I have a function called bounds1Accum. this function works great
bounds1Accum :: [Integer] -> Integer -> Integer -> (Integer, Integer)
bounds1Accum l min max = (minimum (min:l), maximum (max:l))
What I am having problems with is I need another function called bounds1 which will take in just a list and return a min max pair. This function is supposed to wrap around bounds1accum and check to first see if the list being passed to it is empty or not. if the list is empty it will return Nothing or it will Just return the result of calling the bounds1Accum function.
I need two cases and for my signature I have
bounds1 :: [l] -> Maybe(min,max)
bounds1 [] = Nothing (which I am unsure if this will be correct)
then the 2nd case is where I am stumped
I originally had
bounds1 l min max = if null l then Nothing else Just $ bounds1 bounds1Accum min max
but this does not even compile so if anyone can offer suggestions or another way I can look at this problem it would be great. This is for an assignment so I do not want the answer but more so just guidance or help solving the problem.
Thank you
In the second case, you do not have min and max as arguments. You have to take those values from the list.
The if is unnecessary, since you already handle the empty list in the first case.
And why is there bounds1 before the use of bounds1Accum?
Related
I'm currently working on an assignment. I have a function called gamaTipo that converts the values of a tuple into a data type previously defined by my professor.
The problem is: in order for gamaTipo to work, it needs to receive some preceding element. gamaTipo is defined like this: gamaTipo :: Peca -> (Int,Int) -> Peca where Peca is the data type defined by my professor.
What I need to do is to create a funcion that takes a list of tuples and converts it into Peca data type. The part that im strugling with is taking the preceding element of the list. i.e : let's say we have a list [(1,2),(3,4)] where the first element of the list (1,2) always corresponds to Dirt Ramp (data type defined by professor). I have to create a function convert :: [(Int,Int)] -> [Peca] where in order to calculate the element (3,4) i need to first translate (1,2) into Peca, and use it as the previous element to translate (3,4)
Here's what I've tried so far:
updateTuple :: [(Int,Int)] -> [Peca]
updateTuple [] = []
updateTuple ((x,y):xs) = let previous = Dirt Ramp
in (gamaTipo previous (x,y)): updateTuple xs
Although I get no error messages with this code, the expected output isn't correct. I'm also sorry if it's not easy to understand what I'm asking, English isn't my native tongue and it's hard to express my self. Thank you in advance! :)
If I understand correctly, your program needs to have a basic structure something like this:
updateTuple :: [(Int, Int)] -> [Peca]
updateTuple = go initialValue
where
go prev (xy:xys) =
let next = getNextValue prev xy
in prev : (go next xys)
go prev [] = prev
Basically, what’s happening here is:
updateTuple is defined in terms of a helper function go. (Note that ‘helper function’ isn’t standard terminology, it’s just what I’ve decided to call it).
go has an extra argument, which is used to store the previous value.
The implementation of go can then make use of the previous value.
When go recurses, the recursive call can then pass the newly-calculated value as the new ‘previous value’.
This is a reasonably common pattern in Haskell: if a recursive function requires an extra argument, then a new function (often named go) can be defined which has that extra argument. Then the original function can be defined in terms of go.
I'd greatly appreciate if you could tell me how to make a single string from a range between two ints. Like [5..10] i would need to get a "5678910". And then I'd have to calculate how many (zeroes, ones ... nines) there are in a string.
For example: if i have a range from [1..10] i'd need to print out
1 2 1 1 1 1 1 1 1 1
For now i only have a function to search for a element in string.
`countOfElem elem list = length $ filter (\x -> x == elem) list`
But the part how to construct such a string is bugging me out, or maybe there is an easier way? Thank you.
I tried something like this, but it wouldn't work.
let intList = map (read::Int->String) [15..22]
I tried something like this, but it wouldn't work. let intList = map (read::Int->String) [15..22]
Well... the purpose of read is to parse strings to read-able values. Hence it has a type signature String -> a, which obviously doesn't unify with Int -> String. What you want here is the inverse1 of read, it's called show.
Indeed map show [15..22] gives almost the result you asked for – the numbers as decimal-encoded strings – but still each number as a seperate list element, i.e. type [String] while you want only String. Well, how about asking Hoogle? It gives the function you need as the fifth hit: concat.
If you want to get fancy you can then combine the map and concat stages: both the concatMap function and the >>= operator do that. The most compact way to achieve the result: [15..22]>>=show.
1show is only the right inverse of read, to be precise.
I'm trying to write a function in Haskell that calculates all factors of a given number except itself.
The result should look something like this:
factorlist 15 => [1,3,5]
I'm new to Haskell and the whole recursion subject, which I'm pretty sure I'm suppoused to apply in this example but I don't know where or how.
My idea was to compare the given number with the first element of a list from 1 to n div2
with the mod function but somehow recursively and if the result is 0 then I add the number on a new list. (I hope this make sense)
I would appreciate any help on this matter
Here is my code until now: (it doesn't work.. but somehow to illustrate my idea)
factorList :: Int -> [Int]
factorList n |n `mod` head [1..n`div`2] == 0 = x:[]
There are several ways to handle this. But first of all, lets write a small little helper:
isFactorOf :: Integral a => a -> a -> Bool
isFactorOf x n = n `mod` x == 0
That way we can write 12 `isFactorOf` 24 and get either True or False. For the recursive part, lets assume that we use a function with two arguments: one being the number we want to factorize, the second the factor, which we're currently testing. We're only testing factors lesser or equal to n `div` 2, and this leads to:
createList n f | f <= n `div` 2 = if f `isFactorOf` n
then f : next
else next
| otherwise = []
where next = createList n (f + 1)
So if the second parameter is a factor of n, we add it onto the list and proceed, otherwise we just proceed. We do this only as long as f <= n `div` 2. Now in order to create factorList, we can simply use createList with a sufficient second parameter:
factorList n = createList n 1
The recursion is hidden in createList. As such, createList is a worker, and you could hide it in a where inside of factorList.
Note that one could easily define factorList with filter or list comprehensions:
factorList' n = filter (`isFactorOf` n) [1 .. n `div` 2]
factorList'' n = [ x | x <- [1 .. n`div` 2], x `isFactorOf` n]
But in this case you wouldn't have written the recursion yourself.
Further exercises:
Try to implement the filter function yourself.
Create another function, which returns only prime factors. You can either use your previous result and write a prime filter, or write a recursive function which generates them directly (latter is faster).
#Zeta's answer is interesting. But if you're new to Haskell like I am, you may want a "simple" answer to start with. (Just to get the basic recursion pattern...and to understand the indenting, and things like that.)
I'm not going to divide anything by 2 and I will include the number itself. So factorlist 15 => [1,3,5,15] in my example:
factorList :: Int -> [Int]
factorList value = factorsGreaterOrEqual 1
where
factorsGreaterOrEqual test
| (test == value) = [value]
| (value `mod` test == 0) = test : restOfFactors
| otherwise = restOfFactors
where restOfFactors = factorsGreaterOrEqual (test + 1)
The first line is the type signature, which you already knew about. The type signature doesn't have to live right next to the list of pattern definitions for a function, (though the patterns themselves need to be all together on sequential lines).
Then factorList is defined in terms of a helper function. This helper function is defined in a where clause...that means it is local and has access to the value parameter. Were we to define factorsGreaterOrEqual globally, then it would need two parameters as value would not be in scope, e.g.
factorsGreaterOrEqual 4 15 => [5,15]
You might argue that factorsGreaterOrEqual is a useful function in its own right. Maybe it is, maybe it isn't. But in this case we're going to say it isn't of general use besides to help us define factorList...so using the where clause and picking up value implicitly is cleaner.
The indentation rules of Haskell are (to my tastes) weird, but here they are summarized. I'm indenting with two spaces here because it grows too far right if you use 4.
Having a list of boolean tests with that pipe character in front are called "guards" in Haskell. I simply establish the terminal condition as being when the test hits the value; so factorsGreaterOrEqual N = [N] if we were doing a call to factorList N. Then we decide whether to concatenate the test number into the list by whether dividing the value by it has no remainder. (otherwise is a Haskell keyword, kind of like default in C-like switch statements for the fall-through case)
Showing another level of nesting and another implicit parameter demonstration, I added a where clause to locally define a function called restOfFactors. There is no need to pass test as a parameter to restOfFactors because it lives "in the scope" of factorsGreaterOrEqual...and as that lives in the scope of factorList then value is available as well.
I have an Entry data type
data Entry = Entry {
count :: Integer,
name :: String }
Then I want to write a function, that takes the name and a list of Entrys as arguments an give me the Entrys with the highest count. What I have so far is
searchEntry :: String -> [Entry] -> Maybe Integer
searchEntry _ [] = Nothing
searchEntry name1 (x:xs) =
if name x == name1
then Just (count x)
else searchEntry name xs
That gives me the FIRST Entry that the function finds, but I want the Entry with the highest count. How can I implement that?
My suggestion would be to break the problem into two parts:
Find all entries matching a given name
Find the entry with the highest count
You could set it up as
entriesByName :: String -> [Entry] -> [Entry]
entriesByName name entries = undefined
-- Use Maybe since the list might be empty
entryWithHighestCount :: [Entry] -> Maybe Entry
entryWithHighestCount entries = undefined
entryByNameWithHighestCount :: String -> [Entry] -> Maybe Entry
entryByNameWithHighestCount name entires = entryWithHighestCount $ entriesByName name entries
All you have to do is implement the relatively simple functions that are used to implement getEntryByNameWithHighestCount.
You need to add an inner method that takes a current result as a parameter and returns that instead of Nothing when reaching the end of the method.
Also you would need to update your result found logic to compare a potentially existing function and the found value.
I would consider changing the signature of the function to String->Maybe Entry (or String->[Entry]) if you indeed want to return the "Entry" items with the highest count.
Otherwise, you can actually do what you want as a oneliner using some pretty common Haskell functions....
As Bheklilr mentioned, the name filter can be done first, and it is really easy to do this using the filter function....
filter (hasName theName) entries
Note that hasName can be written out fully as a separate function, but Haskell also offers you the following shortcut.
hasName = (== theName) . name
Now you just need the maximum value.... Haskell has a maximum function, but it only works on the Ord class. You can make Entry an instance of Ord, or you can just use the related maximumBy function, that takes an extra ordering function
maximumBy orderFunction entries2
Again, you can write orderFunction yourself (which you might want to do as an excercise), but haskell again offers a shortcut.
orderFunction = compare `on` count
You will need to import some libs to get this all to work (Data.Function, Data.List). You also will need to put in some extra code to account for the Nothing case.
It might be worth it to write out the functions longhand first, but I recommend that you use Hoogle to lookup and understand compare, on, and maximumBy.... Using tricks like this can really shorten your code.
Putting it all together, you can get the entry with the maximum count like this
maxEntry = maximumBy (compare `on` count) $ filter ((theName ==) . name) $ entries
You will need to modify this to account for the Nothing case, or if you want to return all max Entries (this just chooses one), or if you really wanted to return count, and not the entry.
I am trying to convert a given decimal value its corresponding binary form. I am using Ocaml about which I don't know much and am quite confused. So far I have the following code
let dec_to_bin_helper function 1->'T' | 0->'F'
let dec_to_bin x =
List.fold_left(fun a z -> z mod 2 dec_to_bin_helper a) [] a ;;
I must include here that I want my output to be in the form of a list of T's and F's where T's represent the binary 1's and F's represent binary 0's
If I try to run the above code it gives me an error saying "Error: This expression is not a function; it cannot be applied"
I understand that the part where I am calling the helper function is wrong... Any help in the matter would be appreciated!
I don't really understand your second function at all. You are folding an empty list, and your function takes an argument x which it never uses. Am I correct in assuming that you want to take a number and return a list of 'T's and 'F's which represent the binary? If that is the case, this code should work:
let dec_to_bin x =
let rec d2b y lst = match y with 0 -> lst
| _ -> d2b (y/2) ((dec_to_bin_helper (y mod 2))::lst)
in
d2b x [];;
This function inserts (x mod 2) converted into a T/F into a list, then recursively calls the function on x/2 and the list. When x = 0 the list is returned. If call it on 0 an empty list will be returned (I'm not sure if that's what you want or not).
I think the problem that you had is that you are treating lists as if they are mutable and thinking that fold mutates the list. That is not the case, fold just goes through each element in a list and applies a function to it. Since your list is empty it didn't do anything.