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.
Related
I'm beginner in haskell and I tried to add a number in a 2D list with specific index in haskell but I don't know how to do
example i have this:
[[],[],[]]
and I would like to put a number (3) in the index 1 like this
[[],[3],[]]
I tried this
[array !! 1] ++ [[3]]
but it doesn't work
As you may have noticed in your foray so far, Haskell isn't like many other languages in that it is generally immutable, so trying to change a value, especially in a deeply nested structure like that, isn't the easiest thing. [array !! 1] would give you a nested list [[]] but this is not mutable, so any manipulations you do this structure won't be reflected in the original array, it'll be a separate copy.
(There are specialized environments where you can do local mutability, as with e.g. Vectors in the ST monad, but these are an exception.)
For what you're trying to do, you'll have to deconstruct the list to get it to a point where you can easily make the modification, then reconstruct the final structure from the (modified) parts.
The splitAt function looks like it will help you with this: it takes a list and separates it into two parts at the index you give it.
let array = [[],[],[]]
splitAt 1 array
will give you
([[]], [[],[]])
This helps you by getting you closer to the list you want, the middle nested list.
Let's do a destructuring bind to be able to reconstruct your final list later:
let array = [[],[],[]]
(beginning, end) = splitAt 1 array
Next, you'll need to get at the sub-list you want, which is the first item in the end list:
desired = head end
Now you can make your modification -- note, this will produce a new list, it won't modify the one that's there:
desired' = 3:desired
Now we need to put this back into the end list. Unfortunately, the end list is still the original value of [[],[]], so we'll have to replace the head of this with our desired' to make it right:
end' = desired' : (tail end)
This drops the empty sub-list at the beginning and affixes the modified list in its place.
Now all that's left is to recombine the modified end' with the original beginning:
in beginning ++ end'
making the whole snippet:
let array = [[],[],[]]
(beginning, end) = splitAt 1 array
desired = head end
desired' = 3:desired
end' = desired' : (tail end)
in beginning ++ end'
or, if you're entering all these as commands in the REPL:
let array = [[],[],[]]
let (beginning, end) = splitAt 1 array
let desired = head end
let desired' = 3:desired
let end' = desired' : (tail end)
beginning ++ end'
As paul mentions, things in Haskell are immutable. What you want to do must be done not be modifying the list in place, but by destructuring the list, transforming one of its parts, and restructuring the list with this changed part. One way of destructuring (via splitAt) is put forth there; I'd like to offer another.
Lists in Haskell are defined as follows:
data [] a = [] | a : [a]
This reads "A list of a is either empty or an a followed by a list of a". (:) is pronounced "cons" for "constructor", and with it, you can create nonempty lists.
1 : [] -> [1]
1 : [2,3] -> [1,2,3]
1 : 2 : 3 : [] -> [1,2,3]
This goes both ways, thanks to pattern matching. If you have a list [1,2,3], matching it to x : xs will bind its head 1 to the name x and its tail [2,3] to xs. As you can see, we've destructured the list into the two pieces that were initially used to create it. We can then operate on those pieces before putting the list back together:
λ> let x : xs = [1,2,3]
λ> let y = x - 5
λ> y : xs
[-4,2,3]
So in your case, we can match the initial list to x : y : z : [], compute w = y ++ [3], and construct our new list:
λ> let x : y : z : [] = [[],[],[]]
λ> let w = y ++ [3]
λ> [x,w,z]
[[],[3],[]]
But that's not very extensible, and it doesn't solve the problem you pose ("with specific index"). What if later on we want to change the thousandth item of a list? I'm not too keen on matching that many pieces. Fortunately, we know a little something about lists—index n in list xs is index n+1 in list x:xs. So we can recurse, moving one step along the list and decrementing our index each step of the way:
foo :: Int -> [[Int]] -> [[Int]]
foo 0 (x:xs) = TODO -- Index 0 is x. We have arrived; here, we concatenate with [3] before restructuring the list.
foo n (x:xs) = x : foo (n-1) xs
foo n [] = TODO -- Up to you how you would like to handle invalid indices. Consider the function error.
Implement the first of those three yourself, assuming you're operating on index zero. Make sure you understand the recursive call in the second. Then read on.
Now, this works. It's not all that useful, though—it performs a predetermined computation on a specified item in a list of one particular type. It's time to generalize. What we want is a function of the following type signature:
bar :: (a -> a) -> Int -> [a] -> [a]
where bar f n xs applies the transformation f to the value at index n in the list xs. With this, we can implement the function from before:
foo n xs = bar (++[3]) n xs
foo = bar (++[3]) -- Alternatively, with partial application
And believe it or not, changing the foo you already wrote into the much more useful bar is a very simple task. Give it a try!
I'm working on this prolog assignment where I must parse an user-inputted list of string characters (specifically "u"), and determine if all the elements are equal to the string "u". If they are, then it returns the number of elements, if not, it returns false. For example:
uA(-Length,+String,+Leftover) //Prototype
?- uA(L,["u","u","u"],[]).
L = 3 .
?- uA(L,["u","u","d"],[]).
false.
I have a decent grasp on how prolog works, but I'm confused about how lists operate. Any help would be greatly appreciated. Thanks!
Edit: I made some headway with the sort function (thank you!) but I've run into a separate problem.
uA(Length, String) :-
sort(String, [_]),
member("u", String),
length(String, Length).
This does mostly what I need it to, however, when I run it:
?- uA(L, ["u", "u", "u"]).
L = 3 ;
L = 3 ;
L = 3.
Is there any way to make it such that it only prints L = 3 once? Thanks!
If you want to state that all list items are equal, there is no need to sort the list first.
Simply use library predicate maplist/2 together with the builtin predicate (=)/2:
?- maplist(=(X), Xs).
Xs = []
; Xs = [X]
; Xs = [X, X]
; Xs = [X, X, X]
; Xs = [X, X, X, X]
… % ... and so on ...
First of all, be careful with double-quoted terms in Prolog. Their interpretation depends on the value of the standard double_quotes flag. The most portable value of this flag is codes, which makes e.g. "123" being interpreted as [49,50,51]. Other possible values of this flag are atom and chars. Some Prolog systems, e.g. SWI-Prolog, also support a string value.
But back to your question. A quick way to check that all elements in a ground list are equal is to use the standard sort/2 predicate (which eliminates duplicated elements). For example:
| ?- sort(["u","u","u"], [_]).
yes
| ?- sort(["u","u","d"], [_]).
no
As [_] unifies with any singleton list, the call only succeeds if the the sorting results in a list with a single element, which only happens for a non-empty ground list if all its elements are equal. Note that this solution is independent of the value of the double_quotes flag. Note also that you need to deal with an empty list separately.
My approach is to check if every element in the list is the same or not (by checking if the head of the list and it's adjacent element is the same or not). If same then return True else false. Then calculate the length of every element is the same in the list.
isEqual([X,Y]):- X == Y , !.
isEqual([H,H1|T]):- H == H1 , isEqual([H1|T]).
len([],0).
len([_|T],L):- len(T,L1) , L is L1+1.
goal(X):- isEqual(X) , len(X,Length) , write('Length = ') , write(Length).
OUTPUT
?- goal(["u","u","u"]).
Length = 3
true
?- goal(["u","u","a"]).
false
you can do it this way. Hope this helps you.
I am trying to write a very simple function that takes a list (for example : [1,2,3,1,5]) and returns a list of elements that are directly after a specific element.
What I have reached so far is:
function element list = filter (\x -> element:x) list
My desired output:
function 1 [1,2,3,1,5]
=> [2,5]
Try this
map snd $ filter ((== x) . fst) $ zip theList (tail theList)
This won't work on an empty list, you will still need extra code to deal with that.
How does this work?
First, note that the values flow from right to left. The ($) operator allows this to happen. So, the first part evaluated is the zip function.
zip theList (tail theList)
For your example above, this would yield
zip [1,2,3,1,5] [2,3,1,5]
equaling
[(1,2), (2, 3), (3, 1), (1,5)]
which is the set of concurrent pairs in the list.
Next, the filter is applied
filter ((== x) . fst) $ ....
In English, what this says is, filter out only the concurrent pairs whose first element equals x. The output is
[(1,2), (1,5)]
Now we have the list of concurrent pairs starting with 1.
Finally, we apply the map
map snd $ ....
This just pulls out the second value of the pair.
map snd [(1,2), (1,5)] = [2,5]
which is the desired value.
Note, my comment above about failing on the empty list.
This is because tail crashes on the empty list
tail [] --error
There are ways to patch this behavior (see the safe package, for instance), but it is mostly bookkeeping at this point, so I left that for you to work out.
Also note that since all of the functions we used are lazy, this approach would work for lists of infinite length as well.
You can do this quite easily with a simple list comprehension, e.g.:
successors xs i = [y | (x,y) <- zip xs (drop 1 xs), x == i]
This will work to your specifications
next x (i:y:ys) -- look at the first two items in the list
| x == i = -- if the first item == x,
y : next x (y:ys) -- take the second, and continue minus the first element
|otherwise = -- not equal,
next x (y:ys) -- so skip that element
next _ [_] = [] -- if there's no second element, then stop
next _ _ = [] -- if the list is empty, stop
I'm currently learning haskell then I came to this exercise where I have to define a function that gets the product of a list of numbers. I was provided with choices and since I am new to haskell there are some notations that I am a bit unclear of.
So I saw this definition on one of the choices:
p [x, xs] = x * product xs
I can understand this quite a bit, it means to get the product of the list and then multiply it with the value of x.
Then I saw this other definition on one of the other choice:
p (x : xs) = x * product xs
Which I totally do not understand. It uses parenthesis and a colon which I am having a hard time looking for their definition. I appreciate if someone could enlighten me with regards the syntax and semantics.
[x, xs] is a list containing two elements. The first element is called x and the second is called xs. So in this case product xs does not calculate the product of the list, it calculates the product of the second element. Since the elements of the list can't be lists themselves (or else multiplying with x wouldn't work), this is a type error.
x : xs is a list that contains at least one element. Its first element is called x and the list containing its remaining elements is called xs.
: is the cons operator, which appends an element to a list
(x : xs) is pattern-matching a list into an element x and the rest of the list xs
let's get a concrete example:
l = [1,2,3]
show_rest_of_list (x:xs) = xs
show_rest_of_list l
-- would return [2,3]
play_with_list (x:xs) = x : x : xs
play_with_list l
-- would return [1,1,2,3]
append([],Xs,Xs).
append([Head|Tail],List2,[Head|Tail2]):-
append(Tail,List2,Tail2).
The upper append method adds elements from first two parameter slots to the third param variable.
?-append([2,1], [3,4], X).
?-X=[2,1,3,4]
The way I see it in steps is (which is propably wrong):
append(2 | [1], [3,4], 2 | X)
append([1], [3,4], X)
append(1 | [], [3,4], 1 | X)
append([], [3,4], [3,4])
And that's it. I can't wrap my head around how it adds together the elements and that's what i could use help with - a clear explanation on how this method works. I just don't understand how the [2,1] array gets added to the final result.
the X in the recursion is not the same X as in the original call if you rename it in the trace you'll see
append(2 | [1], [3,4], 2 | X1) -- X = [2|X1]
append([1], [3,4], X1)
append(1 | [], [3,4], 1 | X2) -- X1 = [1|X2]
append ([], [3,4], [3,4]) -- X2 = [3,4]
so X1 = [1,3,4] and X = [2,1,3,4]
First, you have to understand how a list is implemented in Prolog. It is an essentially recursive data structure.
The empty list is a list of zero items, represented by the atom [].
Non-empty lists are represented by the structure ./2, consisting of the head of the list (a prolog term), and the tail of the list (another list, consisting of all items save the first). So...
[] — a list of zero items, is represented as[]
[a] — a list of 1 item, is represented as.(a,[])
[a,b] — a list of 2 items, is represented as.(a,.(b,[]))
[a,b,c] — a list of 3 items, is represented as .(a,.(b,.(c,[])))
The standard list notation using square brackets is just syntactic sugar on top of this representation. Saying [Head|Tail] is a polite way of saying .(Head,Tail) and saying [X,Y,Z|More] is the polite way of saying .(X,.(Y,.(Z,More))). (You might be noticing a certain....Lisp-ishness...to the internal list notation here.)
Understanding how a list is represented, the naive algorithm for appending (concatenating) one list to another is this:
First, there are two special cases to consider:
The result of appending a non-empty list X to an empty list Y is X.Append [1,2,3] to [] and get [1,2,3].Note. This case is can be (is normally) handled by the ordinary case below, though. It's an opportunity for optimization as there's no point in recursing down the entire list, just to replace [] with [] at the end, right?.
The result of appending an empty list X to a non-empty list Y is Y.Append [] to [1,2,3] and you get [1,2,3].
Otherwise, we have the ordinary case:
Append non-empty list Y to non-empty list X to produce list Z. To do this is trivial:
We simply recurse down on list X, popping its head as we go and and prepending that to list Z, the result. You'll notice that as this happens, list Z is a broken list structure, since its last node is always unbound rather than being []. This gets fixed at the very end when the source list, list X, is exhausted and degenerates into the special case of being an empty list. At that point, the unbound last node gets bound as it unifies with list Y (a list of zero or more nodes), leaving us with a correct list structure.
The Prolog code for append/3 expresses this algorithm directly. As noted earlier, the first clause is an optional optimization, as that can be handled by the 3rd clause (the ordinary case). It wants a cut, though, as without, backtracking would produce two solutions.
append( X , [] , X ) :- !. ; concatenate empty list X to list Y producing Y
append( [] , Y , Y ). ; concatenate list X to empty list Y producing X
append( [X|Xs] , Y , [X|Zs] ) :- ; anything else
append( Xs , Y , Zs )
.