I hadn't considered the types of the functions I call in the function I'm trying to write, something not usually needed in loosely typed languages with coercion, but something I'm going to have to make a habit of doing. One more thing to add to the check list.
Also, I had considered the numbers in the list to be integral if they didn't have decimal points, which inferred, for me, that polymorphic type a needed to be of a class that would accept either Integral or Floating values, i.e., Num. False reasoning.
Thanks, all.
Michael
--- On Mon, 1/31/11, Daniel Fischer
I'm mapping a function over a list of data, where the mapping function is determined from the data.
g f l = map (g l) l
g f l = map (f l) l probably
So
g serialize "prolog" -> [4,5,3,2,3,1]
But I'm having typing problems trying to do a similar thing with a function that statistically normalizes data.
See: http://people.revoledu.com/kardi/tutorial/Similarity/Normalization.html# Statistic
So
g normalize [2,5,3,2] -> [-0.7071067811865475,1.414213562373095,0.0,-0.7071067811865475]
Is my typing for normalize too loose.
You can omit the type signatures and see what the compiler infers as the type. In this case,
normalize :: (Num a, Num b) => [a] -> a -> b normalize l = let (total,len) = sumlen l avg = total/len stdev = sqrt $ ((/) (len-1)) $ sum $ map ((** 2.0) . (subtract avg)) l in ((/) stdev) . (subtract avg)
In the final result, I suppose it should be (/ stdev) and not ((/) stdev) [the latter is (stdev /), i.e. \x -> stdev / x]. by sumlen's type, len has an Integral type. You want to use (/) to divide, which gives a Fractional constraint, (/) :: Fractional a => a -> a -> a Since it is not sensible for a type to be a member of both, the Fractional and Integral classes, you should convert len to the appropriate type with fromIntegral. For stdev, you call sqrt :: Floating a => a -> a and (**) :: Floating a => a -> a which means the list elements must have a type belonging to Floating (you could replace the (** 2.0) with (^ 2), which would probably be better, but the Floating constraint remains due to the sqrt). Finally, the resulting function is \x -> (x - avg) / stdev, hence x must have the same type as abg and stdev, and the final result has the same type. Altogether, normalize :: Floating a => [a] -> a -> a normalize l = let (total, len0) = sumlen l len = fromIntegral len0 avg = total/len stdev = sqrt $ sum [(x-avg)^2 | x <- l] / (len-1) in (/ stdev) . subtract avg but that gives nonsense if you pass a complex-valued list, so it might be better to restrict the type to normalize :: RealFloat a => [a] -> a -> a
Should I be using Floating rather than Num?
You have to, and one number type only (well, you could use two or three types if you compose with conversion functions, realToFrac for example).