I need to write a function that takes a list of strings and finds the largest string in the list. The catch is it needs to iterate through the list using List.foldl and cannot use recursive calls except for those in the library function of List,foldl.
I wrote
fun longest_string1(xs)=
case xs of
[] => ""
| x::xs' => List.foldl((fn (s,x) => if String.size s > String.size x then s else x) "" x,)
with my interpretation being as follows:
-take in xs, if xs is empty return an empty string
-otherwise for the first item of xs call List.foldl
-List.foldl passes in an anonymous function that checks the length of s, which should represent the accumulator against the head item of the list. -Set the initial accumulator to be the empty string and the initial compare value to be the head of the initial list passed in by the higher order function
However, it does not type check.
I think my issue is in the understanding of the List.foldl function itself and how exactly it reads in its parameters. Can someone please provide some clarification?
So, for the code you posted:
You don't need the case for the empty list. foldl will take care of that for you. Just pass xs to foldl instead of x.
foldl is curried, so you shouldn't have parentheses around the parameters.
Other than that, it actually looks correct. Anyway, if you're still not sure how foldl works, here's a really long and thorough explanation ;)
Okay, let's start with List.foldl.
val foldl : ('a * 'b -> 'b) -> 'b -> 'a list -> 'b
So, there are three parameters. One is a function which we'll worry about later, the second is a value of the same type as the return type, and the last is the list.
So, let's take a simple example - say we have a list of ints, and we want to sum all the numbers. We could do this:
fun sum [] = 0
| sum (x::xs) = x + sum xs
Or, we can use foldl (I'll write foldl instead of List.foldl from now, because I'm lazy).
So, we know that the list is the third parameter. The second should be some sort of starting value, or accumulator, something that would make sense if the list was empty. For a sum, that would be 0.
The first parameter is a function, and this is the tricky part. The type is:
fn : 'a * 'b -> 'b
Okay, so 'a is also the type of the elements in the list, so it makes sense if this is an item from the list. 'b is the type of the starting value and the return value.
What actually happens is that foldl calls the function with the first element in the list, and the accumulator. It then calls itself with the result as the new accumulator, and the rest of the list. So if we do this:
foldl foo 0 [1,2,3]
It'll do
foo (1,0)
And then
foldl foo (foo (1,0)) [2,3]
And so on.
So for summing a list, we'll make the following function:
fn (x,acc) => x + acc
So we can do this:
fun sum xs = foldl (fn (x,acc) => x + acc) 0 xs
Or, even simpler
val sum = foldl op+ 0
(op+ does the same as the anonymous function I used earlier)
Let's walk through it with the list [1,2,3]
foldl op+ 0 [1,2,3]
foldl op+ (op+ (1,0)) [2,3] -> foldl op+ 1 [2,3]
foldl op+ (op+ (2,1)) [3] -> foldl op+ 3 [3]
foldl op+ (op+ (3,3)) [] -> foldl op+ 6 []
6
Related
I need to write a function that takes a list of strings and finds the largest string in the list. In case of tie, it should return string closest to the end of the list. The catch is it needs to iterate through the list using List.foldl and cannot use recursive calls except for those in the library function of List,foldl.
The following code works fine.
fun longest_string2 str_list =
List.foldl(fn (x, acc) => if String.size x >= String.size acc then x else acc) "” str_list
If I run longest_string2 ["Apple", "ball", "Bpple”]; in REPL, I get val it = "Bpple" : string
However, if I reverse the arguments of anonymous function as below I get val it = "Apple" : string.
Since the anonymous function is accessing the elements by name and not position, why does this makes difference?
The definition of List.foldl is
fun foldl (f: 'a*'b->'b) (acc: 'b) (l: 'a list): 'b =
case l of
[] => acc
| x::xs => foldl f (f(x,acc)) xs
If you reverse arguments of the anonymous function, your function becomes the following: (correct me if I misinterpreted your question)
fun longest_string2 str_list =
List.foldl(fn (acc, x) => if String.size x >= String.size acc then x else acc) "” str_list
If your pass ["Apple", "ball", "Bpple”] as argument to longest_string2 now, the foldl function will pattern match your list with x::xs, where x is “Apple” and xs is ["ball", "Bpple”]. When you compute the updated accumulator using f(x, acc), x and acc gets swapped. In other words, in your anonymous function (with reversed arguments), you would be assuming that the first argument would be ”” and the second argument would be Apple but the implementation of List.foldl will pass f(“Apple”, “”). In this case, your anonymous function will label “Apple” as “acc” and “” as “x”.
#3123 has most answered the question, but has not directly addressed this statement in the question.
Since the anonymous function is accessing the elements by name and not
position, why does this makes difference?
foldl takes a function which takes a tuple as its argument, which is positional.
We could define a fold function which takes a record as its argument instead if we really wanted to achieve this:
fun namedFold _ acc [] = acc
| namedFold f acc (x::xs) =
namedFold f (f {acc=acc, x=x}) xs;
And then call it as:
namedFold (fn { acc, x } => acc + x) 0 [1,2,3,4]
Or as:
namedFold (fn { x, acc } => acc + x) 0 [1,2,3,4]
And get the same result.
But then the type of namedFold is fn :({acc: 'a, x: 'b} -> 'a) -> 'a -> 'b list -> 'a and it becomes essentially impossible to easily pass existing functions to it. With foldl defined the way it is, we can easily rewrite that previous called to namedFold as:
foldl op+ 0 [1,2,3,4]
Because op+ has the same signature as the first argument to foldl.
I have this list of tuples
[(4,'a'), (1,'b'), (2,'c'), (2,'a'), (1,'d'), (4,'e')]
I want to get the first elements of every tuple then replicate it to make the following: "aaaabccaadeeee"
I came up with this code, but it only gives me the replicate of the first tuple.
replicate (fst ( head [(4,'a'), (1,'b')])) ( snd ( head [(4,'a'), (1,'b')]))
--output is: "aaaa"
I was thinking to use map for to get the replicate of every tuple, but I didn't succeed.
Since you already know how to find the correct answer for a single element, all you need is a little recursion
func :: [(Int, a)] -> [a]
func [] = []
func ((n, elem):rest) = (replicate n elem) ++ (func rest)
Mapping the values should also work. You just need to concatenate the resulting strings into one.
func :: [(Int, a)] -> [a]
func xs = concat $ map func2 xs where
func2 (n, elem) = replicate n elem
Or, if you are familiar with currying:
func :: [(Int, a)] -> [a]
func xs = concat $ map (uncurry replicate) xs
Finally, if you are comfortable using function composition, the definition becomes:
func :: [(Int, a)] -> [a]
func = concat . map (uncurry replicate)
Using concat and map is so common, there is a function to do just that. It's concatMap.
func :: [(Int, a)] -> [a]
func = concatMap (uncurry replicate)
Let
ls = [(4,'a'), (1,'b'), (2,'c'), (2,'a'), (1,'d'), (4,'e')]
in
concat [replicate i x | (i, x) <- ls]
will give
"aaaabccaadeeee"
The point-free version
concat . map (uncurry replicate)
You are correct about trying to use map. But first lets see why your code did not work
replicate (fst ( head [(4,'a'), (1,'b')])) ( snd ( head [(4,'a'), (1,'b')]))
Your first parameter to replicate is the head of your list which is (4, 'a'). Then you are calling fst on this, thus the first parameter is 4. Same things happens with second parameter and you get 'a'. The result of which you see.
Before using map lets try to do this with recursion. You want to take one element of list and apply replicate to it and then combine it with the result of applying replicate on the second element.
generate [] = []
generate (x:xs) = replicate (fst x) (snd x) ++ generate xs
Do note I am using pattern matching to get the first element of list. You can us the pattern matching to get the element inside the tuple as well, and then you would not need to use the fst/snd functions. Also note I am using pattern matching to define the base case of empty list.
generate [] = []
generate ((x,y):xs) = replicate x y ++ generate xs
Now coming to map, so map will apply your function to every element of the list, here's the first try
generate (x,y) = replicate x y
map generate xs
The result of the above will be slightly different from recursion. Think about it, map is going to apply generate to every element and store the result in a list. generate creates a list. So when you apply map you are creating a list of list. You can use concat to flatten it if you want, which will give you the same result as recursion.
Last thing, if you can use recursion, then you can use fold as well. Fold will just apply a function to every element of the list and return the accumulated results (broadly speaking).
--first parameter is the function to apply, second is the accumulator, third is your list
foldr step [] xs
where step (x,y) acc =
(replicate x y) ++ acc
Again here I have used pattern matching in the function step to extract the elements of the tuple out.
i'm new in the haskell world and i'd like to know how to insert a value in each position of a list in haskell, and return a lists of sublists containing the value in each position. For example:
insert' :: a -> [a] -> [[a]]
insert' a [] = [[a]]
insert' a list = ??
To get something like:
insert' 7 [1,2,3] = [[7,1,2,3],[1,7,2,3],[1,2,7,3],[1,2,3,7]]
insert' :: a -> [a] -> [[a]]
insert' y [] = [[y]]
insert' y xss#(x:xs) = (y : xss) : map (x :) (insert' y xs)
While the empty list case comes natural, let's take a look at insert' y xss#(x:xs). We essentially have two cases we need to cover:
y appears in front of x. Then we can just use y : xss.
y appears somewhere after x. We therefore just insert it in the rest of our list and make sure that x is the first element with map (x:).
Although #delta's answer is definitely more elegant, here a solution with difference lists. If we insert an element x on every location of list ys = [y1,y2,...,yn], the first time we will insert it as head, so that means we can construct x : ys.
. For the second element of the resulting list, we want to construct a list [y1,x,y2,...,yn]. We can do this like y1 : x : y2s. The next lists will all have a structure y1 : ....
The question is: how can we write a recursive structure that keeps track of the fact that we want to put elements in the head. We can use a function for that: we start with a function id. If we now call id (x:ys) then we will of course generate the list (x:ys).
We can however, based on the id function, construct a new function id2 = \z -> id (y1:z). This function will thus put y1 in the head of the list and then add the list with which we call id2 as tail. Next we can construct id3 = \z -> id2 (y2:z). This will put y1 and y2 as first elements followed by the tail z.
So we can put this into the following recursive format:
insert' :: a -> [a] -> [[a]]
insert' x = go id
where go d [] = [d [x]]
go d ys#(yh:yt) = (d (x : ys)) : go (d . (yh :)) yt
So we redirect insert' to go where the initial difference list is simply the id function. Each time we check if we have reached the end of the given list. If that is the case, we return the basecase: we call [x] (as tail) on the difference list, and thus construct a list where we append x as last element.
In case we have not yet reached the last element, we will first emit d (x : ys): we prepend x to the list and provide this as argument to the difference list d. d will prepend y1 : y2 : ... : yk up to the point where we insert x. Furthermore we call recursively go (d . (yh :)) yt on the tail of the list: we thus construct a new difference list, wehere we insert (yh :) as tail of the list. We thus produce a new function with one argument: the tail after the yh element.
This function produces the expected results:
*Main> insert' 4 []
[[4]]
*Main> insert' 4 [1,2,5]
[[4,1,2,5],[1,4,2,5],[1,2,4,5],[1,2,5,4]]
*Main> insert' 7 [1,2,3]
[[7,1,2,3],[1,7,2,3],[1,2,7,3],[1,2,3,7]]
You may also do as follows;
import Data.List
spread :: a -> [a] -> [[a]]
spread x xs = zipWith (++) (inits xs) ((x:) <$> tails xs)
*Main> spread 7 [1,2,3]
[[7,1,2,3],[1,7,2,3],[1,2,7,3],[1,2,3,7]]
*Main> spread 7 []
[[7]]
So this is about three stages.
(x:) <$> tails xs is all about applying the (x:) function to all elements of tails xs function. So tails [1,2,3] would return [[1,2,3],[2,3],[3],[]] and we are to apply an fmap which is designated by <$> in the inline form. This is going to be the third argument of the zipWith function.
(inits xs) which would return [[],[1],[1,2],[1,2,3]], is going to be the second argument to zipWith.
zipWith (++) is obviously will zip two list of lists by concatenating the list elements.
So we may also express the same functionality with applicative function functors as follows;
spread :: a -> [a] -> [[a]]
spread x = zipWith (++) <$> inits <*> fmap (x:) . tails
In this case we fmap the zipWith (++) function with type [[a]] -> [[a]] -> [[a]] over inits and then apply it over to fmap (x:) . tails.
It could get more pointfree but becomes more complicated to read through (at least for me). In my opinion this is as best as it gets.
I'm using SML recently and I'm trying to solve a problem.
I should create a function that accept an int list and return even int list,
I've already created it :
fun evens [] = [] |
evens [x] = [x] |
evens(x::xs) =
case x mod 2 of
0 => x::evens xs |
1 => evens xs;
which gives the correct result but I need to use foldr
this is what I came up with:
fun evens [] = [] |
evens(x::xs) =
case x mod 2 of
0 => foldr (op ::) evens xs [x] |
1 => evens xs;
but it is not working, I'm still confused with how to use foldr in this case.
any advice?
First of all, with foldr you should be looking for a 1-line definition rather than a recursive definition using patterns and cases. The point of foldr is that it incorporates a common recursion pattern -- you just need to use the pattern.
The type of foldr is
fn : ('a * 'b -> 'b) -> 'b -> 'a list -> 'b
In your case 'b is int list and 'a is int. The 'b between the arrows in the middle of foldr's type is the seed value. It typically corresponds to a basis value. When you are constructing lists this basis value is typically []. Thus -- you need to concentrate on the key question of what should be folded over the list. In other words -- what function of type ('a * 'b -> 'b) should you pass to foldr? In your case you need to pass a function of type
int * int list -> int list
this should be a function which, when given an int and an int list either tacks the int onto the list (if it is even) or leaves the list alone. You could define this function ahead of time, define it using let, or just use an anonymous function.
Under the assumption that this is homework, I don't want to give a complete answer, but here is a function which uses foldr to obtain the positive entries in a list:
fun positives xs =
foldr (fn (x,xs) => if x >= 0 then x::xs else xs) [] xs;
-
- positives [3,~2,4,5,0,~1,~1,5];
val it = [3,4,5,0,5] : int list
I have been working with Haskell for a little over a week now so I am practicing some functions that might be useful for something. I want to compare two lists recursively. When the first list appears in the second list, I simply want to return the index at where the list starts to match. The index would begin at 0. Here is an example of what I want to execute for clarification:
subList [1,2,3] [4,4,1,2,3,5,6]
the result should be 2
I have attempted to code it:
subList :: [a] -> [a] -> a
subList [] = []
subList (x:xs) = x + 1 (subList xs)
subList xs = [ y:zs | (y,ys) <- select xs, zs <- subList ys]
where select [] = []
select (x:xs) = x
I am receiving an "error on input" and I cannot figure out why my syntax is not working. Any suggestions?
Let's first look at the function signature. You want to take in two lists whose contents can be compared for equality and return an index like so
subList :: Eq a => [a] -> [a] -> Int
So now we go through pattern matching on the arguments. First off, when the second list is empty then there is nothing we can do, so we'll return -1 as an error condition
subList _ [] = -1
Then we look at the recursive step
subList as xxs#(x:xs)
| all (uncurry (==)) $ zip as xxs = 0
| otherwise = 1 + subList as xs
You should be familiar with the guard syntax I've used, although you may not be familiar with the # syntax. Essentially it means that xxs is just a sub-in for if we had used (x:xs).
You may not be familiar with all, uncurry, and possibly zip so let me elaborate on those more. zip has the function signature zip :: [a] -> [b] -> [(a,b)], so it takes two lists and pairs up their elements (and if one list is longer than the other, it just chops off the excess). uncurry is weird so lets just look at (uncurry (==)), its signature is (uncurry (==)) :: Eq a => (a, a) -> Bool, it essentially checks if both the first and second element in the pair are equal. Finally, all will walk over the list and see if the first and second of each pair is equal and return true if that is the case.