My basic idea is to implement a bubble sort of type ('a list -> 'a list). I use variables which are sorted and result. If I change some of elements in the list, sorted becomes 1. Otherwise, sorted remains 0. Result is one cycle of the comparison.
I think there is something wrong with my sorted variable. Can anyone figure out what the problem is?
let rec sort (l: int list) : int list =
let sorted=0 in
let result = match l with
| []->[]
| x::xs-> if xs=[] then x
else let y::ys = xs in
if x<y then x::sort(xs)
else let sorted=1 in
y::sort(x::ys)
in
if sorted=0 then result
else sort(result)
It seems to me you're trying to use sorted as a mutable variable. OCaml variables are immutable. Once you bind a variable to a value, the binding can't be changed. Each of your let sorted = statements defines a new variable named sorted. So your last test will always show sorted to be equal to 0. It is testing the first definition of sorted, which can never have any other value than 0.
As Jeffrey said, OCaml values are immutable. That is why your program does not work as expected.
But there are other problems with your code:
sorted should have type bool, not int. One the pros of OCaml is that it has a strong type system, so use it.
To deconstruct list, you should use pattern matching only if xs=[] then x else let y::ys = xs in is not not the good way of doing it (and OCaml should warn you that your pattern matching is not exhaustive). You should add other cases to your pattern matching instead.
Like this:
| [] -> []
| x::[] -> x
| x::y::ys -> ...
Related
I am trying to figure out is there anyway that i can use append to make the three lists of integer inside a list to become a list of a list of integers, for example
[[1];[2];[3]] -> [[1;2;3]]
[] -> [[]]
[[]] -> []
but i am not sure how loop really in OCaml.
and the below is what i have tried, but i dont think it work
let rec ls (l : 'a list list) =
match l with
| [] -> []
| x :: y -> l#y
i have tried to use # to do function, but i don't how to remove the bracket.
Note that in your attempt, you never use x which is the head of the list, and the function is not recursive. It never calls itself. Note that # is never necessary in this exercise, which is good because it leads to some ugly performance implications.
Consider that you can use pattern-matching to identify whether a list is empty or not, and to extract elements from the head and the tail of a list. What should the result of flattening an empty list be? An empty list.
let rec flatten =
function
| [] -> []
Now, if the first list in the list of lists is empty, it should be the result of flattening the tail. This seems pretty obvious so far.
let rec flatten =
function
| [] -> []
| []::tl -> flatten tl
Now, if it's not empty then we can cons the first element of the first list onto the result of flattening... I'll leave that as an exercise for you to fill in.
let rec flatten =
function
| [] -> []
| []::tl -> flatten tl
| (x::xs)::tl -> x :: flatten ...
Looping via recursion
While OCaml does have imperative loops, it is much more idiomatic, especially when dealing with lists, to loop via recursion.
In order to use recursion to loop, there must be at least one exit case where the function does not recursively call itself, but there must also be at least one case where it does, and that function call must in some way update the state being passed in so that it converges on the exit case.
If the exit case is passing in an empty list, the recursive calls must get closer to passing in an empty list on each call or the recursion will never end.
If you did want to append...
If you decided you do like #, and don't care about O(n^2) runtime complexity, you can use it with List.fold_left to readily accomplish this goal.
# List.fold_left (#) [] [[1;2]; [3;4]];;
- : int list = [1; 2; 3; 4]
This is equivalent to [] # [1;2] # [3;4].
I've written a function which search through a list of int-list to return the index of the list with an specific length by using pattern-matching:
let rec search x lst i = match lst with
| [] -> raise(Failure "Not found")
| hd :: tl -> if (List.length hd = x) then i else search x tl (i+1)
;;
For example:
utop # search 2 [ [1;2];[1;2;3] ] 0 ;;
- : int = 0
Is there a way to write a function with the same functionality using fold.left ?
What does List.fold_left actually do?
It takes (in reverse order to the order of arguments) a list, an initial value, and a function that works on that initial value and the first element in the list. If the list is empty, it returns the initial value. Otherwise it uses the function to update the initial value by way of recursion and works on the tail of the list.
let rec fold_left f init lst =
match lst with
| [] -> init
| x::xs -> fold_left f (f init x) xs
Now, what information do you need to keep track of as you iterate? The index. Easy enough.
But, what if you don't actually find a list of that length? You need to keep track of whether you've found one. So let's say we use a tuple of the index and a boolean flag.
Your function you pass to fold_left just needs to determine if a match has been found no update is necessary. Essentially we just no-op over the rest of the list. But, if we haven't found a match, then we need to test the current sublist's length and update the init value accordingly.
#glennsl (in a comment) and #Chris already explained that you may use List.fold_left but that it’s not the right tool for the job, because it processes the whole list whereas you want to stop once an occurrence is found. There are solutions but they are not satisfying:
(#Chris’ solution:) use a folding function that ignores the new elements once an occurrence has been found: you’re just wasting time, walking through the remaining tail for nothing;
evade the loop by throwing and catching an exception: better but hacky, you’re working around the normal functioning of List.fold_left.
I just mention that there is a generic function in the standard library that matches your situation almost perfectly:
val find : ('a -> bool) -> 'a list -> 'a
find f l returns the first element of the list l that satisfies the predicate f.
Raises Not_found if there is no value that satisfies f in the list l.
However it does not return the index, unlike what you are asking for. This is a deliberate design choice in the standard library, because list indexing is inefficient (linear time) and you shouldn’t do it. If, after these cautionary words, you still want the index, it is easy to write a generic function find_with_index.
Another remark on your code: you can avoid computing the lengths of inner lists fully, thanks to the following standard function:
val compare_length_with : 'a list -> int -> int
Compare the length of a list to an integer. compare_length_with l len is equivalent to compare (length l) len, except that the computation stops after at most len iterations on the list.
Since 4.05.0
So instead of if List.length hd = x, you can do if List.compare_length_with hd x = 0.
Here's what I've got so far...
fun positive l1 = positive(l1,[],[])
| positive (l1, p, n) =
if hd(l1) < 0
then positive(tl(l1), p, n # [hd(l1])
else if hd(l1) >= 0
then positive(tl(l1), p # [hd(l1)], n)
else if null (h1(l1))
then p
Yes, this is for my educational purposes. I'm taking an ML class in college and we had to write a program that would return the biggest integer in a list and I want to go above and beyond that to see if I can remove the positives from it as well.
Also, if possible, can anyone point me to a decent ML book or primer? Our class text doesn't explain things well at all.
You fail to mention that your code doesn't type.
Your first function clause just has the variable l1, which is used in the recursive. However here it is used as the first element of the triple, which is given as the argument. This doesn't really go hand in hand with the Hindley–Milner type system that SML uses. This is perhaps better seen by the following informal thoughts:
Lets start by assuming that l1 has the type 'a, and thus the function must take arguments of that type and return something unknown 'a -> .... However on the right hand side you create an argument (l1, [], []) which must have the type 'a * 'b list * 'c list. But since it is passed as an argument to the function, that must also mean that 'a is equal to 'a * 'b list * 'c list, which clearly is not the case.
Clearly this was not your original intent. It seems that your intent was to have a function that takes an list as argument, and then at the same time have a recursive helper function, which takes two extra accumulation arguments, namely a list of positive and negative numbers in the original list.
To do this, you at least need to give your helper function another name, such that its definition won't rebind the definition of the original function.
Then you have some options, as to which scope this helper function should be in. In general if it doesn't make any sense to be calling this helper function other than from the "main" function, then it should not be places in a scope outside the "main" function. This can be done using a let binding like this:
fun positive xs =
let
fun positive' ys p n = ...
in
positive' xs [] []
end
This way the helper function positives' can't be called outside of the positive function.
With this take care of there are some more issues with your original code.
Since you are only returning the list of positive integers, there is no need to keep track of the
negative ones.
You should be using pattern matching to decompose the list elements. This way you eliminate the
use of taking the head and tail of the list, and also the need to verify whether there actually is
a head and tail in the list.
fun foo [] = ... (* input list is empty *)
| foo (x::xs) = ... (* x is now the head, and xs is the tail *)
You should not use the append operator (#), whenever you can avoid it (which you always can).
The problem is that it has a terrible running time when you have a huge list on the left hand
side and a small list on the right hand side (which is often the case for the right hand side, as
it is mostly used to append a single element). Thus it should in general be considered bad
practice to use it.
However there exists a very simple solution to this, which is to always concatenate the element
in front of the list (constructing the list in reverse order), and then just reversing the list
when returning it as the last thing (making it in expected order):
fun foo [] acc = rev acc
| foo (x::xs) acc = foo xs (x::acc)
Given these small notes, we end up with a function that looks something like this
fun positive xs =
let
fun positive' [] p = rev p
| positive' (y::ys) p =
if y < 0 then
positive' ys p
else
positive' ys (y :: p)
in
positive' xs []
end
Have you learned about List.filter? It might be appropriate here - it takes a function (which is a predicate) of type 'a -> bool and a list of type 'a list, and returns a list consisting of only the elements for which the predicate evaluates to true. For example:
List.filter (fn x => Real.>= (x, 0.0)) [1.0, 4.5, ~3.4, 42.0, ~9.0]
Your existing code won't work because you're comparing to integers using the intversion of <. The code hd(l1) < 0 will work over a list of int, not a list of real. Numeric literals are not automatically coerced by Standard ML. One must explicitly write 0.0, and use Real.< (hd(l1), 0.0) for your test.
If you don't want to use filter from the standard library, you could consider how one might implement filter yourself. Here's one way:
fun filter f [] = []
| filter f (h::t) =
if f h
then h :: filter f t
else filter f t
I have a list of lists like so:
[["BBBBBBBB",
"BWFFFPFGB",
"BWFFFPFGB",
"BWFFMPFGB",
"BWFFFPF_B",
"BWFFFPF6B",
"BBBBBBB"]]
I've done a little research and have found out how to access individual elements using the !! operator. But when it comes to searching for a certain element 'M' I'm not sure how to go about that. My friend said I need to use something like (x:xs):xss on a list, but when I try this in the WinGHCi haskell program I get this.
Prelude> let list = [["BBBBBBBB",
"BWFFFPFGB",
"BWFFFPFGB",
"BWFFMPFGB",
"BWFFFPF_B",
"BWFFFPF6B",
"BBBBBBB"]]
Prelude> head(x:xs):xss
<interactive>:192:2: Not in scope: `x'
<interactive>:192:4: Not in scope: `xs'
<interactive>:192:8: Not in scope: `xss'
I understand that I declare the name as list and not x:xs but even when I declare it as x:xs I still get the errors. I'm probably still a little new to haskell to really understand what to do so I may be going about this way wrong.
I've looked here Replace individual list elements in Haskell? because eventually I want to replace the M with something different but I'm not completely sure how I would implement that.
Any help/guidance is appreciated, thanks!
First let's see how to replace a W with M
charWM :: Char -> Char
charWM 'W' = 'M' -- If you see W, put M.
charWM x = x -- If you see anything else, put it back as is.
You can rewrite that function how you like by adding other letter transformations.
Now let's make that work over a list. There's a great function map :: (a ->b) -> [a] -> [b] that lets you apply a function on every element on a list.
stringWM :: String -> String
stringWM xs = map charWM xs -- do charWM to everything in xs.
For example stringWM "QWERTY WILL WIN" = "QMERTY MILL MIN"
Next we can do that to a list of lists:
lolWM :: [String] -> [String]
lolWM xss = map stringWM xss
(String is a type synonym for [Char].)
Let's test that out in ghci:
*Main> list'
["BBBBBBBB","BWFFFPFGB","BWFFFPFGB","BWFFMPFGB","BWFFFPF_B","BWFFFPF6B","BBBBBBB"]
*Main> lolWM list'
["BBBBBBBB","BMFFFPFGB","BMFFFPFGB","BMFFMPFGB","BMFFFPF_B","BMFFFPF6B","BBBBBBB"]
All good.
Your example wasn't exactly list', it was [list'] which has 1 element, so to work on that we'd need to map lolWM. Often we wouldn't bother writing stringWM or lolWM and go directly to lists of lists of lists, if that's what we needed:
lololWM = (map.map.map) charWM
map.map.map means map the map of the map. You can allow that to blow your mind a little, or you can just say list of list of list of Char, so map map map - one map per list level.
In the future, maybe you'll want to replace W with Strings instead of characters.
rewriteChar :: Char -> String
rewriteChar 'W' = "--^--"
rewriteChar x = [x] -- put x in a list to make it a string
This time, map isn't enough: map rewriteChar "QWERTY WILL WIN" gives
["Q","--^--","E","R","T","Y"," ","--^--","I","L","L"," ","--^--","I","N"]
We could use concat on that to flatten it into a single list, but it's more fun to do
rewriteString = concatMap rewriteChar
So now rewriteString "QWERTY WILL WIN" give us "Q--^--ERTY --^--ILL --^--IN".
For more mindblowing things to try, there's "QWERTY WILL WIN" >>= rewriteChar and "Hello Mum" >>= \x -> [x,x,x]
First of all, virtually all "variables" in Haskell are immutable, so there's no "changing a list", there are modified copies.
Second, you need to find an element by some criteria. To do that, you need to traverse a list. - This can be done using recursion. Filtering can be done using a function passed as an argument of your traversing function (this function must take an element and return a boolean value).
Try to put the above together and make your own function. Start with a type signature, it shows what you want to do: to take a list of Char (it's better to generalize to a generic type) and a function which possibly changes an element and return a modified list:
replaceFunc :: (Char -> Char) -> String -> String
Also, read http://www.haskell.org/haskellwiki/How_to_work_on_lists , there's a hint there how to apply some function to specific elements only.
I´m a newbie in SML and I´d like to update my function so that it has two outputs: a list AND 1 or 0. The function was proposed here: SML: Remove the entry from the List. It returns an updated list without a row that contains ´elem´.
fun removeElem elem myList = filter (fn x => x <> elem) myList
The function should return a new list AND 1, if a raw has been deleted. Otherwise, it should return an old list AND 0.
Any hint or example is highly appreciated. Thanks.
Note that all SML functions take a single input and return a single output. Instead, think of returning a tuple containing the new list and a flag indicating whether any elements were removed. One possibility is to use a couple of functions from the standard basis to test whether elem is in myList and build up a tuple consisting of that and the results from the filter shown in the question. The test might look like:
Option.isSome (List.find (fn x => x = elem) myList)
There are more concise ways to write that, but it shows the idea. Note that it returns a bool instead of an int; this is more precise, so I won't convert to the integers requested in the question.
A drawback of the above is that it requires traversing the list twice. To avoid that, consider the type that the function must return: a tuple of a list without elem and a flag showing whether any elems have been removed. We can then write a function that take a new value and a (valid) tuple, and returns a valid tuple. One possibility:
fun update(x, (acc, flag)) = if x = elem then (acc, true) else (x :: acc, flag)
We can then apply update to each element of myList one-by-one. Since we want the order of the list to stay the same, apart from the removed elements, we should work through myList from right to left, accumulating the results into an initially empty list. The function foldr will do this directly:
foldr update ([], false) myList
However, there is a lot of logic hidden in the foldr higher-order function.
To use this as a learning exercise, I'd suggest using this problem to implement the function in a few ways:
as a recursive function
as a tail-recursive function
using the higher order functions foldl and foldr
Understanding the differences between these versions will shed a lot of light on how SML works. For each version, let the types guide you.
As has been stated in some of your previous questions; Returning 0 or 1 as an indicator for what happened is a really bad design, as you don't get any guarantees from the types, whether or not you will get -42 as the result. Since you are working with a strongly typed language, you might as well use this to your advantage:
The most obvious thing to do instead would be to return a boolean, as that is actually what you are emulating with 0 and 1. In this case you could return the pair (true, modified_list) or (false, original_list).
Since you want to associate some data with the result, there is another -- perhaps, for some, less -- obvious thing to do; Return the result as an option, indication a change in the list as SOME modified_list and indication no change as NONE.
In either case you would have to "remember" whether or not you actually removed any elements from the original list, and thus you can't use the filter function. Instead you would have to do this for yourself using somewhat the same code as you originally posted.
One way would be like this
fun removeElem _ [] = (false, [])
| removeElem elem (x::xs) =
let
val (b, xs') = removeElem elem xs
in
if elem = x then
(true, xs')
else
(b, x::xs')
end
Another way would be to use a accumulator parameter to store the resulting list
fun removeElem elem xs =
let
fun removeElem' [] true res = SOME (rev res)
| removeElem' [] false _ = NONE
| removeElem' (x::xs) b res =
if elem = x then
removeElem' xs true res
else
removeElem' xs b (x::res)
in
removeElem' xs false []
end
Since the solution is being built in the reverse order, we reverse the result just before we return it. This makes sure that we don't have to use the costly append operation when adding elements to the result list: res # [x]