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 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#Stat... So g normalize [2,5,3,2] -> [-0.7071067811865475,1.414213562373095,0.0,-0.7071067811865475] Is my typing for normalize too loose. Should I be using Floating rather than Num? Michael =======Code============== {- See Problem 42, pg. 63, Prolog by Example, Coelho & Cotta Generate a list of serial numbers for the items of a given list, the members of which are to be numbered in alphabetical order. *Main> serialize "prolog" [4,5,3,2,3,1] *Main> serialize "int.artificial" [5,7,9,1,2,8,9,5,4,5,3,5,2,6] *Main> ["prolog"] >>= serialize [4,5,3,2,3,1] *Main> ["int.artificial"] >>= serialize [5,7,9,1,2,8,9,5,4,5,3,5,2,6] -} import Data.Map hiding (map) import Data.List {- serialize :: [Char] -> [Int] serialize l = map (f l) l where f = ((!) . fromList . ((flip zip) [1..]) . (sort . nub)) -} serialize :: (Ord a, Integral b) => [a] -> a -> b serialize = ((!) . fromList . ((flip zip) [1..]) . (sort . nub)) g f l = map (f l) l 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) sumlen :: (Num a, Integral b) => [a] -> (a,b) sumlen l = sumlen' l 0 0 where sumlen' [] sum len = (sum,len) sumlen' (h:t) sum len = sumlen' t (sum+h) (len+1) ========================= Prelude> :r [1 of 1] Compiling Main ( serialize2.hs, interpreted ) serialize2.hs:34:32: Could not deduce (Integral a) from the context (Num a, Num b) arising from a use of `sumlen' at serialize2.hs:34:32-39 Possible fix: add (Integral a) to the context of the type signature for `normalize' In the expression: sumlen l In a pattern binding: (total, len) = sumlen l In the expression: let (total, len) = sumlen l avg = total / len stdev = sqrt $ ((/) (len - 1)) $ sum $ map ((** 2.0) . (subtract avg)) l in (/ stdev) . (subtract avg) serialize2.hs:36:61: Could not deduce (Floating a) from the context (Num a, Num b) arising from a use of `**' at serialize2.hs:36:61-66 Possible fix: add (Floating a) to the context of the type signature for `normalize' In the first argument of `(.)', namely `(** 2.0)' In the first argument of `map', namely `((** 2.0) . (subtract avg))' In the second argument of `($)', namely `map ((** 2.0) . (subtract avg)) l' serialize2.hs:37:18: Couldn't match expected type `b' against inferred type `a' `b' is a rigid type variable bound by the type signature for `normalize' at serialize2.hs:33:25 `a' is a rigid type variable bound by the type signature for `normalize' at serialize2.hs:33:18 In the expression: (/ stdev) . (subtract avg) In the expression: 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 definition of `normalize': normalize l = let (total, len) = sumlen l avg = total / len .... in (/ stdev) . (subtract avg) Failed, modules loaded: none.
Michael, just leaving out the type declaration for 'normalize', your module complies fine and ghc infers the following type: normalize :: (Integral a, Floating a) => [a] -> a -> a Note that the context (Integral a, Floating a) cannot be met by any of the standard types. (try in ghci: ":i Integral" and ":i Floating") So we have to apply a conversion function like this: (I just replaced len by len' at all occurrences)
normalize l = let (total,len) = sumlen l len' = fromIntegral len avg = total/len' stdev = sqrt $ ((/) (len'-1)) $ sum $ map ((** 2.0) . (subtract avg)) l in ((/) stdev) . (subtract avg)
yielding a type of normalize :: (Floating b) => [b] -> b -> b You could save the conversion by allowing a more liberal type for 'sumlen'. Without the type signature, it is inferred to sumlen :: (Num t, Num t1) => [t] -> (t, t1) -- Steffen On 01/31/2011 06:29 PM, michael rice wrote:
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
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#Stat...
So
g normalize [2,5,3,2] -> [-0.7071067811865475,1.414213562373095,0.0,-0.7071067811865475]
Is my typing for normalize too loose. Should I be using Floating rather than Num?
Michael
=======Code============== {- See Problem 42, pg. 63, Prolog by Example, Coelho & Cotta
Generate a list of serial numbers for the items of a given list, the members of which are to be numbered in alphabetical order.
*Main> serialize "prolog" [4,5,3,2,3,1] *Main> serialize "int.artificial" [5,7,9,1,2,8,9,5,4,5,3,5,2,6]
*Main> ["prolog"] >>= serialize [4,5,3,2,3,1] *Main> ["int.artificial"] >>= serialize [5,7,9,1,2,8,9,5,4,5,3,5,2,6] -}
import Data.Map hiding (map) import Data.List
{- serialize :: [Char] -> [Int] serialize l = map (f l) l where f = ((!) . fromList . ((flip zip) [1..]) . (sort . nub)) -}
serialize :: (Ord a, Integral b) => [a] -> a -> b serialize = ((!) . fromList . ((flip zip) [1..]) . (sort . nub))
g f l = map (f l) l
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)
sumlen :: (Num a, Integral b) => [a] -> (a,b) sumlen l = sumlen' l 0 0 where sumlen' [] sum len = (sum,len) sumlen' (h:t) sum len = sumlen' t (sum+h) (len+1) =========================
Prelude> :r [1 of 1] Compiling Main ( serialize2.hs, interpreted )
serialize2.hs:34:32: Could not deduce (Integral a) from the context (Num a, Num b) arising from a use of `sumlen' at serialize2.hs:34:32-39 Possible fix: add (Integral a) to the context of the type signature for `normalize' In the expression: sumlen l In a pattern binding: (total, len) = sumlen l In the expression: let (total, len) = sumlen l avg = total / len stdev = sqrt $ ((/) (len - 1)) $ sum $ map ((** 2.0) . (subtract avg)) l in (/ stdev) . (subtract avg)
serialize2.hs:36:61: Could not deduce (Floating a) from the context (Num a, Num b) arising from a use of `**' at serialize2.hs:36:61-66 Possible fix: add (Floating a) to the context of the type signature for `normalize' In the first argument of `(.)', namely `(** 2.0)' In the first argument of `map', namely `((** 2.0) . (subtract avg))' In the second argument of `($)', namely `map ((** 2.0) . (subtract avg)) l'
serialize2.hs:37:18: Couldn't match expected type `b' against inferred type `a' `b' is a rigid type variable bound by the type signature for `normalize' at serialize2.hs:33:25 `a' is a rigid type variable bound by the type signature for `normalize' at serialize2.hs:33:18 In the expression: (/ stdev) . (subtract avg) In the expression: 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 definition of `normalize': normalize l = let (total, len) = sumlen l avg = total / len .... in (/ stdev) . (subtract avg) Failed, modules loaded: none.
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On Monday 31 January 2011 18:29:59, michael rice wrote:
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).
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).
participants (3)
-
Daniel Fischer -
michael rice -
Steffen Schuldenzucker