我可以使用SYB的gfoldl一次性完成listify结果的地图吗?
例如考虑以下代码:
extractNums :: Expr -> [Int]
extractNums e = map numVal $ listify isNum e
where isNum :: Expr -> Bool
isNum (Num _) = True
isNum _ = False
numVal :: Expr -> Int
numVal (Num i) = i
numVal _ = error "Somehow filter did not work?"
我不喜欢numVal函数中的那个我必须考虑Expr类型的不同数据构造函数,而我只对Num构造函数感兴趣。我宁愿用下面的vals函数替换isNum和numVals:
vals :: [Int] -> Expr -> [Int]
vals xs (Num x) = x : xs
vals xs _ = xs
这可以用gfoldl完成吗?怎么样?
答案 0 :(得分:2)
也许这不是最优雅的方法,但这是我的尝试:
extractNums :: Expr -> [Int]
extractNums e = everything (++) (mkQ [] q) e
where q (Num n) = [n]
q _ = []
但我希望它的表现低于标准。使用flip (++)
的Maye会更好吗?我现在无法看到它。
另一方面,我刚刚意识到listify
的定义方式类似。所以它至少比你现在的情况更糟糕。
或者,正如下面的@Alex所建议的那样:
import qualified Data.DList as D
extractNums :: Expr -> [Int]
extractNums e = D.toList $ everything (D.append) (mkQ D.empty q) e
where q (Num n) = D.singleton n
q _ = D.empty
答案 1 :(得分:2)
函数listify
定义为
-- | Get a list of all entities that meet a predicate
listify :: Typeable r => (r -> Bool) -> GenericQ [r]
listify p = everything (++) ([] `mkQ` (\x -> if p x then [x] else []))
类似于filter
。我们可以创建一个类似mapMaybe
的替代方案,将您需要的map
和filter
组合成一个:
import Data.Generics
import Data.Generics.Schemes
import Data.Maybe (maybeToList)
import Data.Typeable
listify' :: (Typeable t) => (t -> Maybe r) -> GenericQ [r]
listify' f = everything (++) ([] `mkQ` (maybeToList . f))
然后您的示例可以表示为
numVal :: Expr -> Maybe Int
numVal (Num i) = Just i
numVal _ = Nothing
test :: Expr -> [Int]
test = listify' numVal