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I'm having a hard time getting Haskell and functional programming together in my head. What I am trying to do is manipulate a string so that I am printing/returning specific characters each time based on a number given. For example:
printing "testing" 2 = "etn"
printing "testing" 3 = "sn"
I've read a lot online, and from what I understand I can achieve this with filtering and cycling, but I cannot get/understand the syntax of this language to get a working program.
I'll try to describe my thought process so you can follow. This function fits the pattern of creating an output list (here a string) from an input seed (here a string) by repeated function application (here dropping some elements). Thus I choose an implementation with Data.List.unfoldr.
unfoldr :: (b -> Maybe (a, b)) -> b -> [a]
Okay so, I need to turn the seed b into (Maybe) an output a and the rest of the string. I'll call this subfunction f and pass it into unfoldr.
printing s n = unfoldr f s
where f b = case drop n b of
[] -> Nothing
(x:xs) -> Just (x,xs)
It turns out that attempting to take the head off the front of the list and returning a Maybe is also a common pattern. It's Data.List.uncons, so
printing s n = unfoldr (uncons . drop n) s
Very smooth! So I test it out, and the output is wrong! Your specified output actually eg. for n=2 selects every 2nd character, ie. drops (n-1) characters.
printing s n = unfoldr (uncons . drop (n-1)) s
I test it again and it matches the desired output. Phew!
To demonstrate the Haskell language some alternative solutions to the accepted answer.
Using list comprehension:
printing :: Int -> String -> String
printing j ls = [s | (i, s) <- zip [1 .. ] ls, mod i j == 0]
Using recursion:
printing' :: Int -> String -> String
printing' n ls
| null ls' = []
| otherwise = x : printing' n xs
where
ls' = drop (n - 1) ls
(x : xs) = ls'
In both cases I flipped the arguments so it is easier to do partial application: printing 5 for example is a new function and will give each 5th character when applied to a string.
Note with a minor modification they will work for any list
takeEvery :: Int -> [a] -> [a]
I want to add two positive numbers together without the use of any basic operators like + for addition. I've already worked my way around that (in the add''' function) (i think) may not be efficient but thats not the point right now. I am getting lots of type errors however which i have no idea how to handle, and is very confusing for me as it works on paper and i've come from python.
add 1245 7489
--add :: Int -> Int -> Int
add x y = add'' (zip (add' x) (add' y))
where
add' :: Int -> [Int]
add' 0 = []
add' x = add' (x `div` 10) ++ [x `mod` 10]
conversion [1,2,4,5] [7,4,8,9] then zipping them together [(1,7),(2,4)....]
add'' :: [(Int,Int)] -> [Int]
add'' (x:xs) = [(add''' (head x) (last x))] ++ add'' xs
summary [8,6,...] what happens when the sum reaches 10 is not implemented yet.
where
--add''' :: (Int,Int) -> Int
add''' x y = last (take (succ y) $ iterate succ x)
adding two numbers together
You can't use head and last on tuples. ...Frankly, you should never use these functions at all because they're unsafe (partial), but they can be used on lists. In Haskell, lists are something completely different from tuples.To get at the elements of a tuple, use pattern matching.
add'' ((x,y):xs) = [add''' x y] ++ add'' xs
(To get at the elements of a list, pattern matching is very often the best too.) Alternatively, you can use fst and snd, these do on 2-tuples what you apparently thought head and last would.
Be clear which functions are curried and which aren't. The way you write add''', its type signature is actually Int -> Int -> Int. That is equivalent to (Int, Int) -> Int, but it's still not the same to the type checker.
The result of add'' is [Int], but you're trying to use this as Int in the result of add. That can't work, you need to translate from digits to numbers again.
add'' doesn't handle the empty case. That's fixed easily enough, but better than doing this recursion at all is using standard combinators. In your case, this is only supposed to work element-wise anyway, so you can simply use map – or do that right in the zipping, with zipWith. Then you also don't need to unwrap any tuples at all, because it works with a curried function.
A clean version of your attempt:
add :: Int -> Int -> Int
add x y = fromDigits 0 $ zipWith addDigits (toDigits x []) (toDigits y [])
where
fromDigits :: Int -> [Int] -> Int
fromDigits acc [] = acc
fromDigits acc (d:ds)
= acc `seq` -- strict accumulator, to avoid thunking.
fromDigits (acc*10 + d) ds
toDigits :: Int -> [Int] -> [Int] -- yield difference-list,
toDigits 0 = id -- because we're consing
toDigits x = toDigits (x`div`10) . ((x`mod`10):) -- left-associatively.
addDigits :: Int -> Int -> Int
addDigits x y = last $ take (succ x) $ iterate succ y
Note that zipWith requires both numbers to have the same number of digits (as does zip).
Also, yes, I'm using + in fromDigits, making this whole thing pretty futile. In practice you would of course use binary, then it's just a bitwise-or and the multiplication is a left shift. What you actually don't need to do here is take special care with 10-overflow, but that's just because of the cheat of using + in fromDigits.
By head and last you meant fst and snd, but you don't need them at all, the components are right there:
add'' :: [(Int, Int)] -> [Int]
add'' (pair : pairs) = [(add''' pair)] ++ add'' pairs
where
add''' :: (Int, Int) -> Int
add''' (x, y) = last (take (succ y) $ iterate succ x)
= iterate succ x !! y
= [x ..] !! y -- nice idea for an exercise!
Now the big question that remains is what to do with those big scary 10-and-over numbers. Here's a thought: produce a digit and a carry with
= ([(d, 0) | d <- [x .. 9]] ++ [(d, 1) | d <- [0 ..]]) !! y
Can you take it from here? Hint: reverse order of digits is your friend!
the official answer my professor gave
works on positive and negative numbers too, but still requires the two numbers to be the same length
add 0 y = y
add x y
| x>0 = add (pred x) (succ y)
| otherwise = add (succ x) (pred y)
The other answers cover what's gone wrong in your approach. From a theoretical perspective, though, they each have some drawbacks: they either land you at [Int] and not Int, or they use (+) in the conversion back from [Int] to Int. What's more, they use mod and div as subroutines in defining addition -- which would be okay, but then to be theoretically sound you would want to make sure that you could define mod and div themselves without using addition as a subroutine!
Since you say efficiency is no concern, I propose using the usual definition of addition that mathematicians give, namely: 0 + y = y, and (x+1) + y = (x + y)+1. Here you should read +1 as a separate operation than addition, a more primitive one: the one that just increments a number. We spell it succ in Haskell (and its "inverse" is pred). With this theoretical definition in mind, the Haskell almost writes itself:
add :: Int -> Int -> Int
add 0 y = y
add x y = succ (add (pred x) y)
So: compared to other answers, we can take an Int and return an Int, and the only subroutines we use are ones that "feel" more primitive: succ, pred, and checking whether a number is zero or nonzero. (And we land at only three short lines of code... about a third as long as the shortest proposed alternative.) Of course the price we pay is very bad performance... try add (2^32) 0!
Like the other answers, this only works for positive numbers. When you are ready for handling negative numbers, we should chat again -- there's some fascinating mathematical tricks to pull.
I'm trying to solve the following problem in Haskell: given an integer return the list of its digits. The constraint is I have to only use one of the fold* functions (* = {r,l,1,l1}).
Without such constraint, the code is simple:
list_digits :: Int -> [Int]
list_digits 0 = []
list_digits n = list_digits r ++ [n-10*r]
where
r = div n 10
But how do I use fold* to, essentially grow a list of digits from an empty list?
Thanks in advance.
Is this a homework assignment? It's pretty strange for the assignment to require you to use foldr, because this is a natural use for unfoldr, not foldr. unfoldr :: (b -> Maybe (a, b)) -> b -> [a] builds a list, whereas foldr :: (a -> b -> b) -> b -> [a] -> b consumes a list. An implementation of this function using foldr would be horribly contorted.
listDigits :: Int -> [Int]
listDigits = unfoldr digRem
where digRem x
| x <= 0 = Nothing
| otherwise = Just (x `mod` 10, x `div` 10)
In the language of imperative programming, this is basically a while loop. Each iteration of the loop appends x `mod` 10 to the output list and passes x `div` 10 to the next iteration. In, say, Python, this'd be written as
def list_digits(x):
output = []
while x > 0:
output.append(x % 10)
x = x // 10
return output
But unfoldr allows us to express the loop at a much higher level. unfoldr captures the pattern of "building a list one item at a time" and makes it explicit. You don't have to think through the sequential behaviour of the loop and realise that the list is being built one element at a time, as you do with the Python code; you just have to know what unfoldr does. Granted, programming with folds and unfolds takes a little getting used to, but it's worth it for the greater expressiveness.
If your assignment is marked by machine and it really does require you to type the word foldr into your program text, (you should ask your teacher why they did that and) you can play a sneaky trick with the following "id[]-as-foldr" function:
obfuscatedId = foldr (:) []
listDigits = obfuscatedId . unfoldr digRem
Though unfoldr is probably what the assignment meant, you can write this using foldr if you use foldr as a hylomorphism, that is, building up one list while it tears another down.
digits :: Int -> [Int]
digits n = snd $ foldr go (n, []) places where
places = replicate num_digits ()
num_digits | n > 0 = 1 + floor (logBase 10 $ fromIntegral n)
| otherwise = 0
go () (n, ds) = let (q,r) = n `quotRem` 10 in (q, r : ds)
Effectively, what we're doing here is using foldr as "map-with-state". We know ahead of time
how many digits we need to output (using log10) just not what those digits are, so we use
unit (()) values as stand-ins for those digits.
If your teacher's a stickler for just having a foldr at the top-level, you can get
away with making go partial:
digits' :: Int -> [Int]
digits' n = foldr go [n] places where
places = replicate num_digits ()
num_digits | n > 0 = floor (logBase 10 $ fromIntegral n)
| otherwise = 0
go () (n:ds) = let (q,r) = n `quotRem` 10 in (q:r:ds)
This has slightly different behaviour on non-positive numbers:
>>> digits 1234567890
[1,2,3,4,5,6,7,8,9,0]
>>> digits' 1234567890
[1,2,3,4,5,6,7,8,9,0]
>>> digits 0
[]
>>> digits' 0
[0]
>>> digits (negate 1234567890)
[]
>>> digits' (negate 1234567890)
[-1234567890]
I'm writing a program in Haskell that generates random triads of ints in order to play them as chords using Euterpea. Here's what my list of lists looks like (obviously, it's infinite, but I'll take 10 $ chop 3 $ randInts, where chop is just a function that splits a list up into (ordered) groupings of three and randPitch simply generates ints over the interval 24 - 84):
[[27,33,48],[53,59,82],[31,49,62],[50,60,63],[51,56,79],[57,59,60],[52,63,69],[52,77,79],[32,32,37],[39,77,84]]
To learn a bit about music theory, I'm applying various filters to these triads. The one that's giving me trouble (at least to implement in Haskell, which is a new language for me) is what Dmitri Tymoczko would call "conjunct melodic notation"--i.e., melodies (in this case the top 'note' or pitch of the 3-tuple) should only move short distances.
What I want to do is write a function that recursively filters the endless list of triads such that what results is a list of triads whose top note only moves by <= n between each chord: essentially, a kind of random walk where you are only allowed to move n semitones for any one step. Here are the functions I have so far:
jumpSize :: (Num a, Ord a) => [[a]] -> [[a]]
jumpSize (_:[]) = []
jumpSize (x:y:[]) = (abs (maximum x - maximum y) : x) : []
jumpSize (x:y:zs) = (abs (maximum x - maximum y) : x) : jumpSize (y:zs)
This will prepend the distance from triad x to y at the beginning of list x. So take 5 $ jumpSize $ chop 3 $ randPitch gives me:
[[34,27,33,48],[20,53,59,82],[1,31,49,62],[16,50,60,63],[19,51,56,79]]
I tried to write a recursive function that filters out all the chords with jumps of great than n, recalculates the new jumpSizes and then applies itself again. However, I'm having trouble because this function is either crashing GHCi or giving me all sorts of problems. Structurally, I would like to still produce the chords by filtering over the random array, but I think I'm missing something about how to best do this in a functional language. Maybe I need to supply an initial generator chord? Thanks!
jumpRecur :: (Num a, Ord a) => a -> [[a]] -> [[a]]
jumpRecur n (xs)
| [x | x <- xs, head x > n] == [] = xs
| otherwise = jumpRecur n $ jumpSize $ filter (\x -> head x <= n) xs
Here's a sample output:
*Main> jumpRecur 5 $ jumpSize $ chop 3 $ take 1000 $ randPitch
[[2,2,2,2,2,2,2,2,70,70,76],[1,1,1,1,1,1,1,1,55,74,74],[1,1,1,1,1,1,1,5,26,28,73], [3,3,3,3,3,3,0,1,26,69,74],[5,5,5,5,5,5,5,5,33,43,77],[0,0,0,0,0,0,0,0,47,67,82],[2,2,2,2,2,2,4,4,37,66,82],[3,3,3,3,3,3,3,3,59,69,84],[4,4,4,4,4,4,4,4,59,79,81],[4,4,5,5,5,5,5,5,28,69,77],[5,5,5,5,5,5,5,5,54,68,73],[0,0,0,0,0,0,0,0,32,73,78],[5,5,5,5,5,5,1,4,52,62,78],[0,0,0,0,0,0,0,0,58,71,73],[3,3,3,3,3,3,4,4,25,64,73],[1,1,1,1,1,3,3,3,35,42,76],[4,4,4,4,4,4,4,4,35,39,77]]
It's probably good Haskell idiom to generalize the filter, as follows:
jumpBy :: (a -> a -> Bool) -> [a] -> [a]
jumpBy ok (x:y:zs) | ok x y = x:jumpBy ok (y:zs) -- accept y & step
| otherwise = jumpBy ok (x:zs) -- reject y & retry
jumpBy _ xs = xs
If you are sure all input will be infinite lists, you don't really need that last line which handles the ends of finite lists, but it's good practice to handle all cases.
Then, use your general filter with your specific test:
jumpSize = jumpBy (\x y -> abs(maximum x - maximum y) < n)
The question is to compute the mode (the value that occurs most frequently) of a sorted list of integers.
[1,1,1,1,2,2,3,3] -> 1
[2,2,3,3,3,3,4,4,8,8,8,8] -> 3 or 8
[3,3,3,3,4,4,5,5,6,6] -> 3
Just use the Prelude library.
Are the functions filter, map, foldr in Prelude library?
Starting from the beginning.
You want to make a pass through a sequence and get the maximum frequency of an integer.
This sounds like a job for fold, as fold goes through a sequence aggregating a value along the way before giving you a final result.
foldl :: (a -> b -> a) -> a -> [b] -> a
The type of foldl is shown above. We can fill in some of that already (I find that helps me work out what types I need)
foldl :: (a -> Int -> a) -> a -> [Int] -> a
We need to fold something through that to get the value. We have to keep track of the current run and the current count
data BestRun = BestRun {
currentNum :: Int,
occurrences :: Int,
bestNum :: Int,
bestOccurrences :: Int
}
So now we can fill in a bit more:
foldl :: (BestRun -> Int -> BestRun) -> BestRun -> [Int] -> BestRun
So we want a function that does the aggregation
f :: BestRun -> Int -> BestRun
f (BestRun current occ best bestOcc) x
| x == current = (BestRun current (occ + 1) best bestOcc) -- continuing current sequence
| occ > bestOcc = (BestRun x 1 current occ) -- a new best sequence
| otherwise = (BestRun x 1 best bestOcc) -- new sequence
So now we can write the function using foldl as
bestRun :: [Int] -> Int
bestRun xs = bestNum (foldl f (BestRun 0 0 0 0) xs)
Are the functions filter, map, foldr in Prelude library?
Stop...Hoogle time!
Did you know Hoogle tells you which module a function is from? Hoolging map results in this information on the search page:
map :: (a -> b) -> [a] -> [b]
base Prelude, base Data.List
This means map is defined both in Prelude and in Data.List. You can hoogle the other functions and likewise see that they are indeed in Prelude.
You can also look at Haskell 2010 > Standard Prelude or the Prelude hackage docs.
So we are allowed to map, filter, and foldr, as well as anything else in Prelude. That's good. Let's start with Landei's idea, to turn the list into a list of lists.
groupSorted :: [a] -> [[a]]
groupSorted = undefined
-- groupSorted [1,1,2,2,3,3] ==> [[1,1],[2,2],[3,3]]
How are we supposed to implement groupSorted? Well, I dunno. Let's think about that later. Pretend that we've implemented it. How would we use it to get the correct solution? I'm assuming it is OK to choose just one correct solution, in the event that there is more than one (as in your second example).
mode :: [a] -> a
mode xs = doSomething (groupSorted xs)
where doSomething :: [[a]] -> a
doSomething = undefined
-- doSomething [[1],[2],[3,3]] ==> 3
-- mode [1,2,3,3] ==> 3
We need to do something after we use groupSorted on the list. But what? Well...we should find the longest list in the list of lists. Right? That would tell us which element appears the most in the original list. Then, once we find the longest sublist, we want to return the element inside it.
chooseLongest :: [[a]] -> a
chooseLongest xs = head $ chooseBy (\ys -> length ys) xs
where chooseBy :: ([a] -> b) -> [[a]] -> a
chooseBy f zs = undefined
-- chooseBy length [[1],[2],[3,3]] ==> [3,3]
-- chooseLongest [[1],[2],[3,3]] ==> 3
chooseLongest is the doSomething from before. The idea is that we want to choose the best list in the list of lists xs, and then take one of its elements (its head does just fine). I defined this by creating a more general function, chooseBy, which uses a function (in this case, we use the length function) to determine which choice is best.
Now we're at the "hard" part. Folds. chooseBy and groupSorted are both folds. I'll step you through groupSorted, and leave chooseBy up to you.
How to write your own folds
We know groupSorted is a fold, because it consumes the entire list, and produces something entirely new.
groupSorted :: [Int] -> [[Int]]
groupSorted xs = foldr step start xs
where step :: Int -> [[Int]] -> [[Int]]
step = undefined
start :: [[Int]]
start = undefined
We need to choose an initial value, start, and a stepping function step. We know their types because the type of foldr is (a -> b -> b) -> b -> [a] -> b, and in this case, a is Int (because xs is [Int], which lines up with [a]), and the b we want to end up with is [[Int]].
Now remember, the stepping function will inspect the elements of the list, one by one, and use step to fuse them into an accumulator. I will call the currently inspected element v, and the accumulator acc.
step v acc = undefined
Remember, in theory, foldr works its way from right to left. So suppose we have the list [1,2,3,3]. Let's step through the algorithm, starting with the rightmost 3 and working our way left.
step 3 start = [[3]]
Whatever start is, when we combine it with 3 it should end up as [[3]]. We know this because if the original input list to groupSorted were simply [3], then we would want [[3]] as a result. However, it isn't just [3]. Let's pretend now that it's just [3,3]. [[3]] is the new accumulator, and the result we would want is [[3,3]].
step 3 [[3]] = [[3,3]]
What should we do with these inputs? Well, we should tack the 3 onto that inner list. But what about the next step?
step 2 [[3,3]] = [[2],[3,3]]
In this case, we should create a new list with 2 in it.
step 1 [[2],[3,3]] = [[1],[2],[3,3]]
Just like last time, in this case we should create a new list with 1 inside of it.
At this point we have traversed the entire input list, and have our final result. So how do we define step? There appear to be two cases, depending on a comparison between v and acc.
step v acc#((x:xs):xss) | v == x = (v:x:xs) : xss
| otherwise = [v] : acc
In one case, v is the same as the head of the first sublist in acc. In that case we prepend v to that same sublist. But if such is not the case, then we put v in its own list and prepend that to acc. So what should start be? Well, it needs special treatment; let's just use [] and add a special pattern match for it.
step elem [] = [[elem]]
start = []
And there you have it. All you have to do to write your on fold is determine what start and step are, and you're done. With some cleanup and eta reduction:
groupSorted = foldr step []
where step v [] = [[v]]
step v acc#((x:xs):xss)
| v == x = (v:x:xs) : xss
| otherwise = [v] : acc
This may not be the most efficient solution, but it works, and if you later need to optimize, you at least have an idea of how this function works.
I don't want to spoil all the fun, but a group function would be helpful. Unfortunately it is defined in Data.List, so you need to write your own. One possible way would be:
-- corrected version, see comments
grp [] = []
grp (x:xs) = let a = takeWhile (==x) xs
b = dropWhile (==x) xs
in (x : a) : grp b
E.g. grp [1,1,2,2,3,3,3] gives [[1,1],[2,2],[3,3,3]]. I think from there you can find the solution yourself.
I'd try the following:
mostFrequent = snd . foldl1 max . map mark . group
where
mark (a:as) = (1 + length as, a)
mark [] = error "cannot happen" -- because made by group
Note that it works for any finite list that contains orderable elements, not just integers.