collectF :: (Eq a) => (a -> Bool) -> [a] -> ([a], [a])
collectF f [] = ([], [])
collectF f (x : xs)
| f x = let (ys, zs) = collectF f xs in (x : ys, zs)
| otherwise = ([], x : xs)
groupBy :: (Eq a) => (a -> a -> Bool) -> [a] -> [[a]]
groupBy _ [] = []
groupBy f (x : xs) =
let (ys, zs) = collectF (f x) xs in
(x : ys) : groupBy f zs
*Main> groupBy (\x y -> snd x == snd y) [(1,True),(2,True),(3,False),(4,True),(5,False),(6,False),(7,True)]
[[(1,True),(2,True)],[(3,False)],[(4,True)],[(5,False),(6,False)],[(7,True)]]
我将让您从列表中删除 True 和 False 值。另外,请查看 Haskell 的 List 库 [1]。希望我已经说得很清楚了,但是如果您还有其他问题,请告诉我。