在列表上使用 Haskell 中的翻转和过滤功能

Question

a Haskell newby, so apologies if this is a rather basic question. I want to write a function which takes two lists of variables and returns the first with all variables from the second list removed from it. For example I want the following output

*Main> filterVariables ["y","z","a1","a2"] ["y","a1","a3"]

["z","a2"]

This what I have written so far:

type Var = String

data Term =
    Variable Var
  | Lambda   Var  Term
  | Apply    Term Term


variables :: [Var]
variables =  [l:[] | l <- ['a'..'z']] ++ [l:show x | x <- [1..], l <- ['a'..'z']]

--works but the wrong way around!!!
filterVariables :: [Var] -> [Var] -> [Var]
filterVariables lst = filter ( `notElem` lst)

I get the following ouput:

*Main> filterVariables ["y","z","a1","a2"] ["y","a1","a3"]

["a3"]

Could the flip function be used here?

When I write:

filterVariables lst = flip (filter ( `notElem` lst))

I get this error message:

Couldn't match type ‘[Var]’ with ‘[Var] -> c0’
      Expected type: [Var] -> [Var] -> c0
        Actual type: [Var] -> [Var]
    • Possible cause: ‘filter’ is applied to too many arguments
      In the first argument of ‘flip’, namely ‘(filter (`notElem` lst))’
      In the expression: flip (filter (`notElem` lst))
      In an equation for ‘filterVariables’:
          filterVariables lst = flip (filter (`notElem` lst))

How do I use flip correctly? Or can you suggest another method?

Answer 1

The flip function takes an argument function of two inputs. So if you want to flip the arguments to filterVariables , you have to flip it in a form that takes two arguments. In your version:

filterVariables lst = flip (filter ( `notElem` lst))

you have already accepted one argument, and only one remains to accept. We can fix this by first moving the lst argument to the other side of the equality:

filterVariables lst = filter ( `notElem` lst) -- original, accepts one more argument
filterVariables = \lst -> filter (`notElem` lst) -- now a function which accepts two arguments
filterVariables = flip (\lst -> filter (`notElem` lst)) -- flipped form

Answer 2

I think it might be simpler to write this as:

filterVariables :: Eq => [a] -> [a] -> [a]
filterVariables items lst = filter (`notElem` lst) items

You can not flip on filter (`notElem` lst) since, filter ( notElem lst) has type [a] -> [a] , and flip has type flip :: (b -> a -> c) -> a -> b -> c . This thus should mean that [a] -> [a] is the same type as b -> a -> c , but that implies that [a] and a -> c are the same type.

You can however make use of flip by implementing a point-free variant:

filterVariables :: (Foldable t, Eq a) => [a] -> t a -> [a]
filterVariables =flip ()

Indeed, here filter . flip notElem filter . flip notElem has type filter . flip notElem :: (Eq a, Foldable t) => ta -> [a] -> [a] filter . flip notElem :: (Eq a, Foldable t) => ta -> [a] -> [a] since it is equivalent to \\xs ys -> filter (flip notElem xs) ys or shorter \\xs -> filter (`notElem` xs) , so we can then flip xs and ys .