I'm trying to write the Zipwith function in Haskell.
If I run it with the following values, it should return this result:
Prelude> zipWith (+) [10,20,30] [33,44,94]
[43,64,124]
My code so far is:
Zipwith f [] [] = []
Zipwith f [] _ = []
Zipwith f _ [] = []
Zipwith f (x:xs) (y:ys) = (f x y) : (Zipwith f xs ys)
However, the compiler tells me that I have multiple functions, all Zipwith, that don't have a data definition, but I thought having one wasn't necessary in Haskell.
Also, then it says that I have multiple declarations of f, but it is just an argument, I thought it wouldn't matter that there are multiple definitions of an argument.
Any thoughts?
Haskell functions must start with a lower case. Upper case names are reserved for other stuff, like data types. In this case it would be a good idea to name your function zipWith', because ' is often used to indicate that the function is almost the same, but with a small change.
P.S.
Small critique of your code: You can remove the line zipwith f [] [] = [] because the other lines catch this case already. If you want you can even write it like this:
zipwith f (x:xs) (y:ys) = f x y : zipwith f xs ys
zipwith _ _ _ = []
Since the first one is the only pattern you care about.
Function names must start with a lowercase letter (or symbol). Upper-case letters are reserved for new data types. Your definition would be perfectly correct if it were named zipWith.
To echo the other answers: Function names must start with a lowercase letter. This is not merely a coding convention; it is part of the actual language syntax. As in, it actually stops stuff compiling if you don't do this.
Other programming languages often have conventions on how you should use case, but the Haskell compiler actually uses this to decide what kind of name you're referring to. In this case, it thinks that Zipwith is some sort of data type, which (naturally) doesn't actually exist, which is why the compiler is confused.
Related
Reading about folds on this wonderful book I have a question regarding foldr1 and the head' implementation proposed there, the code in question is:
head' = foldr1 (\x _ -> x)
this code works on infinite list, whereas foldl1 don't. A good visual explanation about why is this answer.
I do not quite understand though why does it work, considering that foldr1 is using the last element as accumulator. For example:
foldr1 (\x _ -> x) [1..]
This works because (I Think) lazy evaluation, even though foldr is starting from the last element of the list (which is infinite), I'm assuming because the function is not making use of any intermediate result, just return the first element.
So, is the compiler smart enough to know that, because inside of the lambda function only x is being used, just returns the first element of the list? even though it should start from the end?
On the contrary, doing
scanr1 (\x _ -> x) [1..]
Will print all elements of the infinite list without ending, which I suppose it's what the foldr is doing, just the compiler is smart enough to not evaluate it and return the head.
Thanks in advance.
Update
I found a really good answer that helped me understand how foldr works more deeply:
https://stackoverflow.com/a/63177677/1612432
foldr1 is using the last element as an initial accumulator value, but the combining function (\x _ -> x) is lazy in its second argument.
So provided the list is non-empty (let alone infinite), the "accumulator" value is never needed, thus never demanded.
foldr does not mean it should start from the right, just that the operations are grouped / associated / parenthesized on the right. If the combining function is strict in its 2nd argument that will entail indeed starting the calculations from the right, but if not -- then not.
So no, this is not about compiler being smart, this is about Haskell's lazy semantics that demand this. foldr is defined so that
foldr g z [x1,x2,...,xn] = g x1 (foldr g z [x2,...,xn])
and
foldr1 g xs = foldr g (last xs) (init xs)
and that's that.
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.
This question already has answers here:
What does the : infix operator do in Haskell?
(4 answers)
Closed 8 years ago.
I've tried looking it up in hoogle and other various haskell dictionaries, but I can't find it. I was under the impression that it prepends, but I'm starting to see it in ways I haven't before and I've started second guessing myself.
For example, this is one of the questions that I don't understand:
(3 points) Fill in the blank with a pattern such that fun1 [(5,6),(7,8)] returns
5 and fun1 [(10,20),(30,40),(50,60)] returns 10:
and the answer is apparently:
((y,_):_)
fun1 _____________ = y
But I am so confused by this. I understand that the underscores mean that you don't really care about what the types of those are, but I don't understand what the (:) does in this answer.
While the other answers correctly explain what : is they don't quite answer the question - in the answer you have in your question : isn't used as a function, but as a constructor to pattern match on. fun (x:xs) = x means "if the argument is of the format (x:xs) give me the x". Pattern matching is used to "pull apart" complex types based on their constructors in Haskell.
In particular, since : is a list constructor you can pull apart lists with :
(conceptually list is defined as data [] a = [] | (:) a [a], although you're not gonna get this to compile because it's builtin syntax).
A non list example: We could define a datatype data F a b = A a | B b. This would create a type F that's parameterized with two types a and b and two constructors A and B with the types a -> F a b and b -> F a b respectively.
You could then write functions that use pattern matching to get at the contained values, like
isA (A _) = True -- this value was constructed with A, so it is an A
isA (B _) = False -- this value was constructed with B so it is not an A
or
getA (A a) = a -- this value was constructed with A so we can get an a out of it
getA (B _) = undefined -- ohps! We can't get an a back here cause we don't have one!
It is a List constructor function. It is used for prepending any value in front of the list.
ghci> 2 : [3,4]
[2,3,4]
It is just another Haskell function. You can also see it's type in ghci:
ghci> :t (:)
(:) :: a -> [a] -> [a]
Regarding your question, the answer is like this ((y,_):_) because it is being used in pattern matching. The first _ is the second element of the pair and the second _ pattern matches a list.
This may help you:
ghci> (5,6):[(7,8)]
[(5,6),(7,8)]
: is the list constructor of type a -> [a] -> [a]. It is usually used infix. but you can use it as prefix if you surround it with parentheses as you did. Just like any infix operation. (E.g. (+) 4 5 == 4 + 5)
So (:) a as is the same as a:as
Every constructor in Haskell can be also used do deconstruct a value of the type if constructs in a pattern match:
f x:xs = xs
would for example define a function that takes a non empty list and returns the tail. It would fail on an empty list because the empty list is constructed by the nullary constructor []. You could make f total by adding that second constructor to the match.
f [] = []
I guess your confusion comes from the fact that in haskell there is syntactic sugar that allows you to write lists in a more convenient way. Instead of (1:(2:(3:[]))) you can write [1,2,3] which is expanded into the former by the compiler.
In addition to the answers of what (:) function does, please, bear in mind that in the context of your question : is used as a deconstructor.
It is better to view (:) as a constructor. Then, just like any other data constructor, it can be used to introspect the contents of the value. Examples are:
f (Just x) = x -- extracts the value wrapped into Maybe a
f (x:_) = x -- extracts the value wrapped into a list, [a]
f ((x,_):_) = x -- extracts the value wrapped into a tuple in the list of tuples
In all these cases Just, : and (,) are constructors. The same syntax can be used to construct or deconstruct the values - depending on the context of the expression. Compare:
f x = Just x -- wraps x into Maybe a
f x xs = x:xs -- wraps x into a list, [a]
f x y z = (x,y):z -- wraps x into a tuple in the list of tuples
To understand what fun1 does, let's first look at another function:
f (x:xs) = x
If you pass this function a list such as [5,12,33], it will match x to 5, and xs to [12,33]. The function just returns x, i.e. the first element. So this function is basically the same as head. Since we don't actually use the value xs, we can rewrite the function as:
f (x:_) = x
Now let's look at fun1, but a slightly modified version.
fun1 ((y,z):xs) = y
If we pass this function the list [(5,6),(7,8)], it will match (y,z) to the pair (5,6) and xs to [(7,8)]. So now y is 5, and that's the value we return. Again, since we don't use z or xs, we can write the function as:
fun1 ((y,_):_) = y
I can't for the life of me figure out why there is a problem with this if statement (haskell noobie.) Can anyone help me out?
fst3 (a,b,c) = a
snd3 (a,b,c) = b
trd3 (a,b,c) = c
fst4 (a,b,c,d) = a
snd4 (a,b,c,d) = b
trd4 (a,b,c,d) = c
qud4 (a,b,c,d) = d
fractionalKnapsack (x:xs) =
fractionalKnapsack (x:xs) []
fractionalKnapsack (x:xs) fracList =
((fst3 x),(snd3 x),(trd3 x),(snd3 x) / (trd3 x)):fracList
if length (x:xs) <= 1
then computeKnapsack sort(fracList)
else fractionalKnapsack xs fracList
computeKnapsack (x:xs) = (x:xs)
There are a few things wrong with this code. You have two different definitions for fractionalKnapsack, each taking a different number of arguments, clearly that causes the compiler some trouble. Also the parse error on the if statement is because there shouldn't actually be an if statement where you are trying to put one, you have already completed the definition of the function before you reach the if statement.
It might help a little bit if you better explained what you are trying to do, or what you expect to be happening with the code you wrote.
: is the "cons" operator. It cons-tructs a list by providing the "head" element on the left, and a "tail" list on the right
ghci> 1 : [2,3,4]
[1,2,3,4]
You can pattern match on lists with more than 0 elements using :.
ghci> let (x:xs) = [1,2,3,4]
ghci> x
1
ghci> xs
[2,3,4]
The way that you are using (x:xs) in your code hints that you do not yet have a firm grasp of the definition of lists nor of pattern matching. Rather than using
if length (x:xs) <= 1
it is more common to simply pattern match. A simple example:
howMany :: [a] -> String
howMany [] = "Zero"
howMany [x] = "One"
howMany (x:xs) = "Many"
Haskell functions can be defined with a sequence of "equations" like this where you pattern match on the possible cases that you are interested in. This brings us to the other issues with your code, which are:
The equations for fractionalKnapsack don't match. One has 1 argument, the other has 2. You probably meant to name the second fractionalKnapsack'.
Neither of the fractionalKnapsack definitions handles the empty list case. I'm not sure about this; this may be acceptable if you know that it will never be given an empty list.
None of your functions have type signatures. Type inference can infer them, but it is usually a good idea to write the type signature first, to express your intent for the function and guide you in its implementation.
The second definition of fractionalKnapsack doesn't make sense. There can only be one expression after the = sign, but you have provided two, separated by a newline. This is invalid Haskell and explains why there is a parse error on "if": because whatever compiler/interpreter you were using did not expect the beginning of another expression!
computeKnapsack sort(fracList)
That's probably an error too. It should be computeKnapsack (sort fracList) (or, equivalently, computeKnapsack $ sort fracList).
When you do computeKnapsack sort(fracList) it's equals to doing computeKnapsack sort (fracList), which is equivalent for doing computeKnapsack sort fracList, which means: "give computeKnapsack two arguments: sort and fracList".
I am reading a tutorial that uses the following example (that I'll generalize somewhat):
f :: Foo -> (Int, Foo)
...
fList :: Foo -> [Int]
fList foo = x : fList bar
where
(x, bar) = f foo
My question lies in the fact that it seems you can refer to x and bar, by name, outside of the tuple where they are obtained. This would seem to act like destructuring parameter lists in other languages, if my guess is correct. (In other words, I didn't have to do the following:)
fList foo = (fst tuple) : fList (snd tuple)
where
tuple = f foo
Am I right about this behavior? I've never seen it mentioned yet in the tutorials/books I've been reading. Can someone point me to more info on the subject?
Edit: Can anything (lists, arrays, etc.) be destructured in a similar way, or can you only do this with tuples?
Seeing your edit, I think what your asking about is Pattern matching.
And to answer your question: Yes, anything you can construct, you can also 'deconstruct' using the constructors. For example, you're probably familiar with this form of pattern matching:
head :: [a] -> a
head (x:xs) = x
head [] = error "Can't take head of empty list"
However, there are more places where you can use pattern matching, other valid notations are:
head xs = case xs of
(y:ys) -> y
[] -> error "Can't take head of empty list"
head xs = let (y:ys) = xs
in y
head xs = y
where
(y:ys) = xs
Note that the last two examples are a bit different from the first to because they give different error messages when you call them with an empty list.
Although these examples are specific to lists, you can do the same with other data types, like so:
first :: (a, b) -> a
first tuple = x
where
(x, y) = tuple
second :: (a, b) -> b
second tuple = let (x, y) = tuple
in y
fromJust :: Maybe a -> a
fromJust ma = x
where
(Just x) = ma
Again, the last function will also crash if you call it with Nothing.
To sum up; if you can create something using constructors (like (:) and [] for lists, or (,) for tuples, or Nothing and Just for Maybe), you can use those same constructors to do pattern matching in a variety of ways.
Am I right about this behavior?
Yes. The names exist only in the block where you have defined them, though. In your case, this means the logical unit that your where clause is applied to, i.e. the expression inside fList.
Another way to look at it is that code like this
x where x = 3
is roughly equivalent to
let x = 3 in x
Yes, you're right. Names bound in a where clause are visible to the full declaration preceding the where clause. In your case those names are f and bar.
(One of the hard things about learning Haskell is that it is not just permitted but common to use variables in the source code in locations that precede the locations where those variables are defined.)
The place to read more about where clauses is in the Haskell 98 Report or in one of the many fine tutorials to be found at haskell.org.