michael rice nowgate at yahoo.com
Mon Jan 31 20:23:02 CET 2011

```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 <daniel.is.fischer at googlemail.com> wrote:

From: Daniel Fischer <daniel.is.fischer at googlemail.com>
Cc: "michael rice" <nowgate at yahoo.com>
Date: Monday, January 31, 2011, 1:38 PM

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).

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