I am trying to write a simple higher order function in Haskell which takes two arguments:
A list of functions of any type.
A list of (Numbers or Strings or Boolean, etc) --> any type
The function should apply all the functions in the first list to all the elements in the the second list, store the values in a list, and return the list. An example of the program is:
Main> apply [(^2),(^3),(^4)] [2..8]
--result: --[4,8,9,16,25,27,32,36,49,64,81,125,128,216,256,343,512,625,1296,2401,4096]
The type of the function must be:
apply :: Ord u => [v->u]->[v]->[u]
To do that I've used two helper functions and used recursion. My program is this:
apply :: Ord u => [v->u]->[v]->[u]
apply p s = myApply p s [] --line 59--
myApply :: Ord u => [v->u]->[u]->[u]->[u]
myApply f_list num_list starterlist
| null f_list = starterlist
| otherwise = myApply (tail f_list) (num_list) ( applyList (head f_list) num_list starterlist )
applyList :: Ord u => (v->u)->[u]->[u]->[u]
applyList f num_list starterlist
| null num_list = starterlist
| otherwise = applyList f (tail num_list) ( (head num_list) : starterlist )
I get the error:
ERROR "Lab2.hs":59 - Inferred type is not general enough
*** Expression : applyList
*** Expected type : Ord a => (b -> a) -> [a] -> [b] -> [a]
*** Inferred type : Ord a => (a -> a) -> [a] -> [a] -> [a]
Any idea what is wrong the with types?
The reason why you get this error is because there are conflicting type signatures:
apply :: Ord u => [v->u]->[v]->[u]
apply p s = myApply p s [] --line 59--
myApply :: Ord u => [v->u]->[u]->[u]->[u]
myApply f_list num_list starterlist
| null f_list = starterlist
| otherwise = myApply (tail f_list) (num_list) ( applyList (head f_list) num_list starterlist )
As you can see, the apply function immediately calls the myApply function. Since myApply has signature [v -> u] -> [u] -> [u] -> [u], it means that apply can only have signature [v->u] -> [u] -> [u].
A quick fix is to generalize both myApply and myApplyList to [v -> u] -> [v] -> [u] -> [u]. Now the compile will also detect an error you've made in your applyList function: you forgot to call f on the head num_list. So you can fix it and obtain the following code:
apply :: Ord u => [v->u]->[v]->[u]
apply p s = myApply p s [] --line 59--
myApply :: Ord u => [v->u]->[v]->[u]->[u]
myApply f_list num_list starterlist
| null f_list = starterlist
| otherwise = myApply (tail f_list) (num_list) ( applyList (head f_list) num_list starterlist )
applyList :: Ord u => (v->u)->[v]->[u]->[u]
applyList f num_list starterlist
| null num_list = starterlist
| otherwise = applyList f (tail num_list) ( (f (head num_list)) : starterlist )
Nevertheless, this code is quite unelegant and uses way to many functions and arguments. You can replace it entirely with a single list comprehension:
apply :: [v -> u] -> [v] -> [u]
apply fs xs = [f x | f <- fs, x <- xs]
Based on your comment you also have to sort the values later in the process, you can do this by using the sort :: Ord a => [a] -> [a] builtin:
-- sorting variant
apply :: Ord u => [v -> u] -> [v] -> [u]
apply fs xs = sort [f x | f <- fs, x <- xs]
This generates the required result:
Prelude Data.List> (\fs xs -> sort [f x | f <- fs, x <- xs]) [(^2),(^3),(^4)] [2..8]
[4,8,9,16,16,25,27,36,49,64,64,81,125,216,256,343,512,625,1296,2401,4096]
Variation on a theme: taking into Willem's comment indicating that this is a pretty convoluted code (making it difficult to analyze) I'd start with a simpler code, without stating any types at all, to first get a simple working case (that is, a function that produces the expected output), still using a helper function, and recursion:
apply fs [] = []
apply [] xs = []
apply (f:fs) xs = (helper f xs) ++ (apply fs xs)
helper f [] = []
helper f (x:xs) = f x : helper f xs
Then I would ask the Haskell compiler to give me information about what type signature it inferred:
*Main> :t apply
apply :: [t1 -> t] -> [t1] -> [t]
By realizing that t1 -> t maps to your v->u, I can see that the type signature can also be written as:
[v -> u] -> [v] -> [u]
Doing the similar for helper, you will end up with the following:
apply :: [v->u] -> [v] -> [u]
apply fs [] = []
apply [] xs = []
apply (f:fs) xs = (helper f xs) ++ (apply fs xs)
helper :: (v -> u) -> [v] -> [u]
helper f [] = []
helper f (x:xs) = f x : helper f xs
From here, you can work your way by adding Ord constraint, and build up sorting functionality, etc.
Related
Can some one tell me, why i am getting the following error:
error: [-Wincomplete-patterns, -Werror=incomplete-patterns]
Pattern match(es) are non-exhaustive
In a case alternative: Patterns not matched: []
|
54 | case list of
| ^^^^^^^^^^^^...
Thats my test:
testMinBy :: Test
testMinBy = TestCase $ do
assertEqual "test1" (minBy (\x -> -x) [1,2,3,4,5]) 5
assertEqual "test2" (minBy length ["a", "abcd", "xx"]) "a"
minBy :: Ord b => (a -> b) -> [a] -> a
minBy measure list =
case list of
(x:y:xs) -> minBy measure (if measure x > measure y then y:xs else x:xs)
[x] -> x
Your pattern does not matches with the empty list. Indeed, that is what the error is saying. You can match the empty list, for example with:
minBy :: Ord b => (a -> b) -> [a] -> a
minBy measure list =
case list of
(x:y:xs) -> minBy measure (if measure x > measure y then y:xs else x:xs)
[x] -> x
[] -> error "Empty list"
Your function however is not very efficient: it will recalculate measure multiple times if an item is the current minimum, and will also pack and unpack lists. You can work with an accumulator here, like:
minBy :: Ord b => (a -> b) -> [a] -> a
minBy _ [] = error "Empty list"
minBy f (x:xs) = go xs x (f x)
where go [] y _ = y
go (y₁:ys) y₀ fy₀
| fy₁ < fy₀ = go ys y₁ fy₁
| otherwise = go ys y₀ fy₀
where fy₁ = f y₁
This means it only once has to check for an empty list, and then knows for sure that this is a non-empty list if it enumerates. It also will determine the f of each item exactly once, and uses accumulators to avoid packing and unpacking a "cons".
I want to write a function that returns the longest prefix of a list, where applying a function to every item in that prefix produces a strictly ascending list.
For example:
longestAscendingPrefix (`mod` 5) [1..10] == [1,2,3,4]
longestAscendingPrefix odd [1,4,2,6,8,9,3,2,1] == [1]
longestAscendingPrefix :: Ord b => (a -> b) -> [a] -> [a]
longestAscendingPrefix _ [] = []
longestAscendingPrefix f (x:xs) = takeWhile (\y z -> f y <= f z) (x:xs)
This code snippet produces the error message in the title. It seems the problem lies within that lambda function.
takeWhile has type takeWhile :: (a -> Bool) -> [a] -> [a]. The first parameter is thus a function that maps an element of the list to a Bool. Your lambda expression has type Ord b => a -> a -> Bool, which does not make much sense.
You can work with explicit recursion with:
longestAscendingPrefix :: Ord b => (a -> b) -> [a] -> [a]
longestAscendingPrefix f = go
where go [] = []
go [x] = …
go (x1:x2:xs) = …
where you need to fill in the … parts the last one makes a recursive call to go.
I was wondering how to write my own sortOn function.
I made a sortBy function and an on function as shown bellow but can't figure out how to combine them and what additional code to add. sortOn is like sortBy but the given function (in here named comp) is applied only once for every element of the list
sortBy :: (a -> a -> Ordering) -> [a] -> [a]
sortBy comp [] = []
sortBy comp [x] = [x]
sortBy comp (x:xs) = insert x (sortBy comp xs)
where
insert x [] = [x]
insert x (y:ys)
| (comp x y == LT) || (comp x y == EQ) = x:y:ys
| otherwise = y:(insert x ys)
on :: (b -> b -> c) -> (a -> b) -> a -> a -> c
on b f x y = b (f x) (f y)
Here's a hint.
If you have a list [a] and you just sort it, the sort function will implicitly make use of the Ord instance for a and specifically the function:
compare :: a -> a -> Ordering
to figure out the relative ordering of pairs of a elements.
Now, if you have a list [a] and a transformation function b, and you want to use sortOn to sort the list of the transformed values, you'll need to figure out the relative ordering of pairs of b elements. How will you do this? Well, you'll implicitly use the Ord instance for b and specifically the function:
compare :: b -> b -> Ordering
In other words, when you try to define:
sortOn :: (Ord b) => (a -> b) -> [a] -> [a]
sortOn f lst = ...
you'll have arguments of type:
f :: a -> b
lst :: [a]
and additional objects of type:
sortBy :: (a -> a -> Ordering) -> [a] -> [a]
on :: (b -> b -> c) -> (a -> b) -> a -> a -> c
compare :: b -> b -> Ordering
Now, can you see how to put them together to define sortOn?
SPOILERS
Further hint: What's the type of compare `on` f?
Further further hint: It's a -> a -> Ordering.
I am currently taking a class in Haskell and am having a bit of trouble understanding how functions are passed as parameters. For this assignment, we were tasked with creating a program that would evaluate expressions. To reduce boiler plating, I wanted to abstract the function by creating a helper function that would take in an operator as an input and return the result
Main Function:
eval :: EDict -> Expr -> Maybe Double
eval _ (Val x) = Just x
eval d (Var i) = find d i
eval d (Add x y) = evalOp d (+) x y
eval d (Mul x y) = evalOp d (*) x y
eval d (Sub x y) = evalOp d (-) x y
Helper Function:
evalOp:: EDict -> ((Num a) => a -> a -> a) -> Expr -> Expr -> Maybe Double
evalOp d op x y =
let r = eval d x
s = eval d y
in case (r, s) of
(Just m, Just n) -> Just (m `op` n)
_ -> Nothing
Other definitions
data Expr
= Val Double
| Add Expr Expr
| Mul Expr Expr
| Sub Expr Expr
| Dvd Expr Expr
| Var Id
| Def Id Expr Expr
deriving (Eq, Show)
type Dict k d = [(k,d)]
define :: Dict k d -> k -> d -> Dict k d
define d s v = (s,v):d
find :: Eq k => Dict k d -> k -> Maybe d
find [] _ = Nothing
find ( (s,v) : ds ) name | name == s = Just v
| otherwise = find ds name
type EDict = Dict String Double
I looked into how +,-, and * are to be passed into other functions and found that these operators are defined by the following definition:
ghci> :t (*)
(*) :: (Num a) => a -> a -> a
However, when I run my code I get the following compilation error:
Illegal polymorphic or qualified type: Num a => a -> a -> a
Perhaps you intended to use RankNTypes or Rank2Types
In the type signature for ‘evalOp’:
evalOp :: EDict
-> ((Num a) => a -> a -> a) -> Expr -> Expr -> Maybe Double
I am not really sure why this is happening as I gave my function the proper parameters as defined by Haskell. Any help would be greatly appreciated as I am still very new to the language.
Right now, your Expr data type is constrained to Double-valued expressions, so there is no need to deal with polymorphism.
evalOp:: EDict -> (Double -> Double -> Double) -> Expr -> Expr -> Maybe Double
evalOp d op x y =
let r = eval d x
s = eval d y
in case (r, s) of
(Just m, Just n) -> Just (m `op` n)
_ -> Nothing
(+) :: Num a => a -> a -> a is a valid argument for evalOp, because its type can be "restricted" to Double -> Double -> Double.
> let f :: Double -> Double -> Double; f = (+)
> f 3 5
8.0
If your expression type were parameterized, then you would put a Num a constraint on your functions (not just on the arguments that involve a, because you want the same a throughout the function).
data Expr a
= Val a
| Add (Expr a) (Expr a)
| Mul (Expr a) (Expr a)
| Sub (Expr a) (Expr a)
| Dvd (Expr a) (Expr a)
| Var Id
| Def Id (Expr a) (Expr a)
deriving (Eq, Show)
type EDict a = Dict String a
evalOp:: Num a => EDict a -> (a -> a -> a) -> Expr a -> Expr a -> Maybe a
evalOp d op x y =
let r = eval d x
s = eval d y
in case (r, s) of
(Just m, Just n) -> Just (m `op` n)
_ -> Nothing
eval :: Num a => EDict a -> Expr a -> Maybe a
eval _ (Val x) = Just x
eval d (Var i) = find d i
eval d (Add x y) = evalOp d (+) x y
eval d (Mul x y) = evalOp d (*) x y
eval d (Sub x y) = evalOp d (-) x y
The error is telling you that you cannot nest a type qualifier inside one of the types in your function chain. Instead, put all of the qualifiers at the beginning of the type signature:
evalOp:: (Num a) => EDict -> (a -> a -> a) -> Expr -> Expr -> Maybe Double
See Haskell - Illegal Polymorphic type? for a more thorough discussion.
I'm attempting problem 11 of "99 Haskell Problems." The problem description is pretty much:
Write a function encodeModified that groups consecutive equal elements, then counts each group, and separates singles from runs.
For example:
Prelude> encodeModified "aaaabccaadeeee"
[Multiple 4 'a',Single 'b',Multiple 2 'c',
Multiple 2 'a',Single 'd',Multiple 4 'e']
Here's my working code:
module Batch2 where
import Data.List -- for `group`
data MultiElement a = Single a | Multiple Int a deriving (Show)
encodeModified :: (Eq a) => [a] -> [MultiElement a]
encodeModified = map f . group
where f xs = case length xs of 1 -> Single (head xs)
_ -> Multiple (length xs) (head xs)
I'd like to take out that pesky repeated (head xs) in the final two lines. I figured I could do so by treating the result of the case clause as a partially applied data constructor, as follows, but no luck:
encodeModified :: (Eq a) => [a] -> [MultiElement a]
encodeModified = map f . group
where f xs = case length xs of 1 -> Single
_ -> Multiple length xs
(head xs)
I also tried putting parenthese around the case clause itself, but to no avail. In that case, the case clause itself failed to compile (throwing an error upon hitting the _ symbol on the second line of the clause).
EDIT: this error was because I added a parenthesis but didn't add an extra space to the next line to make the indentation match. Thanks, raymonad.
I can also solve it like this, but it seems a little messy:
encodeModified :: (Eq a) => [a] -> [MultiElement a]
encodeModified = map (\x -> f x (head x)) . group
where f xs = case length xs of 1 -> Single
_ -> Multiple (length xs)
How can I do this?
The function application operator $ can be used to make this work:
encodeModified = map f . group
where f xs = case length xs of 1 -> Single
_ -> Multiple (length xs)
$ head xs
You could match on xs itself instead:
encodeModified :: (Eq a) => [a] -> [MultiElement a]
encodeModified = map f . group
where f xs = case xs of (x:[]) -> Single x
(x:_) -> Multiple (length xs) x
or more tersely as
encodeModified :: (Eq a) => [a] -> [MultiElement a]
encodeModified = map f . group
where f (x:[]) = Single x
f xs#(x:_) = Multiple (length xs) x
or even
encodeModified :: (Eq a) => [a] -> [MultiElement a]
encodeModified = map f . group
where f as#(x:xs) = case xs of [] -> Single x
_ -> Multiple (length as) x
Admittedly most of these have some repetition, but not of function application.
You could also go with let:
encodeModified :: (Eq a) => [a] -> [MultiElement a]
encodeModified = map f . group
where f xs = let x = head xs
len = length xs in
case len of 1 -> Single x
_ -> Multiple len x