January 2021 - Beginners

GHCi Debugging
by A. Mc. 29 Jan '21

29 Jan '21

Hello, I suspect that the function recursion composition I have used is not working very well, so I'm trying to learn how to use the debugger to understand what is going on, except I don't really understand the output of the GHCi debugger or what is going on very well. I have a main function called 'Analysis' in a library that analyzes a string. I've used :break Analysis and then called :trace and then repeatedly typed :continue, but there are a number of steps between one continue and the next return-and-call of output that I'm not sure I understand. Suggestions on better debugging methods, or even how to fully understand this output, would be very helpful. Thanks in advance. _result :: Bool = _ key :: Int = -18 message :: [Char] = _ [Lib.hs:17:8-69] *Main> :continue Stopped in Main.Analysis, Lib.hs:18:8-84 _result :: Bool = _ key :: Int = -18 message :: [Char] = _ [Lib.hs:18:8-84] *Main> :continue Stopped in Main.Analysis, Lib.hs:19:8-65 _result :: Bool = _ key :: Int = -18 message :: [Char] = "esleespmwfpdvjtdgpcjmwfpzgpcespcp" [Lib.hs:19:8-65] *Main> :continue Stopped in Main.Analysis, Lib.hs:19:70-127 _result :: (Int, Text) = _ key :: Int = -18 message :: [Char] = "esleespmwfpdvjtdgpcjmwfpzgpcespcp" [Lib.hs:19:70-127] *Main> :continue Stopped in Main.Analysis, Lib.hs:15:8-14 _result :: Bool = _ key :: Int = _ [Lib.hs:15:8-14] *Main> :continue Stopped in Main.Analysis, Lib.hs:16:8-70

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As pattern, @
by Lawrence Bottorff 27 Jan '21

27 Jan '21

I'm looking at this <https://wiki.haskell.org/99_questions/Solutions/8> and wondering how the "as pattern" actually is working compress (x:ys@(y:_)) | x == y = compress ys | otherwise = x : compress ys compress ys = ys I'm sure it's just some version of my stab at eliminating consecutive duplicates in a list compress :: Eq a => [a] -> [a] compress [] = [] compress [x] = [x] compress (x:y:xs) = if x == y then compress (y:xs) else x : compress (y:xs) only smarter. Could someone walk me through the (x:ys@(y:_)) part? LB

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Defined list data type compared to Haskell List
by Lawrence Bottorff 26 Jan '21

26 Jan '21

I'm following this <https://wiki.haskell.org/99_questions/Solutions/7> and yet I see this solution data NestedList a = Elem a | List [NestedList a] deriving (Show) flatten1 :: NestedList a -> [a] flatten1 (Elem a ) = [a] flatten1 (List (x:xs)) = flatten1 x ++ flatten1 (List xs) flatten1 (List []) = [] What I find puzzling is this line flatten1 (List (x:xs)) = flatten1 x ++ flatten1 (List xs) where I see (List (x:xs)) as an argument. How is the NestedList type also able to be expressed as a normal consed list with x:xs argument? How is (:) interacting with NestedList? LB

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$ versus .
by Lawrence Bottorff 26 Jan '21

26 Jan '21

I've got this > init $ tail [1,2,3] [2] and this > chopEnds = init $ tail > chopEnds [1,2,3] [1,2] What happened? Why is it not just init $ tail [1,2,3] ? This works fine > chopEnds2 = init . tail > chopEnds2 [1,2,3] [2] What am I missing? LB

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Viewing a Definition
by A. Mc. 25 Jan '21

25 Jan '21

Hello, It looks like the best way for me to tackle my problem is to create a modified form of Data.Text isInfixOf where the return is an int rather than a bool, to be able to sum words that appear twice or more, but ignoring the whitespace delimitor. Does anyone know where to go to view the actual definition rather than the type signature? Thanks in advance and thank you for your time.

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Re: [Haskell-beginners] Count Words In File
by A. Mc. 22 Jan '21

22 Jan '21

I had thought about doing something like that, except about 80% of the time the words in the file won't have a delimitor such as in "theblueskythegreenearth" but I still need to be able to count "the" or the ordered sequence of [ t, h, e ] within such a character string (or [ b, l, u, e ] or whatever) Again, thank you for your time & thanks in advance. On Fri, Jan 22, 2021, 4:03 AM <beginners-request(a)haskell.org> wrote: > Send Beginners mailing list submissions to > beginners(a)haskell.org > > To subscribe or unsubscribe via the World Wide Web, visit > http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners > or, via email, send a message with subject or body 'help' to > beginners-request(a)haskell.org > > You can reach the person managing the list at > beginners-owner(a)haskell.org > > When replying, please edit your Subject line so it is more specific > than "Re: Contents of Beginners digest..." > > > Today's Topics: > > 1. Count Words from File (A. Mc.) > 2. Re: Count Words from File (Francesco Ariis) > > > ---------------------------------------------------------------------- > > Message: 1 > Date: Thu, 21 Jan 2021 09:49:03 -0800 > From: "A. Mc." <47dragonfyre(a)gmail.com> > To: beginners(a)haskell.org > Subject: [Haskell-beginners] Count Words from File > Message-ID: > <CAOsti3mkiq+z= > cC3ZSGsKUmZsF7jivpTkNcyACRfYC01ExPPtg(a)mail.gmail.com> > Content-Type: text/plain; charset="utf-8" > > Hello, > > I apologize in advance if this is crossposting. My IRC client did not > appear to be working properly. > > I am new to Haskell and I need to find a way to count specific words in a > file. File could contain spaces between words, no spacing, uppercase, > lowercase, etc so I've standardized it to once the file is taken in, > convert to lowercase and remove the spacing. I've also read the postings > about using ByteString instead of [Char] so I am trying to use that. But, > as it still seems to either view all elements as fused or each letter as > individual, I'm not entirely sure how to tackle this. The input after > transforming would be something like "theblueskyisveryblue" for uniformity > and would need to count "the" and "blue". Feels like I should be able to > do a map and foldr(?) but I'm not sure how to get Haskell to recognize > 'the' for example and not count all the t's, h's, e's etc in the file, nor > am I entirely sure how to properly compose a map-fold for character arrays > like this. > > Thanks in advance and thank you for your time. >

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Count Words from File
by A. Mc. 21 Jan '21

21 Jan '21

Hello, I apologize in advance if this is crossposting. My IRC client did not appear to be working properly. I am new to Haskell and I need to find a way to count specific words in a file. File could contain spaces between words, no spacing, uppercase, lowercase, etc so I've standardized it to once the file is taken in, convert to lowercase and remove the spacing. I've also read the postings about using ByteString instead of [Char] so I am trying to use that. But, as it still seems to either view all elements as fused or each letter as individual, I'm not entirely sure how to tackle this. The input after transforming would be something like "theblueskyisveryblue" for uniformity and would need to count "the" and "blue". Feels like I should be able to do a map and foldr(?) but I'm not sure how to get Haskell to recognize 'the' for example and not count all the t's, h's, e's etc in the file, nor am I entirely sure how to properly compose a map-fold for character arrays like this. Thanks in advance and thank you for your time.

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list, map, sequence - stack overflow and performance issues
by Julian Ong 21 Jan '21

21 Jan '21

Hi Haskellers - I'm learning Haskell and attempting to solve the Advent of Code 2020 puzzles using Haskell. I'm stuck on part 2 of Day 15 and have been for a while now, so I'm reaching out. The puzzle asks you to find the nth element in a list of integers. Here's how the list is constructed: Start with a seed list of integers, like [0,3,6]. Then, referring to the last element (6), the next element is given by these rules: - If the last element was the first time the element has appeared in the list, then the next element is 0. - Otherwise, the next element is the age, or distance in the number of index positions, between the last element and when it last appeared before that. For example, starting with [0,3,6], the next elements are 0, 3, 3, 1, 0, 4, 0, etc. Part 1 of the puzzle asks you to find the 2020th element in the list. You can do this by constructing increasingly longer lists like this (using Data.List): nextNum :: [Int] -> [Int]nextNum l@(x:xs) = if not (x `elem` xs) then 0 : l else age l : l where age (x:xs) = let Just i = elemIndex x xs in i+1 Then: head $ (iterate nextNum [6,3,0]) !! 2017 will give you the 2020th element of 436. Note that you provide the starting list in reverse order and iterate so that it will keep adding new elements to the head of the list, which is more efficient than adding to the end. You can also use unfoldr to generate the list element by element like this: nextNum' :: [Int] -> IntnextNum' (x:xs) = if not (x `elem` xs) then 0 else age x xs where age x xs = let Just i = elemIndex x xs in i+1 Then: (unfoldr (\l -> Just (nextNum' l, nextNum' l : l)) slist) !! 2016 will give you the 2020th element of 436. --- Part 2 of the puzzle asks you to find the 30000000th element given starting list [9,3,1,0,8,4]. I cannot find a way to do this without stack overflow and performance issues (I've run my attempts overnight with no answer generated). I've tried using Data.Map and Data.Sequence because my Stack Overflow searching suggested these might be more efficient data structures for this sort of task. Here are my attempts: -- Uses Data.Map to avoid duplicate numbers thereby shortening the list. The dictionary entry (k, v) gives the element and the last position of that element. nextNum'' :: (IntMap Int, (Int, Int)) -> (IntMap Int, (Int, Int))nextNum'' (mp, (k, v)) = case IntMap.lookup k mp of Nothing -> (IntMap.insert k v mp, (0, v+1)) Just pos -> (IntMap.insert k v mp, (v-pos, v+1)) Then: snd $ (iterate nextNum'' (IntMap.fromList [(0,1),(3,2)],(6,3))) !! 2017 provides the answer for the 2020th element but either stack overflows or runs for hours (if I use a strict version of iterate) trying to figure out the 30000000th element. Similarly, using Data.Sequence, I tried: nextNum''' :: Seq Int -> IntnextNum'''' (xs :|> x) = if not (x `elem` xs) then 0 else age x xs where age x xs = let Just i = Seq.elemIndexR x xs in Seq.length xs - i aoc15b' :: Seq Int -> Int -> Intaoc15b' slist tnum = (\(xs :> x) -> x) $ Seq.viewr (Seq.unfoldr (\l -> if Seq.length l == tnum then Nothing else let nnum = force (nextNum'''' l) in Just (nnum, force (l |> nnum))) slist) I found that I needed to fix stack overflow problems by using "force" from Control.DeepSeq. Despite seemingly fixing stack overflow issues though, the calculation just takes too long, and in fact, I have never been able to actually output a solution. I thought that using Data.Map or Data.Sequence would speed things up based on my Stack Overflow searching, but I'm unable to come up with a Haskell solution that runs in reasonable time. I'm at a loss for different strategies at this point and would appreciate any ideas from the community. Thanks, Julian

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Dependent and independent variables in foldl and foldr
by Lawrence Bottorff 17 Jan '21

17 Jan '21

I have this myLength1 = foldl (\n _ -> n + 1) 0 and this myLength2 = foldr (\_ n -> n + 1) 0 I am guessing that foldl knows to assign the accumulator-seed argument to the dependent variable and the list argument's elements recursively to the independent variable; and with foldr to do the opposite. Is this a fair assumption? BTW, where can I get a look at the code for fold functions; or does the type definition answer my original question? Not really able to decipher it so well :t foldl foldl :: Foldable t => (b -> a -> b) -> b -> t a -> b LB

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foldr with ($) as the passed function
by Lawrence Bottorff 14 Jan '21

14 Jan '21

With 99 questions <https://wiki.haskell.org/99_questions/Solutions/3> Problem 3 wants a function elementAt that will take a list and an index and return the element for that index. One very odd version in the solutions is elementAt xs n = head $ foldr ($) xs $ replicate (n - 1) tail So the function "passed" is ($) and the accumulator "seed" is the incoming list xs and the list to be worked on is (replicate (n-1) tail) which . . . and I can't fathom what's happening -- other than perhaps (replicate (n-1) (tail xs)) elementAt [1,2] 2 would be foldr ($) [1,2] (replicate 1 (tail [1,2]) foldr ($) [1,2] ([2]) . . . now I'm lost. Can someone walk me through this? LB

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Explanation of composite with foldl
by Lawrence Bottorff 13 Jan '21

13 Jan '21

I see this on the Haskell 99 questions <https://wiki.haskell.org/99_questions/Solutions/2> which will return the second-from-last element of a list lastbut1 :: Foldable f => f a -> a lastbut1 = fst . foldl (\(a,b) x -> (b,x)) (err1,err2) where err1 = error "lastbut1: Empty list" err2 = error "lastbut1: Singleton" I understand how the code works, but not the significance of the type declaration. That looks like a type class. It works without it, I believe. Why have we used Foldable type class? Likewise with this lastbut1safe :: Foldable f => f a -> Maybe a lastbut1safe = fst . foldl (\(a,b) x -> (b,Just x)) (Nothing,Nothing) What's happening with the type definition? LB

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Haskell equivalent of Python's prompt-toolkit
by Guru Devanla 09 Jan '21

09 Jan '21

Hello Cafe, Can someone point me to any packages that are an equivalent to the prompt-toolkit (https://github.com/prompt-toolkit/python-prompt-toolkit) I would like to use functionality in this library, to create a powerful terminal application which will support syntax highlighting and auto completion. Perhaps, I can use a mix of haskell packages for this. For now, the only relevant package I found was http://hackage.haskell.org/package/readline-1.0.3.0/docs/System-Console-Rea… Any pointers into this would be greatly appreciated. Thanks!

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newEmptyMVar return type
by Velichko Lefterov 06 Jan '21

06 Jan '21

Hello, could someone explain me simple how return-type polymorphism works in the following case: newEmptyMVar :: IO (MVar a) ?

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No ghci return for homemade types
by Lawrence Bottorff 03 Jan '21

03 Jan '21

I've tried this data MyMaybe a = Just a | Nothing maybeHead :: [t] -> MyMaybe t maybeHead [] = Nothing maybeHead (x:_) = Just x and it evaluates fine. But this maybeHead [1,2,3] produces error: : • No instance for (Show (MyMaybe Integer)) : arising from a use of ‘show’ : • In the expression: show (maybeHead [1, 2, 3]) : In an equation for ‘it’: it = show (maybeHead [1, 2, 3]) Similarly, data Color a = Blue a | Green a | Red a myFavoriteColor :: Color Int myFavoriteColor = Green 50 but now I'm stuck. How can I access the "Green 50" inside myFavoriteColor? LB

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Type explanation of foldl?
by Lawrence Bottorff 03 Jan '21

03 Jan '21

I've got this myFoldl f init [] = init myFoldl f init (x:xs) = myFoldl f newInit xs where newInit = f init x which when evaluated gives this for its type > :t myFoldl myFoldl :: (t -> a -> t) -> t -> [a] -> t . . . not totally sure what that all means, e.g., how the t as a generic class is behaving is very mysterious. Obviously, the final result is of type t . . . and that must relate to the incoming argument init, correct? (BTW, does a type declaration like this one reflect in any way the recursion going on?) But then with Prelude foldl I'm really clueless > :t foldl foldl :: Foldable t => (b -> a -> b) -> b -> t a -> b I'm guessing the t is again a generic type of the Foldable class, but then. . . Can someone walk me through this? LB

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Type variables
by Lawrence Bottorff 01 Jan '21

01 Jan '21

I put this into ghci makeTriple :: a -> b -> c -> (a,b,c) makeTriple x y z = (x,y,z) then as expected > :t makeTriple : makeTriple :: a -> b -> c -> (a, b, c) but then on a whim I try this > :t (makeTriple 123 "Hi" 234.0) (makeTriple 123 "Hi" 234.0) :: (Fractional c, Num a) => (a, [Char], c) I retry > :t (123,"Hi",234.0) (123,"Hi",234.0) :: (Fractional c, Num a) => (a, [Char], c) What is this telling me? I'm not experienced enough to decode this. LB

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Lambda calc version of Haskell lambda function
by Lawrence Bottorff 01 Jan '21

01 Jan '21

Here is a function declaration makeAddress :: Int -> String -> String -> (Int, String, String) makeAddress number street town = (number,street,town) and here is a lambda function version makeAddressLambda = (\number -> (\street -> (\town -> (number, street, town)))) How would this lambda version look in lambda calculus? Like this? \number.\street.\town.(number street town) then (\number.\street.\town.(number street town) (123 "Sunny St." "Fergus") (\street.\town.(123 street town) ("Sunny St." "Fergus") (\town.(123 "Sunny St.") ("Fergus") (123 "Sunny St." "Fergus") Not always sure. LB

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