I´m trying to write a function that produces a new list containing the given list without the element x.
Moscow ML says that some cases are unused in this match.
fun delete (x,list) = delete(x,[])
|delete(x,(first::rest)) = if first = x then delete(x,rest) else first::delete(x,rest)
Here's how I'd do it on Standard ML::
fun delete (item, list) =
case list of
[]=>[]
| xs::ys => if item = xs then delete(item,ys)
else xs::delete(item,ys)
Without using cases:
fun delete (item, list) = List.filter(fn x => x <> item) list
Nevermind the polyequal signs.
When a call to delete is performed, the patterns defining the function are (essentially) tried in order. Since the first pattern already matches every list, the second pattern will never be reached. That's why the compiler complains.
In other words, you either have to reorder your cases, or better, make them disjoint (e.g. by replacing list with [] in the first case).
Extra hint: The right-hand side of the first case also seems wrong. This will always go into an infinite recursion.
That's because your first case matches any list, so the second case will never be used.
Remember that the case are tried in the order they're written, not selected based on which is the "best" match.
You also have a slight problem of infinite recursion with
delete (x,list) = delete(x,[])
Since list will match [], it will just recurse forever.
If you delete something from an empty list, the result should be the empty list.
You can do one of two things.
Either move the non-empty case first:
fun delete (x, y:ys) = if y = x then delete(x, ys) else y::delete(x, ys)
| delete (x, list) = []
Or, more common, make the first case only match the empty list:
fun delete (x, []) = []
| delete (x, y:ys) = if y = x then delete(x, ys) else y::delete(x, ys)
Related
I am trying to insert a number x into a sorted list l using Ocaml's List.fold_right and return the list with the inserted element. I have figured out a way to insert it if the element is to go at the front of the list or in the middle of the list, however I cannot figure out how to code the case where the element is larger than every element in the list and thus must go at the end.
Here is what I have so far:
let insert_number (x: int) (l: int list): int list =
List.fold_right l ~f:(
fun cur -> fun acc ->
if x < cur then cur::x::accum
else cur::accum
) ~init: []
Using this with a test case like:
insert_number (3) ([1; 2; 4]);;
- : int list = [1; 2; 3; 4]
gives the correct answer. However, with a test case like this:
insert_number (3) ([1; 2]);;
- : int list = [1; 2]
the number is not inserted because it should be added to the end of the list.
Could someone help me understand how I am supposed to integrate this case into the function used with List.fold_right.
A fold works by passing along a set of state as it iterates over each element in a list (or other foldable data structure). The function passed in takes both the current element and that state.
I think you're really really close, but you need as Jeffrey suggests a boolean flag to indicate whether or not the value has been inserted. This will prevent multiple insertions and if the flag is still false when the fold is done, we can detect that and add the value to insert.
This match also serves the purpose of giving us an opportunity to discard the no longer needed boolean flag.
let insert v lst =
match List.fold_right
(fun x (inserted, acc) ->
if v > x && not inserted then (true, x::v::acc)
else (inserted, x::acc))
lst
(false, []) with
| (true, lst) -> lst
| (_, lst) -> v::lst
One way to look at List.fold_right is that it looks at each element of the list in turn, but in reverse order. For each element it transforms the current accumulated result to a new one.
Thinking backward from the end of the list, what you want to do, in essence, is look for the first element of the list that's less than x, then insert x at that point.
So the core of the code might look something like this:
if element < x then element :: x :: accum else element :: accum
However, all the earlier elements of the list will also be less than x. So (it seems to me) you need to keep track of whether you've inserted x into the list or not. This makes the accumulated state a little more complicated.
I coded this up and it works for me after fixing up the case where x goes at the front of the list.
Perhaps there is a simpler way to get it to work, but I couldn't come up with one.
As I alluded to in a comment, it's possible to avoid the extra state and post-processing by always inserting the element and effectively doing a "local sort" of the last two elements:
let insert_number x l =
List.fold_right (
fun cur -> function
| [] when x > cur -> [cur; x]
| [] -> [x; cur]
| x::rest when x > cur -> cur::x::rest
| x::rest -> x::cur::rest
) l []
Also, since folding doesn't seem to actually be a requirement, here's a version using simple recursion instead, which I think is far more comprehensible:
let rec insert_number x = function
| [] -> [x]
| cur::rest when cur > x -> x::cur::rest
| cur::rest -> cur::insert_number x rest
I have a problem with writing a function which result is as shown below:
split([7;1;4;3;6;8;2], 4) = ([1;3;2;4], [7;6;8])
my current code i managed to write:
let split(list, number)=
let split1(list, number, lesser, greater)=
if list = [] then lesser::greater
else if List.hd list <= element then (List.hd list)::(lesser)
else (List.hd list)::(greater)
in
(List.tl lista, element, [], []);;
Thanks in advance for your help.
For the future, it helps to be more specific when asking a question on SO. What exactly is your problem? SO users will be skeptical of someone who wants others to help them, but won't help themselves.
Your code has nearly the correct structure, but there are a few errors in there that seem to be getting in your way.
Firstly lesser::greater is wrong, since the left hand side of a cons operator must be a list itself, but what you really want is a list where both of these are elements. So instead try [lesser;greater].
Secondly, if you think through your code, you will notice that it suddenly stops. You checked the first element, but you didn't look at the rest of the list. Since you want to keep splitting the list, you need your code to keep executing till the end of the list. To achieve this, we use recursion. Recursion mean that your function split1 will call itself again. It can be very confusing the first time you see it - each time split1 runs it will take the first element off, and then split the remainder of the list.
What does (List.hd list)::(lesser) actually mean? The lesser here really means all of the lesser elements in the rest of the list. You need to keep taking an element out of the list and putting it in either lesser or greater.
Finally avoid using List.hd excessively - it is neater to find the head and tail using pattern matching.
Here's a working version of the code:
let split(list, number)=
let rec split1(list, number, lesser, greater)=
match list with
| [] -> [List.rev lesser;List.rev greater]
| head::tail ->
match (head <= number) with
true -> split1(tail,number,head::lesser,greater)
| false -> split1(tail,number,lesser,head::greater)
in split1(list, number, [], []);;
split([1;2;3;4;5],3);;
The split1 function takes the elements off one at a time, and adds them to the lists.
Maybe my comments on the following code snippet would help:
let split list num =
let rec loop lesser greater list =
match list with
| [] -> (lesser, greater)
(* when your initial list is empty, you have to return the accumulators *)
| x :: xs ->
if x <= num then
(* x is lesser than num, so add x in the lesser list and
call loop on the tail of the list (xs) *)
else
(* x is greater than num, so add x in the greater list and
call loop on the tail of the list (xs) *)
in
(* here you make the first call to loop by initializing
your accumulators with empty list*)
loop [] [] list
I am looking for a way of walking through a list, grabbing an element (in that order its given I suppose), using it with another function, then returning to that list and continuing the operation WITHOUT losing that element from the list.
In examples I've seen the only way to accomplish this would be to do the following:
counter :: (Eq a1, Num a) => a1 -> [a1] -> a
counter a [] = 0
counter a [x] = if a == x then 1 else 0
counter a (x:xs) = if a == x then counter a xs + 1 else counter a xs
permut :: Eq a => [a] -> [a] -> Bool
permut [] [] = True
permut [x] [] = False
But this gets rid of the element x from the list xs when called again both within permut. I know that this also serves to end the recursive calls/terminate the function, but I need to have the whole list available in my counter function in order for it to work. I want to simply walk through the list and keep the whole thing intact. Is this possible?
EDIT: Updated use case. I am trying to work on checking if one list is a permutation of another. My thought process is within each list two properties will hold if they are a permutation:
They will have the same number of total elements (will implement this piece later)
They will have the same number of each element
Right now my counter function works, but I am losing elements as I iterate recursively through the permut function. I don't care about efficiency, I don't mind counting the same element again and comparing it if the number shows up in the list multiple times.
Your pattern match is a bit too exhaustive, and you need to check your condition for all elements. It should look like:
permut [] ys = null ys
permut xs ys = all condition xs
where
condition x = ....
Bonus suggestion: Whenever it is the case that it is true that you find yourself writing
if ..... then True else False
you write to much.
----- Addendum:
So you found all, it does walk through the list (xs in our case) and it checks if the condition is True for all elements (of xs in our case). For example:
all even [1,2,3]
is a short way to say:
even 1 && even 2 && even 3
So, in your where clause you have condition, and it has access to the original xs and ys lists, and it is called for each element in xs. Hence, all you need to check is if the argument x of condition occurs the same number of times in xs and ys. For this, you have already counter, so it'll be a one liner.
I want to make a program insertAt where z is the place in the list, and y is the number being inserted into the list xs. Im new to haskell and this is what I have so far.
insertAt :: Int-> Int-> [Int]-> [Int]
insertAt z y xs
| z==1 = y:xs
but I'm not sure where to go from there.
I have an elementAt function, where
elementAt v xs
| v==1 = head xs
| otherwise = elementAt (v-1) (tail xs)
but I'm not sure how I can fit it in or if I even need to. If possible, I'd like to avoid append.
If this isn't homework: let (ys,zs) = splitAt n xs in ys ++ [new_element] ++ zs
For the rest of this post I'm going to assume you're doing this problem as homework or to teach yourself how to do this kind of thing.
The key to this kind of problem is to break it down into its natural cases. You're processing two pieces of data: the list you're inserting into, and the position in that list. In this case, each piece of data has two natural cases: the list you're procssing can be empty or not, and the number you're processing can be zero or not. So the first step is to write out all four cases:
insertAt 0 val [] = ...
insertAt 0 val (x:xs) = ...
insertAt n val [] = ...
insertAt n val (x:xs) = ...
Now, for each of these four cases, you need to think about what the answer should be given that you're in that case.
For the first two cases, the answer is easy: if you want to insert into the front of a list, just stick the value you're interested in at the beginning, whether the list is empty or not.
The third case demonstrates that there's actually an ambiguity in the question: what happens if you're asked to insert into, say, the third position of a list that's empty? Sounds like an error to me, but you'll have to answer what you want to do in that case for yourself.
The fourth case is most interesting: Suppose you want to insert a value into not-the-first position of a list that's not empty. In this case, remember that you can use recursion to solve smaller instances of your problem. In this case, you can use recursion to solve, for instance, insertAt (n-1) val xs -- that is, the result of inserting your same value into the tail of your input list at the n-1th position. For example, if you were trying to insert 5 into position 3 (the fourth position) of the list [100,200,300], you can use recursion to insert 5 into position 2 (the third position) of the list [200,300], which means the recursive call would produce [200,300,5].
We can just assume that the recursive call will work; our only job now is to convert the answer to that smaller problem into the answer to the original problem we were given. The answer we want in the example is [100,200,300,5] (the result of inserting 5 into position 4 of the list [100,200,300], and what we have is the list [200,300,5]. So how can we get the result we want? Just add back on the first element! (Think about why this is true.)
With that case finished, we've covered all the possible cases for combinations of lists and positions to update. Since our function will work correctly for all possibilities, and our possibilities cover all possible inputs, that means our function will always work correctly. So we're done!
I'll leave it to you to translate these ideas into Haskell since the point of the exercise is for you to learn it, but hopefully that lets you know how to solve the problem.
You could split the list at index z and then concatenate the first part of the list with the element (using ++ [y]) and then with the second part of the list. However, this would create a new list as data is immutable by default. The first element of the list by convention has the index 0 (so adjust z accordingly if you want the meaning of fist elemnt is indexed by 1).
insertAt :: Int -> Int-> [Int] -> [Int]
insertAt z y xs = as ++ (y:bs)
where (as,bs) = splitAt z xs
While above answers are correct, I think this is more concise:
insertAt :: Int -> Int-> [Int]-> [Int]
insertAt z y xs = (take z xs) ++ y:(drop z xs)
Im trying to swap the first and last element of a list in haskell. I've tried pattern matchnig, expressions, functions, etc. This is my last attempt:
cambio xs = [ cabeza++([x]++cola)|x<-xs, cabeza <- init x, cola <- last x, drop 1 x, drop 0 ([init x])]
My compiler throws the next error:
Couldn't match expected type `Bool' with actual type `[a0]'
In the return type of a call of `drop'
In the expression: drop 1 x
In a stmt of a list comprehension: drop 1 x
Can anyone help me? I've tried to do this for 2 days
Here are a few hints:
You can't solve this with list comprehension.
Identify the base (trivial) cases - empty list and list of one element. Write equations that cover those cases.
In all other cases the length of the input list will be >= 2. The list you want is
[z] ++ xs ++ [a]
where z is the last element, a the first element of the input list and xs the middle part of the input.
Now tell me (or yourself), how long will xs be, if the length of the input string was k?
Write the equation that covers the case of lists with more than 1 elements. You can use functions like head, last, length, drop or take.
I think that lists aren't the best data structure for doing this, but here it goes:
swap list = last list : (init . tail $ list) ++ [head list]
This is going to require traversing the list and will be slow on long lists. This is the nature of linked lists.
Updated with base cases from question asker:
swap [] = []
swap [a] = [a]
swap list = last list : (init . tail $ list) ++ [head list]
This is a fairly straightforward thing to do, especially with the standard list functions:
swapfl [] = []
swapfl [x] = [x]
swapfl (x:xs) = (last xs : init xs) ++ [x]
Or without them (although this is less readable and usually not done, and not recommended):
swapfl' [] = []
swapfl' [x] = [x]
swapfl' (x:xs) = let (f, g) = sw x xs in f:g
where sw k [y] = (y, [k])
sw k (y:ys) = let (n, m) = sw k ys in (n, y:m)
Or one of many other ways.
I hope that helps ... I know I didn't do much explaining, but frankly, it's hard to tell exactly what you were having trouble with as far as this function is concerned, seeing as you also seem to completely misunderstand list comprehensions. I think it might be most beneficial if I explain those instead?
And why this cant be solved with a list comprehension? I tough they were like functions but with a different form
Not really. List comprehensions are useful for easily defining lists, and they're very closely related to set-builder notation in mathematics. That would not be useful for this particular application, because, while they're very good at modifying the elements of a list, comprehensions are not very good at reordering lists.
In a comprehension, you have three parts: the definition of an element in the list, one or more input lists, and zero or more predicates:
[ definition | x <- input1, y <- input2, predicate1, predicate2 ]
The definition describes a single element of the list we're making, in terms of the variables the arrows in the inputs are pointing at (x and y in this case). Each input has a list on the right of the arrow, and a variable on the left. Each element in the list we're making is built by extracting each combination of elements from the input lists into those variables, and evaluating the definition part using those values. For example:
[ x + y | x <- [1, 3], y <- [2, 4] ]
This generates:
[1 + 2, 1 + 4, 3 + 2, 3 + 4] == [3, 5, 5, 7]
Also, you can include predicates, which are like filters. Each predicate is a boolean expression defined in terms of the input elements, and each is evaluated whenever a new list element is. If any of the predicates come out to be false, those elements aren't put in the list we're making.
Let's look at your code:
cambio xs = [ cabeza++([x]++cola) | x<-xs, cabeza <- init x, cola <- last x,
drop 1 x, drop 0 ([init x])]
The inputs for this comprehension are x <- xs, cabeza <- init x, and cola <- last x. The first one means that every element in xs is going to be used to define elements for the new list, and each element is going to be named x. The other two don't make any sense, because init and last are type [a] -> a, but are on the right side of the arrow and so must be lists, and x must be an element of a list because it's on the left side of its arrow, so in order for this to even compile, xs would have to be type [[[a]]], which I'm sure is not what you want.
The predicates you used are drop 1 x and drop 0 [init x]. I kind of understand what you were trying to do with the first one, dropping the first element of the list, but that wouldn't work because x is just an element of the list, not the list itself. In the second one, drop 0 means "remove zero elements from the beginning of the following list", which would do absolutely nothing. In either case, putting something like that in a predicate wouldn't work because the predicate needs to be a boolean value, which is why you got the compiler error. Here's an example:
pos xs = [ x | x <- xs, x >= 0 ]
This function takes a list of numbers, removes all the negative numbers, and returns the result. The predicate is the x >= 0, which is a boolean expression. If the expression evaluates to false, the element being evaluated is filtered out of the resulting list.
The element definition you used is cabeza ++ [x] ++ cola. This means "Each element in the resulting list is itself a list, made up of all elements in the list cabeza, followed by a single element that contains x, followed by all elements in the list cola", which seems like the opposite of what you were going for. Remember that the part before the pipe character defines a single element, not the list itself. Also, note that putting square brackets around a variable creates a new list that contains that variable, and only that variable. If you say y = [x], this means that y contains a single element x, and doesn't say anything about whether x is a list or not.
I hope that helps clear some things up.