(Why we did this.)
Revision as of 12:02, 16 February 2007
I just wrote this function:
apply :: (Ord k) => k -> v -> (v -> v) -> [(k,v)] -> [(k,v)] apply k v f ds = let (p1,px) = span ( (k >) . fst) ds (p2,p3) = case px of  -> ((k,v),) (x:xs) -> if fst x == k then ((k, f $ snd x), xs) else ((k, v), x:xs) in p1 ++ (p2 : p3)
As you can see (?!), this takes a list of key/value pairs and processes it as follows:
- The function is given a key to look for.
- If the key is found, a function is applied to the associated value.
- If the key is not found, it is inserted (at the correct place) with a specified 'default value'.
1 Data.MapWhen you are making excessive use of (key,value) pairs it is usually time to switch to
insertWith :: Ord k => (a -> a -> a) -> k -> a -> Map k a -> Map k a
Here the update function receives the new value as well. --Twanvl
Thanks for the tip! A whole new module for me to learn. My oh my... I do love the way Haskell type signatures almost tell you what the whole function does!
MathematicalOrchid 19:27, 15 February 2007 (UTC)
2 Outer Context
This function actually occurred in the definition of a larger utility. It has the type
decode :: (Eq k) => [([k],[v])] -> [k] -> [v]
This takes a large input, searches the lookup table for the longest possible matching key, and returns the corresponding value. It then processes the rest of the input the same way.The