So i've got a list of tuples like this one :
xs = [("a","b"),("a","c"),("b","d")
and i want a function that counts the number of times a certain value appears in the first position of the tuple. If i used the list xs and the letter 'a', it would return the value 2, because the letter 'a' appears two times in the first position of the tuple. This function shouldn't be recursive.
So what i've got is this:
f xs = (fst $ unzip xs) / length(xs)
Now i have all the elements down on a list. this would be easy if it was recursive, but if i don't want it that way, how can i do it ?
If we're not using recursion, we need to use some higher order functions. In particular, filter looks helpful, it removes elements who don't satisfy some condition.
Well if we use filter we can get a list of all elements with the first element being the correct thing.
count :: Eq a => [(a, b)] -> Int
count x = length . filter ((== x) . fst)
I suppose since you're studying, you should work to understand some folds, start with
count x = foldr step 0
where step (a, b) r | a == x = 1 + r
| otherwise = r
If you map the first elements into a list find all occurences of your value and count the length of the resulting list:
countOccurences :: Eq a => a -> [(a, b)] -> Int
countOccurences e = length . filter ((==)e) . map fst
Related
I'm trying to make it so that on a tuple input (n,m) and a list of tuples xs , if the first item in the tuple in xs is in (n,m) then keep it that way in the new list otherwise add the a tuple consisting of some value k from n to m as a first element and as second element it should be 0.My question is:how can i say "repeat 0" using guards ? Since clearly my code won't run since my code says "repeat = 0"
expand :: (Int,Int) -> Profile ->Profile
expand (n,m) [] = zip [n..m] (repeat 0)
expand (n,m) (x:xs) = zip [n..m] (repeat (|(fst (x) `elem` [n..m]) == False = 0
|otherwise = snd (x))
You can use a helper function here that converts a number in the [ n .. m ] range to a 2-tuple. Here we thus try to find an element in the list xs that matches with the first item of that tuple, if we do not find such element, we use 0:
import Data.List(find)
expand :: (Int,Int) -> Profile -> Profile
expand (n,m) xs = map go [n .. m]
where go i | Just l <- find (\(f, _) -> f == i) xs = l
| otherwise = (i, 0)
For a list, find was implemented as [src]:
find :: (a -> Bool) -> [a] -> Maybe a
find p = listToMaybe . filter p
filter thus will make a list that contains the elements that satisfy the predicate p, and listToMaybe :: [a] -> Maybe a will convert an empty list [] to Nothing, and for a non-empty list (x:_) it will wrap the first element x in a Just data constructor. Due to Haskell's laziness, it will thus look for the first element that satisfies the predicate.
this thus gives us:
Prelude Data.List> expand (2,7) [(4, 2.3), (6, 3)]
[(2,0.0),(3,0.0),(4,2.3),(5,0.0),(6,3.0),(7,0.0)]
I have the following Haskell code, which returns the an instance of the requested element from the list (just the first instance) or returns 0 if the element is not present. I am trying to return the position of the element in the list instead but don't know how to (I am very new to programming).
e.g. 'm' 'some' is returning m but I want it to return 3.
findMe x [] = 0
findMe x (y:ys) = if x == y
then x
else findMe x ys
Try this. If 0 is returned because element not present, I assume that it is 1 index start.
findMe x list = findMeP list 1
where
findMeP [] _ = 0
findMeP (y:ys) n
| x == y = n
| otherwise = findMeP ys (n + 1)
As a demo, take a look at this.
Although the previous answer is correct, here is a solution that I'd prefer since it makes use of basic building blocks from the base library and avoids explicit recursion.
import Data.List (find) {- for find :: (a -> Bool) -> [a] -> Maybe a -}
findMe :: Eq a => a -> [a] -> Maybe (Int, a)
findMe x = find ((== x).snd) . zip [1..]
The function returns both the element and its index, counting from zero, wrapped in a Maybe type. If the element is not found, Nothing is returned.
To extract the element or the index alone one can map the fst and snd functions over the Maybe result:
findMeIndex = fmap fst . findMe
findMeElem = fmap snd . findMe
Example:
findMe 'c' ['a','b','c','d'] == Just (2,'c')
findMe 'z' ['a','b','c','d'] == Nothing
findMeIndex 'c' ['a','b','c','d'] == Just 2
findMeIndex 'z' ['a','b','c','d'] == Nothing
If you need to start counting from 1, you can replace [0..] with [1..]. In either case, wrapping the result in a Maybe is preferred to returning some special value signaling the absence of the element (say -1 or 0), because then the user cannot ever mistake to interpret your result. For example, if you count from 1 and return 0 on failure, and someone uses your function erroneously thinking that you count from 0, they may interpret a failure as if the element was found at the first position. Here instead, the user is forced to handle the failure case explicitly.
The function works as follows. zip [0..] produces a list of pairs coupling each element with its index, starting from zero ([0..] is the infinite list [0,1,2,3,..]). Then, find scans the list (in exactly the same way as OP's original code), returning the first element for which the function ((== x).snd) returns True, wrapped in a Maybe type, if found, or Nothing otherwise. Which element does find look for? Remember that find is fed with a list of pairs. So by composing the snd function with (== x) we find the pair whose second component is equal to x
In Haskell I need to perform a function, whose declaration of types is as follows:
split ::[Integer] -> Maybe ([Integer],[Integer])
Let it work as follows:
split [1,2,3,4,5,15] = Just ([1,2,3,4,5],[15])
Because, 1 + 2 + 3 + 4 + 5 = 15
split [1,3,3,4,3] = Just ([1,3,3],[4,3])
Because, 1 + 3 + 3 = 7 = 4 + 3
split [1,5,7,8,0] = Nothing
I have tried this, but it doesn't work:
split :: [Integer] -> ([Integer], [Integer])
split xs = (ys, zs)
where
ys <- subsequences xs, ys isInfixOf xs, sum ys == sum zs
zs == xs \\ ys
Determines whether the list of positive integers xs can be divided into two parts (without rearranging its elements) with the same sum. If possible, its value is the pair formed by the two parts. If it's not, its value is Nothing.
How can I do it?
Not a complete answer, since this is a learning exercise and you want hints, but if you want to use subsequences from Data.List, you could then remove each element of the subsequence you are checking from the original list with \\, to get the difference, and compare the sums. You were on the right track, but you need to either find the first subsequence that works and return Just (ys, zs), or else Nothing.
You can make the test for some given subsequence a predicate and search with find.
What you could also do is create a function that gives all possible splittings of a list:
splits :: [a] -> [([a], [a])]
splits xs = zipWith splitAt [1..(length xs)-1] $ repeat xs
Which works as follows:
*Main> splits [1,2,3,4,5,15]
[([1],[2,3,4,5,15]),([1,2],[3,4,5,15]),([1,2,3],[4,5,15]),([1,2,3,4],[5,15]),([1,2,3,4,5],[15])]
Then you could just use find from Data.List to find the first pair of splitted lists that have equal sums:
import Data.List
splitSum :: [Integer] -> Maybe ([Integer], [Integer])
splitSum xs = find (\(x, y) -> sum x == sum y) $ splits xs
Which works as intended:
*Main> splitSum [1,2,3,4,5,15]
Just ([1,2,3,4,5],[15])
Since find returns Maybe a, the types automatically match up.
The full practice exam question is:
Using anonymous functions and mapping functions, define Haskell
functions which return the longest String in a list of Strings, e.g.
for [“qw”, “asd”,”fghj”, “kl”] the function should return “fghj”.
I tried doing this and keep failing and moving onto others, but I would really like to know how to tackle this. I have to use mapping functions and anonymous functions it seems, but I don't know how to write code to make each element check with each to find the highest one.
I know using a mapping function like "foldr" can make you perform repeating operations to each element and return one result, which is what we want to do with this question (check each String in the list of Strings for the longest, then return one string).
But with foldr I don't know how to use it to make checks between elments to see which is "longest"... Any help will be gladly appreciated.
So far I've just been testing if I can even use foldr to test the length of each element but it doesn't even work:
longstr :: [String] -> String
longstr lis = foldr (\n -> length n > 3) 0 lis
I'm quite new to haskell as this is a 3 month course and it's only been 1 month and we have a small exam coming up
I'd say they're looking for a simple solution:
longstr xs = foldr (\x acc -> if length x > length acc then x else acc) "" xs
foldr is like a loop that iterates on every element of the list xs. It receives 2 arguments: x is the element and acc (for accumulator) in this case is the longest string so far.
In the condition if the longest string so far is longer than the element we keep it, otherwise we change it.
Another idea:
Convert to a list of tuples: (length, string)
Take the maximum of that list (which is some pair).
Return the string of the pair returned by (2).
Haskell will compare pairs (a,b) lexicographically, so the pair returned by (2) will come from the string with largest length.
Now you just have to write a maximum function:
maximum :: Ord a => [a] -> a
and this can be written using foldr (or just plain recursion.)
To write the maximum function using recursion, fill in the blanks:
maximum [a] = ??? -- maximum of a single element
maximum (a:as) = ??? -- maximum of a value a and a list as (hint: use recursion)
The base case for maximum begins with a single element list since maximum [] doesn't make sense here.
You can map the list to a list of tuples, consisting of (length, string). Sort by length (largest first) and return the string of the first element.
https://stackoverflow.com/a/9157940/127059 has an answer as well.
Here's an example of building what you want from the bottom up.
maxBy :: Ord b => (a -> b) -> a -> a -> a
maxBy f x y = case compare (f x) (f y) of
LT -> y
_ -> x
maximumBy :: Ord b => (a -> b) -> [a] -> Maybe a
maximumBy _ [] = Nothing
maximumBy f l = Just . fst $ foldr1 (maxBy snd) pairs
where
pairs = map (\e -> (e, f e)) l
testData :: [String]
testData = ["qw", "asd", "fghj", "kl"]
test :: Maybe String
test = maximumBy length testData
main :: IO ()
main = print test
How do I manually split [1,2,4,5,6,7] into [[1],[2],[3],[4],[5],[6],[7]]? Manually means without using break.
Then, how do I split a list into sublists according to a predicate? Like so
f even [[1],[2],[3],[4],[5],[6],[7]] == [[1],[2,3],[4,5],[6,7]]
PS: this is not homework, and I've tried for hours to figure it out on my own.
To answer your first question, this is rather an element-wise transformation than a split. The appropriate function to do this is
map :: (a -> b) -> [a] -> [b]
Now, you need a function (a -> b) where b is [a], as you want to transform an element into a singleton list containing the same type. Here it is:
mkList :: a -> [a]
mkList a = [a]
so
map mkList [1,2,3,4,5,6,7] == [[1],[2],...]
As for your second question: If you are not allowed (homework?) to use break, are you then allowed to use takeWhile and dropWhile which form both halves of the result of break.
Anyway, for a solution without them ("manually"), just use simple recursion with an accumulator:
f p [] = []
f p (x:xs) = go [x] xs
where go acc [] = [acc]
go acc (y:ys) | p y = acc : go [y] ys
| otherwise = go (acc++[y]) ys
This will traverse your entire list tail recursively, always remembering what the current sublist is, and when you reach an element where p applies, outputting the current sublist and starting a new one.
Note that go first receives [x] instead of [] to provide for the case where the first element already satisfies p x and we don't want an empty first sublist to be output.
Also, this operates on the original list ([1..7]) instead of [[1],[2]...]. But you can use it on the transformed one as well:
> map concat $ f (odd . head) [[1],[2],[3],[4],[5],[6],[7]]
[[1,2],[3,4],[5,6],[7]]
For the first, you can use a list comprehension:
>>> [[x] | x <- [1,2,3,4,5,6]]
[[1], [2], [3], [4], [5], [6]]
For the second problem, you can use the Data.List.Split module provided by the split package:
import Data.List.Split
f :: (a -> Bool) -> [[a]] -> [[a]]
f predicate = split (keepDelimsL $ whenElt predicate) . concat
This first concats the list, because the functions from split work on lists and not list of lists. The resulting single list is the split again using functions from the split package.
First:
map (: [])
Second:
f p xs =
let rs = foldr (\[x] ~(a:r) -> if (p x) then ([]:(x:a):r) else ((x:a):r))
[[]] xs
in case rs of ([]:r) -> r ; _ -> rs
foldr's operation is easy enough to visualize:
foldr g z [a,b,c, ...,x] = g a (g b (g c (.... (g x z) ....)))
So when writing the combining function, it is expecting two arguments, 1st of which is "current element" of a list, and 2nd is "result of processing the rest". Here,
g [x] ~(a:r) | p x = ([]:(x:a):r)
| otherwise = ((x:a):r)
So visualizing it working from the right, it just adds into the most recent sublist, and opens up a new sublist if it must. But since lists are actually accessed from the left, we keep it lazy with the lazy pattern, ~(a:r). Now it works even on infinite lists:
Prelude> take 9 $ f odd $ map (:[]) [1..]
[[1,2],[3,4],[5,6],[7,8],[9,10],[11,12],[13,14],[15,16],[17,18]]
The pattern for the 1st argument reflects the peculiar structure of your expected input lists.