[Haskell-cafe] HMatrix Vector/Matrix interpolation ala Matlab interp/interp2 ??

[gutti]

Hi Henning,

thanks for the code review -- reason that I don't use the type declaration a
lot -- It causes trouble , because I don't yet fully understand it. 

When I declare what I think is right is fails - see Message at bottom -- so
what's wrong ?

By the way I just used lists so far - no arrays -- Why is !! inefficient ? -
what is better - Cheers Phil

1 import Data.List
2
3 -- Input Data
4 xi :: [Double]
5 xi = [0 .. 10]
6 yi :: [Double]
7 yi = [2, 3, 5, 6, 7, 8, 9, 10 , 9, 8, 7]	
8 x = 11 :: Double
9 
10 -- Functions
11 limIndex :: [a] -> Int -> Int
12 limIndex xi idx 
13    | idx < 0 = 0
14     | idx > (length xi)-2 = (length xi)-2
15    | otherwise = idx 				
16
17 getIndex :: [a] -> Double -> Int
18 getIndex xi x = limIndex xi (maybe (length xi) id (findIndex (>x) xi)-1)
19 
20 getPnts :: [a] -> [a] -> Int -> [a]
21 getPnts xi yi idx = [xi !! idx, xi !! (idx+1), yi !! idx, yi !! (idx+1)]
22
23 interp :: [a] -> [a] -> Double -> Double
24 interp xi yi x = 
25 	let pts = getPnts xi yi (getIndex xi x) 		
26 	in (pts!!3-pts!!2)/(pts!!1-pts!!0)*(x-pts!!0)+pts!!2
27 
28 -- Calc 
29 y = interp xi yi x
30 
31 main = do 
32	-- Output Data
33   	print (y)

 === Compiler Error Message ===

interp_v4.hs:18:66:
    Couldn't match expected type `Double' against inferred type `a'
      `a' is a rigid type variable bound by
          the type signature for `getIndex' at interp_v4.hs:17:13
      Expected type: [Double]
      Inferred type: [a]
    In the second argument of `findIndex', namely `xi'
    In the third argument of `maybe', namely `(findIndex (> x) xi)'

interp_v4.hs:26:39:
    Couldn't match expected type `Double' against inferred type `a'
      `a' is a rigid type variable bound by
          the type signature for `interp' at interp_v4.hs:23:11
    In the second argument of `(-)', namely `pts !! 0'
    In the second argument of `(*)', namely `(x - pts !! 0)'
    In the first argument of `(+)', namely
        `(pts !! 3 - pts !! 2) / (pts !! 1 - pts !! 0) * (x - pts !! 0)'


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