How to filter list elements in Haskell based on previous value in the list?

问题

I'm working on creating a function in Haskell that filters the numbers of a list on a condition based on the previous element in the list.

Example

the previous number is a multiple of 2

myFunction [1, 2, 5, 6, 3]
# expected output:
[5,3]

I know how to apply filter but so far I have seen that the filters take only one argument at a time.

I tried with scanl1, foldl1, and map but I'm new to Haskell and I have not been able to do so; any clue?

回答1:

Edit

It should be:

myFunction []       =  []
myFunction [x]      =  []
myFunction [x,y]    =  if (x `mod` 2) == 0 then [y] else []
myFunction (x:y:xs) =  if (x `mod` 2) == 0 
                      then y : (myFunction xs)
                      else myFunction (y:xs)

because for the input:

myFuntion [1, 2, 5, 6, 3]

the correct output should be:

[5,3]

回答2:

If you prefer to use library functions, there is a known trick for this sort of situations, which consists in zip'ing the input with its own tail, that is the initial list minus its first element.

 λ> 
 λ> inls = [1, 2, 5, 6, 3]
 λ> 
 λ> let pairs = zip (tail inls) inls
 λ> pairs
 [(2,1),(5,2),(6,5),(3,6)]
 λ> 

and the resulting pair list is then an easy target for map and filter. As in:

λ> let myFunction ls = let pairs = zip (tail ls) ls  in  map fst $ filter (even . snd) pairs
λ> 
λ> ls
 [1,2,5,6,3]
λ> 
λ> myFunction ls
[5,3]
λ> 

来源：https://stackoverflow.com/questions/58687161/how-to-filter-list-elements-in-haskell-based-on-previous-value-in-the-list