So, I'm new here, and I would like to ask 2 questions about some code:
Duplicate each element in list by n times. For example, duplicate [1,2,3] should give [1,2,2,3,3,3]
duplicate1 xs = x*x ++ duplicate1 xs
What is wrong in here?
Take positive numbers from list and find the minimum positive subtraction. For example, [-2,-1,0,1,3] should give 1 because (1-0) is the lowest difference above 0.
For your first part, there are a few issues: you forgot the pattern in the first argument, you are trying to square the first element rather than replicate it, and there is no second case to end your recursion (it will crash). To help, here is a type signature:
replicate :: Int -> a -> [a]
For your second part, if it has been covered in your course, you could try a list comprehension to get all differences of the numbers, and then you can apply the minimum function. If you don't know list comprehensions, you can do something similar with concatMap.
Don't forget that you can check functions on http://www.haskell.org/hoogle/ (Hoogle) or similar search engines.
Tell me if you need a more thorough answer.
To your first question:
Use pattern matching. You can write something like duplicate (x:xs). This will deconstruct the first cell of the parameter list. If the list is empty, the next pattern is tried:
duplicate (x:xs) = ... -- list is not empty
duplicate [] = ... -- list is empty
the function replicate n x creates a list, that contains n items x. For instance replicate 3 'a' yields `['a','a','a'].
Use recursion. To understand, how recursion works, it is important to understand the concept of recursion first ;)
1)
dupe :: [Int] -> [Int]
dupe l = concat [replicate i i | i<-l]
Theres a few problems with yours, one being that you are squaring each term, not creating a new list. In addition, your pattern matching is off and you would create am infinite recursion. Note how you recurse on the exact same list as was input. I think you mean something along the lines of duplicate1 (x:xs) = (replicate x x) ++ duplicate1 xs and that would be fine, so long as you write a proper base case as well.
2)
This is pretty straight forward from your problem description, but probably not too efficient. First filters out negatives, thewn checks out all subtractions with non-negative results. Answer is the minumum of these
p2 l = let l2 = filter (\x -> x >= 0) l
in minimum [i-j | i<-l2, j<-l2, i >= j]
Problem here is that it will allow a number to be checkeed against itself, whichwiull lend to answers of always zero. Any ideas? I'd like to leave it to you, commenter has a point abou t spoon-feeding.
1) You can use the fact that list is a monad:
dup = (=<<) (\x -> replicate x x)
Or in do-notation:
dup xs = do x <- xs; replicate x x; return x
2) For getting only the positive numbers from a list, you can use filter:
filter (>= 0) [1,-1,0,-5,3]
-- [1,0,3]
To get all possible "pairings" you can use either monads or applicative functors:
import Control.Applicative
(,) <$> [1,2,3] <*> [1,2,3]
[(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)]
Of course instead of creating pairs you can generate directly differences when replacing (,) by (-). Now you need to filter again, discarding all zero or negative differences. Then you only need to find the minimum of the list, but I think you can guess the name of that function.
Here, this should do the trick:
dup [] = []
dup (x:xs) = (replicate x x) ++ (dup xs)
We define dup recursively: for empty list it is just an empty list, for a non empty list, it is a list in which the first x elements are equal to x (the head of the initial list), and the rest is the list generated by recursively applying the dup function. It is easy to prove the correctness of this solution by induction (do it as an exercise).
Now, lets analyze your initial solution:
duplicate1 xs = x*x ++ duplicate1 xs
The first mistake: you did not define the list pattern properly. According to your definition, the function has just one argument - xs. To achieve the desired effect, you should use the correct pattern for matching the list's head and tail (x:xs, see my previous example). Read up on pattern matching.
But that's not all. Second mistake: x*x is actually x squared, not a list of two values. Which brings us to the third mistake: ++ expects both of its operands to be lists of values of the same type. While in your code, you're trying to apply ++ to two values of types Int and [Int].
As for the second task, the solution has already been given.
HTH
Related
I want to create a function that returns every third int from a list of ints without using any predefined functions. For example, everyThird [1,2,3,4,5] --> [1,4]
everyThird:: [a] -> [a]
Could I just continue to iterate over the list using tail and appending to a new list every third call? I am new to Haskell and very confused with all of this
One other way of doing this is to handle three different base cases, in all of which we're at the end of the list and the list is less than three elements long, and one recursive case, where the list is at least three elements long:
everyThird :: [a] -> [a]
everyThird [] = []
everyThird [x] = [x]
everyThird [x, _] = [x]
everyThird (x:_:_:xs) = x:everyThird xs
You want to do exactly what you said: iterate over the list and include the element only on each third call. However, there's a problem. Haskell is a funny language where the idea of "changing" a variable doesn't make sense, so the usual approach of "have a counter variable i which tells us whether we're on the third element or not" won't work in the usual way. Instead, we'll create a recursive helper function to maintain the count for us.
everyThird :: [Int] -> [Int]
everyThird xs = helper 0 xs
where helper _ [] = []
helper 0 (x : xs) = x : helper 2 xs
helper n (_ : xs) = helper (n - 1) xs
We have three cases in the helper.
If the list is empty, stop and return the empty list.
If the counter is at 0 (that is, if we're on the third element), make a list starting with the current element and ending with the rest of the computation.
If the counter is not at zero, count down and continue iteration.
Because of the way pattern matching works, it will try these three statements in order.
Notice how we use an additional argument to be the counter variable since we can't mutate the variable like we would in an imperative language. Also, notice how we construct the list recursively; we never "append" to an existing list because that would imply that we're mutating the list. We simply build the list up from scratch and end up with the correct result on the first go round.
Haskell doesn't have classical iteration (i.e. no loops), at least not without monads, but you can use similar logic as you would in a for loop by zipping your list with indexes [0..] and applying appropriate functions from Data.List.
E.g. What you need to do is filter every third element:
everyThirdWithIndexes list = filter (\x -> snd x `mod` 3 == 0) $ zip list [0..]
Of course you have to get rid of the indexes, there are two elegant ways you can do this:
everyThird list = map (fst) . everyThirdWithIndexes list
-- or:
everyThird list = fst . unzip . everyThirdWithIndexes list
If you're not familiar with filter and map, you can define a simple recursion that builds a list from every first element of a list, drops the next two and then adds another from a new function call:
everyThird [] = [] -- both in case if the list is empty and the end case
everyThird (x:xs) = x : everyThird (drop 2 xs)
EDIT: If you have any questions about these solutions (e.g. some syntax that you are not familiar with), feel free to ask in the comments. :)
One classic approach:
everyThird xs = [x | (1,x) <- zip (cycle [1..3]) xs]
You can also use chunksOf from Data.List.Split to seperate the lists into chunks of 3, then just map the first element of each:
import Data.List.Split
everyThird :: [a] -> [a]
everyThird xs = map head $ chunksOf 3 xs
Which works as follows:
*Main> everyThird [1,2,3,4,5]
[1,4]
Note: You may need to run cabal install split to use chunksOf.
Doing the third of the 99-Haskell problems (I am currently trying to learn the language) I tried to incorporate pattern matching as well as recursion into my function which now looks like this:
myElementAt :: [a] -> Int -> a
myElementAt (x ++ xs) i =
if length (x ++ xs) == i && length xs == 1 then xs!!0
else myElementAt x i
Which gives me Parse error in pattern: x ++ xs. The questions:
Why does this give me a parse error? Is it because Haskell is no idea where to cut my list (Which is my best guess)?
How could I reframe my function so that it works? The algorithmic idea is to check wether the list has the length as the specified inde; if yes return the last elemen; if not cut away one element at the end of the list and then do the recursion.
Note: I know that this is a really bad algorithm, but it I've set myself the challenge to write that function including recursion and pattern matching. I also tried not to use the !! operator, but that is fine for me since the only thing it really does (or should do if it compiled) is to convert a one-element list into that element.
Haskell has two different kinds of value-level entities: variables (this also includes functions, infix operators like ++ etc.) and constructors. Both can be used in expressions, but only constructors can also be used in patterns.
In either case, it's easy to tell whether you're dealing with a variable or constructor: a constructor always starts with an uppercase letter (e.g. Nothing, True or StateT) or, if it's an infix, with a colon (:, :+). Everything else is a variable. Fundamentally, the difference is that a constructor is always a unique, immediately matcheable value from a predefined collection (namely, the alternatives of a data definition), whereas a variable can just have any value, and often it's in principle not possible to uniquely distinguish different variables, in particular if they have a function type.
Yours is actually a good example for this: for the pattern match x ++ xs to make sense, there would have to be one unique way in which the input list could be written in the form x ++ xs. Well, but for, say [0,1,2,3], there are multiple different ways in which this can be done:
[] ++[0,1,2,3]
[0] ++ [1,2,3]
[0,1] ++ [2,3]
[0,1,2] ++ [3]
[0,1,2,3]++ []
Which one should the runtime choose?
Presumably, you're trying to match the head and tail part of a list. Let's step through it:
myElementAt (x:_) 0 = x
This means that if the head is x, the tail is something, and the index is 0, return the head. Note that your x ++ x is a concatenation of two lists, not the head and tail parts.
Then you can have
myElementAt(_:tl) i = myElementAt tl (i - 1)
which means that if the previous pattern was not matched, ignore the head, and take the i - 1 element of the tail.
In patterns, you can only use constructors like : and []. The append operator (++) is a non-constructor function.
So, try something like:
myElementAt :: [a] -> Int -> a
myElementAt (x:xs) i = ...
There are more issues in your code, but at least this fixes your first problem.
in standard Haskell pattern matches like this :
f :: Int -> Int
f (g n 1) = n
g :: Int -> Int -> Int
g a b = a+b
Are illegal because function calls aren't allowed in patterns, your case is just a special case as the operator ++ is just a function.
To pattern match on lists you can do it like this:
myElementAt :: [a] -> Int -> a
myElementAt (x:xs) i = // result
But in this case x is of type a not [a] , it is the head of the list and xs is its tail, you'll need to change your function implementation to accommodate this fact, also this function will fail with the empty list []. However that's the idiomatic haskell way to pattern match aginst lists.
I should mention that when I said "illegal" I meant in standard Haskell, there are GHC extensions that give something similar to that , it's called ViewPatterns But I don't think you need it especially that you're still learning.
I want to get the first element of every tuple stored in a list like so:
INPUT : [(1,2,3), (1,4,5), (1,6,7)]
Wanted OUTPUT : [(1,1,1)] % Note the tuple notation
This is the function I have written so far:
f [] = []
f ((x,y,z):xs) = x:(f xs)
But it gives me this output:
[1,1,1]
instead of
[(1,1,1)]
Where is my misunderstanding?
As yet stated in the comment by #pdexter, you cannot create a tuple with arbitrary length. So what you're trying to do will only work when the tuple exactly pattern matches a tuple with 3 Ints. Let me show you an example to clarify what I mean:
helper :: [(Int,Int,Int)] -> (Int,Int,Int)
helper xs = (x,y,z)
where [x,y,z] = f xs
f :: [(Int,Int,Int)] -> [Int]
f [] = []
f ((x,y,z):xs) = x : (f xs)
Here we manage to create (x,y,z) by pattern matching on [x,y,z], but what if there were more than three tuples, our pattern matching would no longer be valid?
So you should ask yourself the question what it is you're really trying to accomplish with this code and what scenarios might occur (e.g.: empty tuples, tuples with varying lengths mutually, .. ?) and also, how should your program handle these different cases?
If there is no functional reason as to why you like the output to be in a tuple, I suggest you keep the function as is.
Good luck!
I am currently working through H-99 Questions after reading Learn You a Haskell. So far I felt like I had a pretty good grasp of the concepts, and I didn't have too much trouble solving or understanding the previous problems. However, this one stumped me and I don't understand the solution.
The problem is:
Generate the combinations of K distinct objects chosen from the N elements of a list
In how many ways can a committee of 3 be chosen from a group of 12 people? We all know that there are C(12,3) = 220 possibilities (C(N,K) denotes the well-known binomial coefficients). For pure mathematicians, this result may be great. But we want to really generate all the possibilities in a list.
The solution provided:
import Data.List
combinations :: Int -> [a] -> [[a]]
combinations 0 _ = [ [] ]
combinations n xs = [ y:ys | y:xs' <- tails xs, ys <- combinations (n-1) xs']
The main point of confusion for me is the y variable. according to how tails works it should be getting assigned the entire list at the beginning and then that list will be preppend to ys after it is generate. However, when the function run it return a list of lists no longer than the n value passed in. Could someone please help me understand exactly how this works?
Variable y is not bound to the whole xs list. For instance, assume xs=[1,2,3]. Then:
y:xs' is matched against [1,2,3] ==> y=1 , xs'=[2,3]
y:xs' is matched against [2,3] ==> y=2 , xs'=[3]
y:xs' is matched against [3] ==> y=3 , xs'=[]
y:xs' is matched against [] ==> pattern match failure
Note that y is an integer above, while xs' is a list of integers.
The Haskell code can be read a a non-deterministic algorithm, as follows. To generate a combination of n elements from xs, get any tail of xs (i.e., drop any number of elements from the beginning). If the tail is empty, ignore it. Otherwise, let the tail be y:xs', where y is the first element of the tail and xs' the remaining (possibly empty) part. Take y and add it to the combination we are generating (as the first element). Then recursively choose other n-1 arguments from the xs' remaining part, and add those to the combination as well. When n drops to zero, we know there is only one combination, namely the empty combination [], so take that.
y is not appended to ys. That would involve the (++) :: [a] -> [a] -> [a] operator.
For that matter the types would not match if you tried to append y and ys. y has type a, while ys has type [a].
Rather, y is consed to ys using (:) :: a -> [a] -> [a] (the cons operator).
The length of the returned list is equal to n because combinations recurses from n to 0 so it will produce exactly n inner lists.
I am new to Haskell and am trying to learn the basics. I am having a hard time understanding how to manipulate the contents of a list.
Assume I have the following list and I would like to create a function to subtract 1 from every element in the list, where I can simply pass x to the function, how would this be done?
Prelude>let x = 1:2:3:4:5:[]
Something like:
Prelude>subtractOne(x)
(You can write 1:2:3:4:5:[] more simply as [1,2,3,4,5] or even [1..5].)
Comprehensions
You'd like to use list comprehensions, so here it is:
subtractOne xs = [ x-1 | x <- xs ]
Here I'm using xs to stand for the list I'm subtracting one from.
The first thing to notice is x <- xs which you can read as "x is taken from xs". This means we're going to take each of the numbers in xs in turn, and each time we'll call the number x.
x-1 is the value we're calculating and returning for each x.
For more examples, here's one that adds one to each element [x+1|x<-xs] or squares each element [x*x|x<-xs].
More than one list
Let's take list comprehension a little further, to write a function that finds the squares then the cubes of the numbers we give it, so
> squaresAndCubes [1..5]
[1,4,9,16,25,1,8,27,64,125]
We need
squaresAndCubes xs = [x^p | p <- [2,3], x <- xs]
This means we take the powers p to be 2 then 3, and for each power we take all the xs from xs, and calculate x to the power p (x^p).
What happens if we do that the other way around?
squaresAndCubesTogether xs = = [x^p | x <- xs, p <- [2,3]]
We get
> squaresAndCubesTogether [1..5]
[1,1,4,8,9,27,16,64,25,125]
Which takes each x and then gives you the two powers of it straight after each other.
Conclusion - the order of the <- bits tells you the order of the output.
Filtering
What if we wanted to only allow some answers?
Which numbers between 2 and 100 can be written as x^y?
> [x^y|x<-[2..100],y<-[2..100],x^y<100]
[4,8,16,32,64,9,27,81,16,64,25,36,49,64,81]
Here we allowed all x and all y as long as x^y<100.
Since we're doing exactly the same to each element, I'd write this in practice using map:
takeOne xs = map (subtract 1) xs
or shorter as
takeOne = map (subtract 1)
(I have to call it subtract 1 because - 1 would be parsed as negative 1.)
You can do this with the map function:
subtractOne = map (subtract 1)
The alternative solution with List Comprehensions is a little more verbose:
subtractOne xs = [ x - 1 | x <- xs ]
You may also want to add type annotations for clarity.
You can do this quite easily with the map function, but I suspect you want to roll something yourself as a learning exercise. One way to do this in Haskell is to use recursion. This means you need to break the function into two cases. The first case is usually the base case for the simplest kind of input. For a list, this is an empty list []. The result of subtracting one from all the elements of the empty list is clearly an empty list. In Haskell:
subtractOne [] = []
Now we need to consider the slightly more complex recursive case. For any list other than an empty list, we can look at the head and tail of the input list. We will subtract one from the head and then apply subtractOne to the rest of the list. Then we need to concatenate the results together to form a new list. In code, this looks like this:
subtractOne (x:xs) = (x - 1) : subtractOne xs
As I mentioned earlier, you can also do this with map. In fact, it is only one line and the preferred Haskellism. On the other hand, I think it is a very good idea to write your own functions which use explicit recursion in order to understand how it works. Eventually, you may even want to write your own map function for further practice.
map (subtract 1) x will work.
subtractOne = map (subtract 1)
The map function allows you to apply a function to each element of a list.