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6















I built a function that verifies that all elements of a foldable structure are equal.



Compared to a similar function on the lists, it seems to me that the more general function is disproportionately complex, but I have not been able to simplify it.



Do you have any suggestions?



import Data.Monoid
import Data.Sequence as SQ
import Data.Matrix as MT

allElementsEqualL :: Eq a => [a] -> Bool
allElementsEqualL [] = True
allElementsEqualL (x:ns) = all (== x) ns
-- allElementsEqualL [1,1,1] -> True

allElementsEqualF :: (Foldable t, Eq a) => t a -> Bool
allElementsEqualF xs = case (getFirst . foldMap (First . Just) $ xs) of
Nothing -> True
Just x -> all (== x) xs

-- allElementsEqualF [1,1,1] -> True

-- allElementsEqualF $ SQ.fromList [1,1,1] -> True

-- allElementsEqualF $ MT.fromLists [[1,1],[1,1]] -> True









share|improve this question


















  • 1





    Of course, you can always do allElementsEqualF = allElementsEqualL . toList.

    – Alexey Romanov
    5 hours ago













  • @AlexeyRomanov I recently thought of this solution, but I thought it could be very expensive from the point of view of conversion between types. If instead everything happened in a "lazy" way, maybe it would be the most convenient and fastest solution. Is it correct?

    – Alberto Capitani
    4 hours ago











  • @AlexeyRomanov I thought also a mixed solution: allElementsEqualF2 xs -- | F.null xs = True -- | otherwise = all (== x) xs -- where -- x = head $ F.toList xs --- so if goList is lazy, the test is carried out upon the original type (with all).

    – Alberto Capitani
    4 hours ago











  • I decided it was worth a separate answer after all :)

    – Alexey Romanov
    2 hours ago


















6















I built a function that verifies that all elements of a foldable structure are equal.



Compared to a similar function on the lists, it seems to me that the more general function is disproportionately complex, but I have not been able to simplify it.



Do you have any suggestions?



import Data.Monoid
import Data.Sequence as SQ
import Data.Matrix as MT

allElementsEqualL :: Eq a => [a] -> Bool
allElementsEqualL [] = True
allElementsEqualL (x:ns) = all (== x) ns
-- allElementsEqualL [1,1,1] -> True

allElementsEqualF :: (Foldable t, Eq a) => t a -> Bool
allElementsEqualF xs = case (getFirst . foldMap (First . Just) $ xs) of
Nothing -> True
Just x -> all (== x) xs

-- allElementsEqualF [1,1,1] -> True

-- allElementsEqualF $ SQ.fromList [1,1,1] -> True

-- allElementsEqualF $ MT.fromLists [[1,1],[1,1]] -> True









share|improve this question


















  • 1





    Of course, you can always do allElementsEqualF = allElementsEqualL . toList.

    – Alexey Romanov
    5 hours ago













  • @AlexeyRomanov I recently thought of this solution, but I thought it could be very expensive from the point of view of conversion between types. If instead everything happened in a "lazy" way, maybe it would be the most convenient and fastest solution. Is it correct?

    – Alberto Capitani
    4 hours ago











  • @AlexeyRomanov I thought also a mixed solution: allElementsEqualF2 xs -- | F.null xs = True -- | otherwise = all (== x) xs -- where -- x = head $ F.toList xs --- so if goList is lazy, the test is carried out upon the original type (with all).

    – Alberto Capitani
    4 hours ago











  • I decided it was worth a separate answer after all :)

    – Alexey Romanov
    2 hours ago














6












6








6


1






I built a function that verifies that all elements of a foldable structure are equal.



Compared to a similar function on the lists, it seems to me that the more general function is disproportionately complex, but I have not been able to simplify it.



Do you have any suggestions?



import Data.Monoid
import Data.Sequence as SQ
import Data.Matrix as MT

allElementsEqualL :: Eq a => [a] -> Bool
allElementsEqualL [] = True
allElementsEqualL (x:ns) = all (== x) ns
-- allElementsEqualL [1,1,1] -> True

allElementsEqualF :: (Foldable t, Eq a) => t a -> Bool
allElementsEqualF xs = case (getFirst . foldMap (First . Just) $ xs) of
Nothing -> True
Just x -> all (== x) xs

-- allElementsEqualF [1,1,1] -> True

-- allElementsEqualF $ SQ.fromList [1,1,1] -> True

-- allElementsEqualF $ MT.fromLists [[1,1],[1,1]] -> True









share|improve this question














I built a function that verifies that all elements of a foldable structure are equal.



Compared to a similar function on the lists, it seems to me that the more general function is disproportionately complex, but I have not been able to simplify it.



Do you have any suggestions?



import Data.Monoid
import Data.Sequence as SQ
import Data.Matrix as MT

allElementsEqualL :: Eq a => [a] -> Bool
allElementsEqualL [] = True
allElementsEqualL (x:ns) = all (== x) ns
-- allElementsEqualL [1,1,1] -> True

allElementsEqualF :: (Foldable t, Eq a) => t a -> Bool
allElementsEqualF xs = case (getFirst . foldMap (First . Just) $ xs) of
Nothing -> True
Just x -> all (== x) xs

-- allElementsEqualF [1,1,1] -> True

-- allElementsEqualF $ SQ.fromList [1,1,1] -> True

-- allElementsEqualF $ MT.fromLists [[1,1],[1,1]] -> True






haskell






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 6 hours ago









Alberto CapitaniAlberto Capitani

534823




534823








  • 1





    Of course, you can always do allElementsEqualF = allElementsEqualL . toList.

    – Alexey Romanov
    5 hours ago













  • @AlexeyRomanov I recently thought of this solution, but I thought it could be very expensive from the point of view of conversion between types. If instead everything happened in a "lazy" way, maybe it would be the most convenient and fastest solution. Is it correct?

    – Alberto Capitani
    4 hours ago











  • @AlexeyRomanov I thought also a mixed solution: allElementsEqualF2 xs -- | F.null xs = True -- | otherwise = all (== x) xs -- where -- x = head $ F.toList xs --- so if goList is lazy, the test is carried out upon the original type (with all).

    – Alberto Capitani
    4 hours ago











  • I decided it was worth a separate answer after all :)

    – Alexey Romanov
    2 hours ago














  • 1





    Of course, you can always do allElementsEqualF = allElementsEqualL . toList.

    – Alexey Romanov
    5 hours ago













  • @AlexeyRomanov I recently thought of this solution, but I thought it could be very expensive from the point of view of conversion between types. If instead everything happened in a "lazy" way, maybe it would be the most convenient and fastest solution. Is it correct?

    – Alberto Capitani
    4 hours ago











  • @AlexeyRomanov I thought also a mixed solution: allElementsEqualF2 xs -- | F.null xs = True -- | otherwise = all (== x) xs -- where -- x = head $ F.toList xs --- so if goList is lazy, the test is carried out upon the original type (with all).

    – Alberto Capitani
    4 hours ago











  • I decided it was worth a separate answer after all :)

    – Alexey Romanov
    2 hours ago








1




1





Of course, you can always do allElementsEqualF = allElementsEqualL . toList.

– Alexey Romanov
5 hours ago







Of course, you can always do allElementsEqualF = allElementsEqualL . toList.

– Alexey Romanov
5 hours ago















@AlexeyRomanov I recently thought of this solution, but I thought it could be very expensive from the point of view of conversion between types. If instead everything happened in a "lazy" way, maybe it would be the most convenient and fastest solution. Is it correct?

– Alberto Capitani
4 hours ago





@AlexeyRomanov I recently thought of this solution, but I thought it could be very expensive from the point of view of conversion between types. If instead everything happened in a "lazy" way, maybe it would be the most convenient and fastest solution. Is it correct?

– Alberto Capitani
4 hours ago













@AlexeyRomanov I thought also a mixed solution: allElementsEqualF2 xs -- | F.null xs = True -- | otherwise = all (== x) xs -- where -- x = head $ F.toList xs --- so if goList is lazy, the test is carried out upon the original type (with all).

– Alberto Capitani
4 hours ago





@AlexeyRomanov I thought also a mixed solution: allElementsEqualF2 xs -- | F.null xs = True -- | otherwise = all (== x) xs -- where -- x = head $ F.toList xs --- so if goList is lazy, the test is carried out upon the original type (with all).

– Alberto Capitani
4 hours ago













I decided it was worth a separate answer after all :)

– Alexey Romanov
2 hours ago





I decided it was worth a separate answer after all :)

– Alexey Romanov
2 hours ago












4 Answers
4






active

oldest

votes


















8














I don't know about less complicated, but I think this is the "cleanest" way to do it. By "clean," I mean it's one traversal over the structure using a single, special Monoid.



data Same a = Vacuous | Fail | Same a
instance Eq a => Semigroup (Same a) where
Vacuous <> x = x
Fail <> _ = Fail
s@(Same l) <> Same r = if l == r then s else Fail
x <> Vacuous = x
_ <> Fail = Fail
instance Eq a => Monoid (Same a) where
mempty = Vacuous

allEq :: (Foldable f, Eq a) => f a -> Bool
allEq xs = case foldMap Same xs of
Fail -> False
_ -> True





share|improve this answer


























  • I think Same is isomorphic to Success a from the zero package.

    – Rein Henrichs
    5 hours ago











  • @ReinHenrichs I kind of doubt it. Same a has two extra constructors compared to a, whereas Success a has just one. Now it might be Success (Maybe a) or something... but at that point I would say having a custom type is more readable.

    – Daniel Wagner
    5 hours ago











  • It is, however, true that Zero (Same a) with zero = Fail.

    – HTNW
    5 hours ago











  • Same a has n + 2 elements. Success a is Maybe (Maybe a), which also has n + 2 elements. (Ignoring bottoms.)

    – Rein Henrichs
    5 hours ago











  • @ReinHenrichs Hm. When I follow your link, I see newtype Success a = Success { getSuccess :: Maybe a }, which has just one Maybe wrapper, not two.

    – Daniel Wagner
    5 hours ago





















4














The convenient thing about your first function that doesn't exist in your second is that we have a convenient way of getting the "head" of a list. Fortunately, we can do the same for a Foldable. Let's write a head' that works on any Foldable (and for the sake of type safety we'll have our head' return a Maybe)



head' :: (Foldable t, Eq a) => t a -> Maybe a
head' = foldr (a _ -> Just a) Nothing


Now we can write basically the same code as the list case for the general one.



allElementsEqualF :: (Foldable t, Eq a) => t a -> Bool
allElementsEqualF f = case head' f of
Nothing -> True
Just a -> all (== a) f


Syntactically, it looks different, but it's the exact same thing you did in your list case: check if the structure is empty and, if it's not, then see if every element is equal to the first.



Note that, technically, this is not quite equivalent to the code you posted, as it compares the first element against itself. So if your == operator is for some reason not reflexive, you'll get different results (try running my code and yours on the list [read "NaN" :: Double])






share|improve this answer































    2














    Silvio's answer is syntactically small and easy to understand; however, it may do extra work associated with doing two folds if the Foldable instance can't compute head' cheaply. In this answer I will discuss how to perform the calculation in just one pass whether the underlying Foldable can compute head' cheaply or not.



    The basic idea is this: instead of tracking just "are all the elements equal so far", we'll also track what they're all equal to. So:



    data AreTheyEqual a
    = Empty
    | Equal a
    | Inequal
    deriving Eq


    This is a Monoid, with Empty as the unit and Inequal as an absorbing element.



    instance Eq a => Semigroup (AreTheyEqual a) where
    Empty <> x = x
    x <> Empty = x
    Equal a <> Equal b | a == b = Equal a
    _ <> _ = Inequal

    instance Eq a => Monoid (AreTheyEqual a) where
    mempty = Empty


    Now we can use foldMap to summarize an entire Foldable, like so:



    allElementsEqual :: (Eq a, Foldable f) => f a -> Bool
    allElementsEqual = (Inequal /=) . foldMap Equal





    share|improve this answer


























    • I think this is the same as my answer?

      – HTNW
      5 hours ago











    • @HTNW Yep! Seems we overlapped in answering.

      – Daniel Wagner
      5 hours ago











    • It doesn't necessarily do two traversals. For structures where foldr can implement guarded recursion, it may only look at the top-most constructor the first time.

      – Rein Henrichs
      5 hours ago











    • @ReinHenrichs Good point. I'll amend my answer appropriately.

      – Daniel Wagner
      5 hours ago



















    2














    A rather trivial option, and I would generally prefer one of the other answers, is reusing allElementsEqualL:



    allElementsEqualF = allElementsEqualL . toList


    or after inlining



    allElementsEqualF xs = case toList xs of
    [] -> True
    x:xs' -> all (== x) xs'


    It's laziness which makes it reasonable. The all call doesn't demand the entire xs', but only until it finds the first one different from x. So toList will also not demand the entire xs. And at the same time, already examined elements don't need to be kept in memory.



    You could write a Foldable instance for which toList is less lazy than necessary, but except for those cases I think it should do exactly as much work as Daniel Wagner's and HTNW's answer (with slight overhead not depending on input size).




    I thought also a mixed solution:



    allElementsEqualF2 xs | F.null xs = True 
    | otherwise = all (== x) xs
    where x = head $ F.toList xs


    so if goList is lazy, the test is carried out upon the original type (with all).




    This does slightly more work in the non-empty case than Silvio's answer, because F.null duplicates exactly as much of F.toList's work as head' does. So Silvio's code has to get to the first element 2 times (one for head' and another inside all), and yours does it 3 times (null, head $ toList xs and all again).






    share|improve this answer


























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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      8














      I don't know about less complicated, but I think this is the "cleanest" way to do it. By "clean," I mean it's one traversal over the structure using a single, special Monoid.



      data Same a = Vacuous | Fail | Same a
      instance Eq a => Semigroup (Same a) where
      Vacuous <> x = x
      Fail <> _ = Fail
      s@(Same l) <> Same r = if l == r then s else Fail
      x <> Vacuous = x
      _ <> Fail = Fail
      instance Eq a => Monoid (Same a) where
      mempty = Vacuous

      allEq :: (Foldable f, Eq a) => f a -> Bool
      allEq xs = case foldMap Same xs of
      Fail -> False
      _ -> True





      share|improve this answer


























      • I think Same is isomorphic to Success a from the zero package.

        – Rein Henrichs
        5 hours ago











      • @ReinHenrichs I kind of doubt it. Same a has two extra constructors compared to a, whereas Success a has just one. Now it might be Success (Maybe a) or something... but at that point I would say having a custom type is more readable.

        – Daniel Wagner
        5 hours ago











      • It is, however, true that Zero (Same a) with zero = Fail.

        – HTNW
        5 hours ago











      • Same a has n + 2 elements. Success a is Maybe (Maybe a), which also has n + 2 elements. (Ignoring bottoms.)

        – Rein Henrichs
        5 hours ago











      • @ReinHenrichs Hm. When I follow your link, I see newtype Success a = Success { getSuccess :: Maybe a }, which has just one Maybe wrapper, not two.

        – Daniel Wagner
        5 hours ago


















      8














      I don't know about less complicated, but I think this is the "cleanest" way to do it. By "clean," I mean it's one traversal over the structure using a single, special Monoid.



      data Same a = Vacuous | Fail | Same a
      instance Eq a => Semigroup (Same a) where
      Vacuous <> x = x
      Fail <> _ = Fail
      s@(Same l) <> Same r = if l == r then s else Fail
      x <> Vacuous = x
      _ <> Fail = Fail
      instance Eq a => Monoid (Same a) where
      mempty = Vacuous

      allEq :: (Foldable f, Eq a) => f a -> Bool
      allEq xs = case foldMap Same xs of
      Fail -> False
      _ -> True





      share|improve this answer


























      • I think Same is isomorphic to Success a from the zero package.

        – Rein Henrichs
        5 hours ago











      • @ReinHenrichs I kind of doubt it. Same a has two extra constructors compared to a, whereas Success a has just one. Now it might be Success (Maybe a) or something... but at that point I would say having a custom type is more readable.

        – Daniel Wagner
        5 hours ago











      • It is, however, true that Zero (Same a) with zero = Fail.

        – HTNW
        5 hours ago











      • Same a has n + 2 elements. Success a is Maybe (Maybe a), which also has n + 2 elements. (Ignoring bottoms.)

        – Rein Henrichs
        5 hours ago











      • @ReinHenrichs Hm. When I follow your link, I see newtype Success a = Success { getSuccess :: Maybe a }, which has just one Maybe wrapper, not two.

        – Daniel Wagner
        5 hours ago
















      8












      8








      8







      I don't know about less complicated, but I think this is the "cleanest" way to do it. By "clean," I mean it's one traversal over the structure using a single, special Monoid.



      data Same a = Vacuous | Fail | Same a
      instance Eq a => Semigroup (Same a) where
      Vacuous <> x = x
      Fail <> _ = Fail
      s@(Same l) <> Same r = if l == r then s else Fail
      x <> Vacuous = x
      _ <> Fail = Fail
      instance Eq a => Monoid (Same a) where
      mempty = Vacuous

      allEq :: (Foldable f, Eq a) => f a -> Bool
      allEq xs = case foldMap Same xs of
      Fail -> False
      _ -> True





      share|improve this answer















      I don't know about less complicated, but I think this is the "cleanest" way to do it. By "clean," I mean it's one traversal over the structure using a single, special Monoid.



      data Same a = Vacuous | Fail | Same a
      instance Eq a => Semigroup (Same a) where
      Vacuous <> x = x
      Fail <> _ = Fail
      s@(Same l) <> Same r = if l == r then s else Fail
      x <> Vacuous = x
      _ <> Fail = Fail
      instance Eq a => Monoid (Same a) where
      mempty = Vacuous

      allEq :: (Foldable f, Eq a) => f a -> Bool
      allEq xs = case foldMap Same xs of
      Fail -> False
      _ -> True






      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited 16 mins ago









      Joseph Sible

      7,52031337




      7,52031337










      answered 6 hours ago









      HTNWHTNW

      10.5k1832




      10.5k1832













      • I think Same is isomorphic to Success a from the zero package.

        – Rein Henrichs
        5 hours ago











      • @ReinHenrichs I kind of doubt it. Same a has two extra constructors compared to a, whereas Success a has just one. Now it might be Success (Maybe a) or something... but at that point I would say having a custom type is more readable.

        – Daniel Wagner
        5 hours ago











      • It is, however, true that Zero (Same a) with zero = Fail.

        – HTNW
        5 hours ago











      • Same a has n + 2 elements. Success a is Maybe (Maybe a), which also has n + 2 elements. (Ignoring bottoms.)

        – Rein Henrichs
        5 hours ago











      • @ReinHenrichs Hm. When I follow your link, I see newtype Success a = Success { getSuccess :: Maybe a }, which has just one Maybe wrapper, not two.

        – Daniel Wagner
        5 hours ago





















      • I think Same is isomorphic to Success a from the zero package.

        – Rein Henrichs
        5 hours ago











      • @ReinHenrichs I kind of doubt it. Same a has two extra constructors compared to a, whereas Success a has just one. Now it might be Success (Maybe a) or something... but at that point I would say having a custom type is more readable.

        – Daniel Wagner
        5 hours ago











      • It is, however, true that Zero (Same a) with zero = Fail.

        – HTNW
        5 hours ago











      • Same a has n + 2 elements. Success a is Maybe (Maybe a), which also has n + 2 elements. (Ignoring bottoms.)

        – Rein Henrichs
        5 hours ago











      • @ReinHenrichs Hm. When I follow your link, I see newtype Success a = Success { getSuccess :: Maybe a }, which has just one Maybe wrapper, not two.

        – Daniel Wagner
        5 hours ago



















      I think Same is isomorphic to Success a from the zero package.

      – Rein Henrichs
      5 hours ago





      I think Same is isomorphic to Success a from the zero package.

      – Rein Henrichs
      5 hours ago













      @ReinHenrichs I kind of doubt it. Same a has two extra constructors compared to a, whereas Success a has just one. Now it might be Success (Maybe a) or something... but at that point I would say having a custom type is more readable.

      – Daniel Wagner
      5 hours ago





      @ReinHenrichs I kind of doubt it. Same a has two extra constructors compared to a, whereas Success a has just one. Now it might be Success (Maybe a) or something... but at that point I would say having a custom type is more readable.

      – Daniel Wagner
      5 hours ago













      It is, however, true that Zero (Same a) with zero = Fail.

      – HTNW
      5 hours ago





      It is, however, true that Zero (Same a) with zero = Fail.

      – HTNW
      5 hours ago













      Same a has n + 2 elements. Success a is Maybe (Maybe a), which also has n + 2 elements. (Ignoring bottoms.)

      – Rein Henrichs
      5 hours ago





      Same a has n + 2 elements. Success a is Maybe (Maybe a), which also has n + 2 elements. (Ignoring bottoms.)

      – Rein Henrichs
      5 hours ago













      @ReinHenrichs Hm. When I follow your link, I see newtype Success a = Success { getSuccess :: Maybe a }, which has just one Maybe wrapper, not two.

      – Daniel Wagner
      5 hours ago







      @ReinHenrichs Hm. When I follow your link, I see newtype Success a = Success { getSuccess :: Maybe a }, which has just one Maybe wrapper, not two.

      – Daniel Wagner
      5 hours ago















      4














      The convenient thing about your first function that doesn't exist in your second is that we have a convenient way of getting the "head" of a list. Fortunately, we can do the same for a Foldable. Let's write a head' that works on any Foldable (and for the sake of type safety we'll have our head' return a Maybe)



      head' :: (Foldable t, Eq a) => t a -> Maybe a
      head' = foldr (a _ -> Just a) Nothing


      Now we can write basically the same code as the list case for the general one.



      allElementsEqualF :: (Foldable t, Eq a) => t a -> Bool
      allElementsEqualF f = case head' f of
      Nothing -> True
      Just a -> all (== a) f


      Syntactically, it looks different, but it's the exact same thing you did in your list case: check if the structure is empty and, if it's not, then see if every element is equal to the first.



      Note that, technically, this is not quite equivalent to the code you posted, as it compares the first element against itself. So if your == operator is for some reason not reflexive, you'll get different results (try running my code and yours on the list [read "NaN" :: Double])






      share|improve this answer




























        4














        The convenient thing about your first function that doesn't exist in your second is that we have a convenient way of getting the "head" of a list. Fortunately, we can do the same for a Foldable. Let's write a head' that works on any Foldable (and for the sake of type safety we'll have our head' return a Maybe)



        head' :: (Foldable t, Eq a) => t a -> Maybe a
        head' = foldr (a _ -> Just a) Nothing


        Now we can write basically the same code as the list case for the general one.



        allElementsEqualF :: (Foldable t, Eq a) => t a -> Bool
        allElementsEqualF f = case head' f of
        Nothing -> True
        Just a -> all (== a) f


        Syntactically, it looks different, but it's the exact same thing you did in your list case: check if the structure is empty and, if it's not, then see if every element is equal to the first.



        Note that, technically, this is not quite equivalent to the code you posted, as it compares the first element against itself. So if your == operator is for some reason not reflexive, you'll get different results (try running my code and yours on the list [read "NaN" :: Double])






        share|improve this answer


























          4












          4








          4







          The convenient thing about your first function that doesn't exist in your second is that we have a convenient way of getting the "head" of a list. Fortunately, we can do the same for a Foldable. Let's write a head' that works on any Foldable (and for the sake of type safety we'll have our head' return a Maybe)



          head' :: (Foldable t, Eq a) => t a -> Maybe a
          head' = foldr (a _ -> Just a) Nothing


          Now we can write basically the same code as the list case for the general one.



          allElementsEqualF :: (Foldable t, Eq a) => t a -> Bool
          allElementsEqualF f = case head' f of
          Nothing -> True
          Just a -> all (== a) f


          Syntactically, it looks different, but it's the exact same thing you did in your list case: check if the structure is empty and, if it's not, then see if every element is equal to the first.



          Note that, technically, this is not quite equivalent to the code you posted, as it compares the first element against itself. So if your == operator is for some reason not reflexive, you'll get different results (try running my code and yours on the list [read "NaN" :: Double])






          share|improve this answer













          The convenient thing about your first function that doesn't exist in your second is that we have a convenient way of getting the "head" of a list. Fortunately, we can do the same for a Foldable. Let's write a head' that works on any Foldable (and for the sake of type safety we'll have our head' return a Maybe)



          head' :: (Foldable t, Eq a) => t a -> Maybe a
          head' = foldr (a _ -> Just a) Nothing


          Now we can write basically the same code as the list case for the general one.



          allElementsEqualF :: (Foldable t, Eq a) => t a -> Bool
          allElementsEqualF f = case head' f of
          Nothing -> True
          Just a -> all (== a) f


          Syntactically, it looks different, but it's the exact same thing you did in your list case: check if the structure is empty and, if it's not, then see if every element is equal to the first.



          Note that, technically, this is not quite equivalent to the code you posted, as it compares the first element against itself. So if your == operator is for some reason not reflexive, you'll get different results (try running my code and yours on the list [read "NaN" :: Double])







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 6 hours ago









          Silvio MayoloSilvio Mayolo

          14.9k22654




          14.9k22654























              2














              Silvio's answer is syntactically small and easy to understand; however, it may do extra work associated with doing two folds if the Foldable instance can't compute head' cheaply. In this answer I will discuss how to perform the calculation in just one pass whether the underlying Foldable can compute head' cheaply or not.



              The basic idea is this: instead of tracking just "are all the elements equal so far", we'll also track what they're all equal to. So:



              data AreTheyEqual a
              = Empty
              | Equal a
              | Inequal
              deriving Eq


              This is a Monoid, with Empty as the unit and Inequal as an absorbing element.



              instance Eq a => Semigroup (AreTheyEqual a) where
              Empty <> x = x
              x <> Empty = x
              Equal a <> Equal b | a == b = Equal a
              _ <> _ = Inequal

              instance Eq a => Monoid (AreTheyEqual a) where
              mempty = Empty


              Now we can use foldMap to summarize an entire Foldable, like so:



              allElementsEqual :: (Eq a, Foldable f) => f a -> Bool
              allElementsEqual = (Inequal /=) . foldMap Equal





              share|improve this answer


























              • I think this is the same as my answer?

                – HTNW
                5 hours ago











              • @HTNW Yep! Seems we overlapped in answering.

                – Daniel Wagner
                5 hours ago











              • It doesn't necessarily do two traversals. For structures where foldr can implement guarded recursion, it may only look at the top-most constructor the first time.

                – Rein Henrichs
                5 hours ago











              • @ReinHenrichs Good point. I'll amend my answer appropriately.

                – Daniel Wagner
                5 hours ago
















              2














              Silvio's answer is syntactically small and easy to understand; however, it may do extra work associated with doing two folds if the Foldable instance can't compute head' cheaply. In this answer I will discuss how to perform the calculation in just one pass whether the underlying Foldable can compute head' cheaply or not.



              The basic idea is this: instead of tracking just "are all the elements equal so far", we'll also track what they're all equal to. So:



              data AreTheyEqual a
              = Empty
              | Equal a
              | Inequal
              deriving Eq


              This is a Monoid, with Empty as the unit and Inequal as an absorbing element.



              instance Eq a => Semigroup (AreTheyEqual a) where
              Empty <> x = x
              x <> Empty = x
              Equal a <> Equal b | a == b = Equal a
              _ <> _ = Inequal

              instance Eq a => Monoid (AreTheyEqual a) where
              mempty = Empty


              Now we can use foldMap to summarize an entire Foldable, like so:



              allElementsEqual :: (Eq a, Foldable f) => f a -> Bool
              allElementsEqual = (Inequal /=) . foldMap Equal





              share|improve this answer


























              • I think this is the same as my answer?

                – HTNW
                5 hours ago











              • @HTNW Yep! Seems we overlapped in answering.

                – Daniel Wagner
                5 hours ago











              • It doesn't necessarily do two traversals. For structures where foldr can implement guarded recursion, it may only look at the top-most constructor the first time.

                – Rein Henrichs
                5 hours ago











              • @ReinHenrichs Good point. I'll amend my answer appropriately.

                – Daniel Wagner
                5 hours ago














              2












              2








              2







              Silvio's answer is syntactically small and easy to understand; however, it may do extra work associated with doing two folds if the Foldable instance can't compute head' cheaply. In this answer I will discuss how to perform the calculation in just one pass whether the underlying Foldable can compute head' cheaply or not.



              The basic idea is this: instead of tracking just "are all the elements equal so far", we'll also track what they're all equal to. So:



              data AreTheyEqual a
              = Empty
              | Equal a
              | Inequal
              deriving Eq


              This is a Monoid, with Empty as the unit and Inequal as an absorbing element.



              instance Eq a => Semigroup (AreTheyEqual a) where
              Empty <> x = x
              x <> Empty = x
              Equal a <> Equal b | a == b = Equal a
              _ <> _ = Inequal

              instance Eq a => Monoid (AreTheyEqual a) where
              mempty = Empty


              Now we can use foldMap to summarize an entire Foldable, like so:



              allElementsEqual :: (Eq a, Foldable f) => f a -> Bool
              allElementsEqual = (Inequal /=) . foldMap Equal





              share|improve this answer















              Silvio's answer is syntactically small and easy to understand; however, it may do extra work associated with doing two folds if the Foldable instance can't compute head' cheaply. In this answer I will discuss how to perform the calculation in just one pass whether the underlying Foldable can compute head' cheaply or not.



              The basic idea is this: instead of tracking just "are all the elements equal so far", we'll also track what they're all equal to. So:



              data AreTheyEqual a
              = Empty
              | Equal a
              | Inequal
              deriving Eq


              This is a Monoid, with Empty as the unit and Inequal as an absorbing element.



              instance Eq a => Semigroup (AreTheyEqual a) where
              Empty <> x = x
              x <> Empty = x
              Equal a <> Equal b | a == b = Equal a
              _ <> _ = Inequal

              instance Eq a => Monoid (AreTheyEqual a) where
              mempty = Empty


              Now we can use foldMap to summarize an entire Foldable, like so:



              allElementsEqual :: (Eq a, Foldable f) => f a -> Bool
              allElementsEqual = (Inequal /=) . foldMap Equal






              share|improve this answer














              share|improve this answer



              share|improve this answer








              edited 5 hours ago

























              answered 5 hours ago









              Daniel WagnerDaniel Wagner

              104k7163288




              104k7163288













              • I think this is the same as my answer?

                – HTNW
                5 hours ago











              • @HTNW Yep! Seems we overlapped in answering.

                – Daniel Wagner
                5 hours ago











              • It doesn't necessarily do two traversals. For structures where foldr can implement guarded recursion, it may only look at the top-most constructor the first time.

                – Rein Henrichs
                5 hours ago











              • @ReinHenrichs Good point. I'll amend my answer appropriately.

                – Daniel Wagner
                5 hours ago



















              • I think this is the same as my answer?

                – HTNW
                5 hours ago











              • @HTNW Yep! Seems we overlapped in answering.

                – Daniel Wagner
                5 hours ago











              • It doesn't necessarily do two traversals. For structures where foldr can implement guarded recursion, it may only look at the top-most constructor the first time.

                – Rein Henrichs
                5 hours ago











              • @ReinHenrichs Good point. I'll amend my answer appropriately.

                – Daniel Wagner
                5 hours ago

















              I think this is the same as my answer?

              – HTNW
              5 hours ago





              I think this is the same as my answer?

              – HTNW
              5 hours ago













              @HTNW Yep! Seems we overlapped in answering.

              – Daniel Wagner
              5 hours ago





              @HTNW Yep! Seems we overlapped in answering.

              – Daniel Wagner
              5 hours ago













              It doesn't necessarily do two traversals. For structures where foldr can implement guarded recursion, it may only look at the top-most constructor the first time.

              – Rein Henrichs
              5 hours ago





              It doesn't necessarily do two traversals. For structures where foldr can implement guarded recursion, it may only look at the top-most constructor the first time.

              – Rein Henrichs
              5 hours ago













              @ReinHenrichs Good point. I'll amend my answer appropriately.

              – Daniel Wagner
              5 hours ago





              @ReinHenrichs Good point. I'll amend my answer appropriately.

              – Daniel Wagner
              5 hours ago











              2














              A rather trivial option, and I would generally prefer one of the other answers, is reusing allElementsEqualL:



              allElementsEqualF = allElementsEqualL . toList


              or after inlining



              allElementsEqualF xs = case toList xs of
              [] -> True
              x:xs' -> all (== x) xs'


              It's laziness which makes it reasonable. The all call doesn't demand the entire xs', but only until it finds the first one different from x. So toList will also not demand the entire xs. And at the same time, already examined elements don't need to be kept in memory.



              You could write a Foldable instance for which toList is less lazy than necessary, but except for those cases I think it should do exactly as much work as Daniel Wagner's and HTNW's answer (with slight overhead not depending on input size).




              I thought also a mixed solution:



              allElementsEqualF2 xs | F.null xs = True 
              | otherwise = all (== x) xs
              where x = head $ F.toList xs


              so if goList is lazy, the test is carried out upon the original type (with all).




              This does slightly more work in the non-empty case than Silvio's answer, because F.null duplicates exactly as much of F.toList's work as head' does. So Silvio's code has to get to the first element 2 times (one for head' and another inside all), and yours does it 3 times (null, head $ toList xs and all again).






              share|improve this answer






























                2














                A rather trivial option, and I would generally prefer one of the other answers, is reusing allElementsEqualL:



                allElementsEqualF = allElementsEqualL . toList


                or after inlining



                allElementsEqualF xs = case toList xs of
                [] -> True
                x:xs' -> all (== x) xs'


                It's laziness which makes it reasonable. The all call doesn't demand the entire xs', but only until it finds the first one different from x. So toList will also not demand the entire xs. And at the same time, already examined elements don't need to be kept in memory.



                You could write a Foldable instance for which toList is less lazy than necessary, but except for those cases I think it should do exactly as much work as Daniel Wagner's and HTNW's answer (with slight overhead not depending on input size).




                I thought also a mixed solution:



                allElementsEqualF2 xs | F.null xs = True 
                | otherwise = all (== x) xs
                where x = head $ F.toList xs


                so if goList is lazy, the test is carried out upon the original type (with all).




                This does slightly more work in the non-empty case than Silvio's answer, because F.null duplicates exactly as much of F.toList's work as head' does. So Silvio's code has to get to the first element 2 times (one for head' and another inside all), and yours does it 3 times (null, head $ toList xs and all again).






                share|improve this answer




























                  2












                  2








                  2







                  A rather trivial option, and I would generally prefer one of the other answers, is reusing allElementsEqualL:



                  allElementsEqualF = allElementsEqualL . toList


                  or after inlining



                  allElementsEqualF xs = case toList xs of
                  [] -> True
                  x:xs' -> all (== x) xs'


                  It's laziness which makes it reasonable. The all call doesn't demand the entire xs', but only until it finds the first one different from x. So toList will also not demand the entire xs. And at the same time, already examined elements don't need to be kept in memory.



                  You could write a Foldable instance for which toList is less lazy than necessary, but except for those cases I think it should do exactly as much work as Daniel Wagner's and HTNW's answer (with slight overhead not depending on input size).




                  I thought also a mixed solution:



                  allElementsEqualF2 xs | F.null xs = True 
                  | otherwise = all (== x) xs
                  where x = head $ F.toList xs


                  so if goList is lazy, the test is carried out upon the original type (with all).




                  This does slightly more work in the non-empty case than Silvio's answer, because F.null duplicates exactly as much of F.toList's work as head' does. So Silvio's code has to get to the first element 2 times (one for head' and another inside all), and yours does it 3 times (null, head $ toList xs and all again).






                  share|improve this answer















                  A rather trivial option, and I would generally prefer one of the other answers, is reusing allElementsEqualL:



                  allElementsEqualF = allElementsEqualL . toList


                  or after inlining



                  allElementsEqualF xs = case toList xs of
                  [] -> True
                  x:xs' -> all (== x) xs'


                  It's laziness which makes it reasonable. The all call doesn't demand the entire xs', but only until it finds the first one different from x. So toList will also not demand the entire xs. And at the same time, already examined elements don't need to be kept in memory.



                  You could write a Foldable instance for which toList is less lazy than necessary, but except for those cases I think it should do exactly as much work as Daniel Wagner's and HTNW's answer (with slight overhead not depending on input size).




                  I thought also a mixed solution:



                  allElementsEqualF2 xs | F.null xs = True 
                  | otherwise = all (== x) xs
                  where x = head $ F.toList xs


                  so if goList is lazy, the test is carried out upon the original type (with all).




                  This does slightly more work in the non-empty case than Silvio's answer, because F.null duplicates exactly as much of F.toList's work as head' does. So Silvio's code has to get to the first element 2 times (one for head' and another inside all), and yours does it 3 times (null, head $ toList xs and all again).







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited 1 hour ago

























                  answered 2 hours ago









                  Alexey RomanovAlexey Romanov

                  112k26217360




                  112k26217360






























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