me try and address some of your issues
You should be using pattern matching instead of using null x, hd x and tl x.
This also apply to decomposing tuples and records. For example
fun number_in_month ((x1, x2, x3) :: xs, m) = ...
or, since we don't ever use x1 and x3
fun number_in_month ((_, x2, _) :: xs, m) = ...
This way it is clearly seen that the first argument is a list of 3-tuples, and no type annotation
is needed
Also when you omit the explicit type annotation, which is the whole idea of having a type system
that can infer them for you (see next point), then this code
fun foo42 xs = map (fn x => #2 x) xs
will give you some nasty errors on "unresolved flex record" (this error message is from SML/NJ)
/tmp/sml20620PlF:105.5-105.44 Error: unresolved flex record
(can't tell what fields there are besides #2)
which is easily fixed by decomposing the 3-tuple
fun foo42 xs = map (fn (_, x2, _) => x2) xs
Speaking of type annotations. They are (almost always) not needed, and they clutter up the
readability of the code. Not to mention that they unnecessarily restricts the types you function
may be used on.
Also the type annotation you have given is erroneous according to what you really wan't. You
should have places parenthesis around the int * int * int. Currently it is interpreted as a
3-tuple of two ints and an int list int * int * (int list).
If you really insist in type annotating your function, then you can do it like this
val number_in_month : (int * int * int) list * int -> int =
fn ([] , m) => 0
| ((_,x2,_) :: xs, m) => 42
This is "almost" like Haskell, where the type is given just before the function declaration.
Try to be more consistent in they way you indent your code. That will give you better clarity.
Here I'm specifically thinking of the way you have indented the else part end the in ... end
part. The below part is clearly still erroneous in so many ways i can't begin to imagine, but it
gives an idea as how to do it
fun number_in_month(x : int*int*int list, m: int) =
if null x then 0
else
let fun inc x = x + 1;
in
val counter = 0;
if m = #2 (hd x) andalso m > 0 then
inc counter
number_in_month((tl x), m)
else
number_in_month((tl x), m)
end
You can't declare a variable val counter = 0 inside the in ... end part of a let-expression.
The semantics of a let-expression is
let
dec
in
exp_1; ...; exp_n
end
thus all declarations (function and value bindings, etc) must go in the let ... in part.
There is no need on earth to have an increment function, it just clutters the readability.
Remember that SML uses single assignment, thus variables are immutable after they are declared.
The sequence-thing inside your nested if-expression
inc counter
number_in_month((tl x), m)
makes absolutely no sense. The only way you can have more than one expression inside the
then ... else part (actually any place, where a single expression is expected), is with a
sequence (exp_1; ...; exp_n). However this is only usable when all but the last expression has
side effect(s), as their results is ignored/thrown away
- (print "Foo\n"; print "Bar\n"; 42);
Foo
Bar
val it = 42 : int
If you search a bit here on SO, you will see that a quite similar question has recently been asked and answered. Though it differs in the the type of the last argument, you might still get some useful pointers.
All in all a solution might look like
fun number_in_month ([], _) = 0
| number_in_month ((_,x2,_) :: xs, m) =
if x2 = m then
1 + number_in_month(xs, m)
else
number_in_month(xs, m)
However since your problem is simpler than the previously stated one, you could easily use some of the higher-order functions from the list module in the basis library
fun number_in_month (xs, m) = length (List.filter (fn (_, x2, _) => x2 = m) xs)
Or even (arguably) simpler, by folding over the list and incrementing a variable along the way each time it matches
fun number_in_month (xs, m) = foldl (fn ((_, x2, _), b) => if x2 = m then b+1 else b) 0 xs
fun number_in_month (L : (int*int*int) list, m : int) =
if L = nil
then 0
else
(if #2 (hd L) = m then 1 else 0) + number_in_month (tl L,m);
TESTING:
number_in_month ([] , 2);
number_in_month ([(1,2,3)] , 2);
number_in_month ([(1,2,3),(2,2,2)] , 2);
number_in_month ([(1,2,3),(2,2,2),(19,11,29)] , 2);
number_in_month ([(1,2,3),(2,2,2),(19,11,29),(10,28,19)] , 2);
number_in_month ([(1,2,3),(2,2,2),(19,11,29),(10,2,19)] , 2);
number_in_month ([(1,2,3),(2,2,2),(19,11,29),(10,28,19)] , 2);
number_in_month ([(1,2,3),(2,2,2),(19,11,29),(10,28,19)] , 2);
number_in_month ([(1,2,3),(2,2,2),(19,11,29),(10,28,19),(16,2,7)] , 2);
Reference:
http://www.cs.sunysb.edu/~leo/CSE215/smllistexamples.txt
http://www.standardml.org/Basis/list.html
Related
I wrote a function that is supposed to receive a list of tuples. I access the components of the tuples with # and the code compiles:
fun recheck ([], n) = []
| recheck (h::t, n) =
if ((#1 h) * (#1 h)) + ((#2 h) * (#2 h)) = n then
h::recheck(t, n)
else
recheck(t, n)
But another function that basically does the same thing, namely receiving a list of tuples and accessing those, causes an error.
fun validate ([]) = true
| validate (h::t) =
if 1 = (#1 h) then
true
else
false
Can't find a fixed record type. Found near #1
What is the difference here and why does the latter cause an error?
Edit
The first function actually does not compile on its own.
But this entire snippet does:
fun drop ([], n) = []
| drop (h::t, 0) = h::t
| drop (h::t, n) =
drop(t, n-1)
fun sts_linear (y, n) =
if y < (Math.sqrt(n)+1.0) then
let
(* x^2 + y^2 = n => x = sqrt(n-y^2) *)
val x = Math.sqrt(n - (y * y));
val xr = Real.realRound(x);
in
if (abs(x - xr) < 0.000000001) then
[(Real.trunc xr, Real.trunc y)]#sts_linear (y+1.0, n)
else
(
[]#sts_linear (y+1.0, n)
)
end
else []
fun recheck ([], n) = []
| recheck (h::t, n) =
if ((#1 h) * (#1 h)) + ((#2 h) * (#2 h)) = n then
h::recheck(t, n)
else
recheck(t, n)
fun sts (n) =
(
let
val pairs = sts_linear(0.0, Real.fromInt n);
in
recheck(drop(pairs, Real.ceil( Real.fromInt (length(pairs))/2.0 ) ), n)
end
)
Your first code doesn't compile, at least with SML/NJ:
If you got it to compile then it must have been in a nonstandard extension of SML.
The problem with both of your definitions is that there is no polymorphic idea of a tuple of arbitrary arity in SML. You can write functions to work on lists of pairs. You can write functions to work on lists of triples. But -- you can't write functions to work simultaneously on lists of pairs and lists of triples (at least if your function tries to do things with these pairs/triples as tuples).
One solution is to get rid of # and use pattern-matching to extract the components:
fun validate [] = true
| validate ((x,y)::t) =
if x = 1 then
true
else
false
But, if you really want to write a function which can polymorphically apply to either lists of pairs or list of triples (or quadruples,...), the easiest thing to do is to represent the pairs, triples, etc. as lists rather than tuples. Lists which contains lists of nonspecified size are not a problem in SML.
Trying to minimize this down, as I have seen the following work in SML/NJ
and i'm not aware of it actually being a compiler extension
val p1 = {x=0, y=0};
val p2 = {x=1, y=1};
val p3 = {x=1, y=1, z=1};
There is an awkward construct from a compiler error perspective
not many languages have errors that work in this fashion,
because the function is valid, but produces a type error
unless an invocation of the function exists to resolve the
type of 'record', thus to resolve the error more code must be added.
fun getFoo(field) = fn record => field record;
Without the following actual calling of the getX
the compiler cannot determine the type of record
of which the complete type information of ALL fields
of the record must be known to the compiler, not just the #x field.
let val getX = getFoo(#x);
val x1 = getX(p1);
val x2 = getX(p2);
val x3 = getFoo(#x)(p3);
in () end;
while the following commented out snippet results in an error because the types of
p1 and p3 are different, and so different invocations of getFoo
are required
(*
let val getX = getFoo(#x);
val x1 = getX(p1);
val x3 = getX(p3);
in () end;
*)
and the following is insufficient since it never resolves the record.
let val getX = getFoo(#x) in () end;
I have a simple function (used for some problems of project Euler, in fact). It turns a list of digits into a decimal number.
fromDigits :: [Int] -> Integer
fromDigits [x] = toInteger x
fromDigits (x:xs) = (toInteger x) * 10 ^ length xs + fromDigits xs
I realized that the type [Int] is not ideal. fromDigits should be able to take other inputs like e.g. sequences, maybe even foldables ...
My first idea was to replace the above code with sort of a "fold with state". What is the correct (= minimal) Haskell-category for the above function?
First, folding is already about carrying some state around. Foldable is precisely what you're looking for, there is no need for State or other monads.
Second, it'd be more natural to have the base case defined on empty lists and then the case for non-empty lists. The way it is now, the function is undefined on empty lists (while it'd be perfectly valid). And notice that [x] is just a shorthand for x : [].
In the current form the function would be almost expressible using foldr. However within foldl the list or its parts aren't available, so you can't compute length xs. (Computing length xs at every step also makes the whole function unnecessarily O(n^2).) But this can be easily avoided, if you re-thing the procedure to consume the list the other way around. The new structure of the function could look like this:
fromDigits' :: [Int] -> Integer
fromDigits' = f 0
where
f s [] = s
f s (x:xs) = f (s + ...) xs
After that, try using foldl to express f and finally replace it with Foldable.foldl.
You should avoid the use of length and write your function using foldl (or foldl'):
fromDigits :: [Int] -> Integer
fromDigits ds = foldl (\s d -> s*10 + (fromIntegral d)) 0 ds
From this a generalization to any Foldable should be clear.
A better way to solve this is to build up a list of your powers of 10. This is quite simple using iterate:
powersOf :: Num a => a -> [a]
powersOf n = iterate (*n) 1
Then you just need to multiply these powers of 10 by their respective values in the list of digits. This is easily accomplished with zipWith (*), but you have to make sure it's in the right order first. This basically just means that you should re-order your digits so that they're in descending order of magnitude instead of ascending:
zipWith (*) (powersOf 10) $ reverse xs
But we want it to return an Integer, not Int, so let's through a map fromIntegral in there
zipWith (*) (powersOf 10) $ map fromIntegral $ reverse xs
And all that's left is to sum them up
fromDigits :: [Int] -> Integer
fromDigits xs = sum $ zipWith (*) (powersOf 10) $ map fromIntegral $ reverse xs
Or for the point-free fans
fromDigits = sum . zipWith (*) (powersOf 10) . map fromIntegral . reverse
Now, you can also use a fold, which is basically just a pure for loop where the function is your loop body, the initial value is, well, the initial state, and the list you provide it is the values you're looping over. In this case, your state is a sum and what power you're on. We could make our own data type to represent this, or we could just use a tuple with the first element being the current total and the second element being the current power:
fromDigits xs = fst $ foldr go (0, 1) xs
where
go digit (s, power) = (s + digit * power, power * 10)
This is roughly equivalent to the Python code
def fromDigits(digits):
def go(digit, acc):
s, power = acc
return (s + digit * power, power * 10)
state = (0, 1)
for digit in digits:
state = go(digit, state)
return state[0]
Such a simple function can carry all its state in its bare arguments. Carry around an accumulator argument, and the operation becomes trivial.
fromDigits :: [Int] -> Integer
fromDigits xs = fromDigitsA xs 0 # 0 is the current accumulator value
fromDigitsA [] acc = acc
fromDigitsA (x:xs) acc = fromDigitsA xs (acc * 10 + toInteger x)
If you're really determined to use a right fold for this, you can combine calculating length xs with the calculation like this (taking the liberty of defining fromDigits [] = 0):
fromDigits xn = let (x, _) = fromDigits' xn in x where
fromDigits' [] = (0, 0)
fromDigits' (x:xn) = (toInteger x * 10 ^ l + y, l + 1) where
(y, l) = fromDigits' xn
Now it should be obvious that this is equivalent to
fromDigits xn = fst $ foldr (\ x (y, l) -> (toInteger x * 10^l + y, l + 1)) (0, 0) xn
The pattern of adding an extra component or result to your accumulator, and discarding it once the fold returns, is a very general one when you're re-writing recursive functions using folds.
Having said that, a foldr with a function that is always strict in its second parameter is a really, really bad idea (excessive stack usage, maybe a stack overflow on long lists) and you really should write fromDigits as a foldl as some of the other answers have suggested.
If you want to "fold with state", probably Traversable is the abstraction you're looking for. One of the methods defined in Traversable class is
traverse :: Applicative f => (a -> f b) -> t a -> f (t b)
Basically, traverse takes a "stateful function" of type a -> f b and applies it to every function in the container t a, resulting in a container f (t b). Here, f can be State, and you can use traverse with function of type Int -> State Integer (). It would build an useless data structure (list of units in your case), but you can just discard it. Here's a solution to your problem using Traversable:
import Control.Monad.State
import Data.Traversable
sumDigits :: Traversable t => t Int -> Integer
sumDigits cont = snd $ runState (traverse action cont) 0
where action x = modify ((+ (fromIntegral x)) . (* 10))
test1 = sumDigits [1, 4, 5, 6]
However, if you really don't like building discarded data structure, you can just use Foldable with somewhat tricky Monoid implementation: store not only computed result, but also 10^n, where n is count of digits converted to this value. This additional information gives you an ability to combine two values:
import Data.Foldable
import Data.Monoid
data Digits = Digits
{ value :: Integer
, power :: Integer
}
instance Monoid Digits where
mempty = Digits 0 1
(Digits d1 p1) `mappend` (Digits d2 p2) =
Digits (d1 * p2 + d2) (p1 * p2)
sumDigitsF :: Foldable f => f Int -> Integer
sumDigitsF cont = value $ foldMap (\x -> Digits (fromIntegral x) 10) cont
test2 = sumDigitsF [0, 4, 5, 0, 3]
I'd stick with first implementation. Although it builds unnecessary data structure, it's shorter and simpler to understand (as far as a reader understands Traversable).
I'm trying to write a function in SML that takes in a list of ints and will output a list of ordered pairs of ints. The ordered pairs first int is the int that occurred in the input list and the second int in the ordered pair is the number of times it occurred in the input list. Also the list returned should be in ascending order according to the first int in the ordered pairs.
For example input list [1, 1, 1, 2, 3, 3, 5] would output as [(1,3), (2, 1), (3, 2), (5, 1)].
So far I have a function that uses foldl
UPDATED the code since original post.
fun turnIntoPairs l = foldl (fn (e, a) => if List.exists (fn (x, _) => x = e) a then x + 1 else a # [(e, 1)]) [] l;
I'm having trouble updating the list where I find the ordered pair that is already in the list - I want to add one to the second int in the ordered pair that was found while it's still in the list.
Any help would be greatly appreciated!
C:\Program Files (x86)\SMLNJ\\bin\.run\run.x86-win32.exe: Fatal error -- Uncaught exception Error with 0
raised at ../compiler/TopLevel/interact/evalloop.sml:66.19-66.27
[autoloading done]
C:\Users\Localadmin\Desktop\CS 671\Program 3\commonFactors.sml:1.87 Error: unbound variable or constructor: x
C:\Users\Localadmin\Desktop\CS 671\Program 3\commonFactors.sml:1.44-1.110 Error: types of if branches do not agree [literal]
then branch: int
else branch: (''Z * int) list
in expression:
if (List.exists (fn <pat> => <exp>)) a
then <errorvar> + 1
else a # (e,1) :: nil
[Finished in 0.5s with exit code 1]
Not really sure how to fix your current program, but you can solve this problem by splitting it in two: grouping equal elements and then ordering the list.
(* Groups successive equal elements into a tuples (value, count) *)
fun group (l as (x :: _)) =
let val (firstGroup, rest) = List.partition (fn y => x = y) l
in
(x, List.length firstGroup) :: (group rest)
end
| group [] = []
(* Now that we have our elements grouped, what's left is to order
them as required. *)
fun turnIntoPairs xs =
ListMergeSort.sort (fn ((x, _), (y, _)) => x >= y) (group xs)
Let's just look at the function you're passing to foldl:
(fn (e, a) => if List.exists (fn (x, _) => x = e) a then x + 1 else a # [(e, 1)])
The first problem (which the type-checker is complaining about) is that your if expression returns either x + 1, or a # [(e, 1)], which seems problematic on account of the former being a value of type int and the latter being of type (int * int) list.
Let's rewrite your code using some helper functions that I won't define and see if it gets clearer:
(fn (e, a) => if List.exists (fn (x, _) => x = e) a then increment a e else a # [(e, 1)])
Where increment has the type (int * int) list -> int -> (int * int) list.
Can you implement increment?
Like Gian, I would prefer to divide this into two functions: One that folds and one helper function that inserts. Incidentally, the insert function would take an element and an existing (int * int) list just as the accumulator function that fold accepts these two arguments.
Normally I would write an insert function curried (i.e. insert x xs) but if I write it uncurried (i.e. insert (x, xs)), I can pass it directly to foldl:
fun insert (x, []) = [(x,1)]
| insert (x, ((y,c)::xs)) =
if x = y then (y,c+1)::xs else (y,c)::insert (x, xs)
fun turnIntoPairs xs = foldl insert [] xs
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.
I'm very new to SML and I am trying a list exercise. The goal is sum up the previous numbers of a list and create a new list. For example, an input list [1, 4, 6, 9] would return [1, 5, 11, 20].
This is my solution so far, but I think the issue is with how I'm defining the function.
fun rec sum:int list -> int list =
if tl(list) = nil then
hd(list)
else
hd :: sum((hd(tail) + hd(tl(list)))::tl(tl(list)));
Besides that you are using rec as a function name, then you have some minor issues to work on.
The explicit type annotation you have made is treated as an annotation of the function result.
Thus, according to what you have written, then it should return a function and not the expected
list. This is clearly seen from the below example:
- fun rec_ sum : int list -> int list = raise Domain;
val rec_ = fn : 'a -> int list -> int list
Your should be careful of using the head and tail functions, when you don't do any checks on the
number of elements in the list. This could be done with either the length function, or (even
easier and often better) by pattern matching the number of elements.
Your code contains sum as a function call and tail as an variable. The variable tail has never
been defined, and using sum as a function call, makes me believe that you are actually using rec
as a keyword, but don't know what it means.
The keyword rec is used, when defining functions using the val keyword. In this case, rec is
needed to be able to define recursive functions (not a big surprise). In reality, the keyword fun
is syntactic sugar (a derived form) of val rec.
The following 3 are examples of how it could have been made:
The first is a simple, straight forward solution.
fun sumList1 (x::y::xs) = x :: sumList1 (x+y::xs)
| sumList1 xs = xs
This second example, uses a helper function, with an added argument (an accumulator). The list is constructed in the reverse order, to avoid using the slow append (#) operator. Thus we reverse the list before returning it:
fun sumList2 xs =
let
fun sumList' [] acc = rev acc
| sumList' [x] acc = rev (x::acc)
| sumList' (x :: y :: xs) acc = sumList' (y+x :: xs) (x :: acc)
in
sumList' xs []
end
The last example, show how small and easy it can be, if you use the standard list functions. Here the fold left is used, to go through all elements. Again note that the list is constructed in the reverse order, thus it is reversed as the last step:
fun sumList3 [] = []
| sumList3 (x::xs) = rev (foldl (fn (a, b) => hd b + a :: b) [x] xs)
try this -
fun recList ([], index, sum) = []
| recList (li, index, sum) =
if index=0 then
hd li :: recList (tl li, index+1, hd li)
else
sum + hd li :: recList (tl li, index+1, sum + hd li)
fun recSum li = recList (li, 0, 0)
In your case -
recSum([1,4,6,9]) ;
will give
val it = [1,5,11,20] : int list
also don't use rec as fun name -it keyword .