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I try to find the center of all the red pixels contained in an image using Haskell and repa.

My problem is, that the code is not fast enough, probably because I read the image with repa-io which returns an Array Z DIM3 Word8, but I was not able to figure out how to filter the repa array by how red the given pixels are.

That's why I convert the array to a normal list, using R.toList. The rest of the calculations are then done using lists.

Here is my code:

import Data.Array.Repa.Repr.ForeignPtr (F, fromForeignPtr, toForeignPtr)
import Data.Array.Repa (Z(..), (:.)(..), DIM0, DIM1, DIM2, DIM3, Array(..),
    Source, Shape)
import qualified Data.Array.Repa as R
import Data.Array.Repa.IO.DevIL
import Data.List

import System.Environment (getArgs)

main :: IO ()
main = do
    [path] <- getArgs
    (RGBA arr) <- runIL $ readImage path
    putStrLn $ "red center " ++ (show . findRedDot $ arr)

-- | find the red dot
findRedDot :: (Ord a, Num a, Source r a) => Array r DIM3 a -> (Int, Int)
findRedDot img = (median . fst $ splitted, median . snd $ splitted)
    where -- function returning the coordinates as a tuple, when the pixel
          -- is red enough, otherwise return (-1, -1)
          convert f (Z :. i :. j) = let r = f (Z :. i :. j :. 0)
                                        g = f (Z :. i :. j :. 1)
                                        b = f (Z :. i :. j :. 2) in
                                    if r > 237 && (g + b) < 50
                                    then (i, j)
                                    else (-1, -1)
          -- turn the 3rd dimension into tuples containing the coordinates
          -- (see convert function)
          packed = R.traverse img
                   (\(Z :. w :. h :. _) -> Z :. w :. h)
                   convert
          -- turns the DIM2 repa array into a default haskell list
          lst = R.toList packed
          p (-1, -1) = False
          p _        = True
          -- split up the list into x and y component
          splitted = splitUp $ filter p lst

-- | split up a list of tuples into two lists
splitUp :: [(Int, Int)] -> ([Int], [Int])
splitUp [] = ([], [])
splitUp ((x, y):rest) = (x:(fst next), y:(snd next))
    where next = splitUp rest

-- | find the median of a numeric list
median :: (Num a, Ord a) => [a] -> a
median [] = -1
median l = sorted !! mid
    where len = length l
          mid = len `quot` 2
          sorted = sort l

I did compile with ghc -O2 -threaded -prof -fprof-auto Main.hs.

Running this with ./Main.hs example.png +RTS -p gives me the correct result after some time, when using a 1920*1080 image which is completely red (This is the worst case, because finding the median is very time consuming).

Content of Main.prof

Mon May  8 21:26 2017 Time and Allocation Profiling Report  (Final)

       test5 +RTS -p -RTS fully-red-extreme.png

    total time  =        0.81 secs   (809 ticks @ 1000 us, 1 processor)
    total alloc = 2,248,665,408 bytes  (excludes profiling overheads)

COST CENTRE          MODULE    %time %alloc

median.sorted        Main       48.2   44.7
splitUp              Main       10.5   13.3
findRedDot.convert   Main       10.0    2.2
findRedDot.lst       Main        7.0   19.2
splitUp.next         Main        6.4    2.2
findRedDot.splitted  Main        3.5    5.2
findRedDot.convert.r Main        3.2    4.4
main                 Main        2.8    0.0
findRedDot.convert.b Main        2.6    4.4
findRedDot.convert.g Main        1.4    4.4
median.len           Main        1.4    0.0


                                                                     individual     inherited
COST CENTRE                MODULE                  no.     entries  %time %alloc   %time %alloc

MAIN                       MAIN                     58           0    0.1    0.0   100.0  100.0
 main                      Main                    117           0    2.8    0.0    99.9  100.0
  findRedDot               Main                    118           1    0.0    0.0    97.0  100.0
   findRedDot.splitted     Main                    123           1    3.5    5.2    21.3   20.7
    splitUp                Main                    130     2073601   10.5   13.3    16.9   15.5
     splitUp.next          Main                    133     2073600    6.4    2.2     6.4    2.2
    findRedDot.p           Main                    124     2073600    0.9    0.0     0.9    0.0
   findRedDot.lst          Main                    122           1    7.0   19.2    25.2   34.7
    findRedDot.packed      Main                    125           0    1.0    0.0    18.2   15.5
     findRedDot.convert    Main                    126     2073600   10.0    2.2    17.2   15.5
      findRedDot.convert.b Main                    129     2073600    2.6    4.4     2.6    4.4
      findRedDot.convert.g Main                    128     2073600    1.4    4.4     1.4    4.4
      findRedDot.convert.r Main                    127     2073600    3.2    4.4     3.2    4.4
   findRedDot.packed       Main                    120           1    0.0    0.0     0.0    0.0
    findRedDot.packed.\    Main                    121           1    0.0    0.0     0.0    0.0
   median                  Main                    119           2    1.0    0.0    50.6   44.7
    median.sorted          Main                    134           2   48.2   44.7    48.2   44.7
    median.len             Main                    132           2    1.4    0.0     1.4    0.0
    median.mid             Main                    131           2    0.0    0.0     0.0    0.0
 CAF                       Main                    115           0    0.0    0.0     0.0    0.0
  main                     Main                    116           1    0.0    0.0     0.0    0.0
 CAF                       GHC.IO.Encoding         102           0    0.0    0.0     0.0    0.0
 CAF                       GHC.IO.Handle.FD        100           0    0.0    0.0     0.0    0.0
 CAF                       GHC.Event.Thread         98           0    0.0    0.0     0.0    0.0
 CAF                       GHC.IO.Encoding.Iconv    94           0    0.0    0.0     0.0    0.0
 CAF                       GHC.Conc.Signal          84           0    0.0    0.0     0.0    0.0
 CAF                       GHC.IO.Handle.Text       79           0    0.0    0.0     0.0    0.0

The execution time is definitely longer than the one given in the .prof file and I don't know whether the sorting of the list can be accelerated in some way.

The most important goal is to spare the conversion from a repa array into a list and to do the whole computation with repa arrays.

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  • \$\begingroup\$ Looking at the docs, Data.Repa.Array.Auto, Data.Repa.Array.Generic and Data.Repa.Array.Material each define filter. \$\endgroup\$
    – Gurkenglas
    May 9 '17 at 6:01
  • \$\begingroup\$ Small remark: That's the alpha of version 4, @Gurkenglas. Erich uses version 3. repa-array is not an official variant (yet). \$\endgroup\$
    – Zeta
    May 9 '17 at 6:30
  • \$\begingroup\$ I would prefer to use the official version, but I will definitely have a look at this. \$\endgroup\$
    – Erich
    May 9 '17 at 7:27
  • \$\begingroup\$ Ok, I looked at the docs, but I think the Array type has completely changed. Could you give me a hint where to look for instructions? Or could you give me a short example? \$\endgroup\$
    – Erich
    May 9 '17 at 15:23
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Small things I noticed:

  • splitUp is already defined in the prelude as unzip.
  • median working on a list makes it slow and your implementation is not semantically correct.
    Generally the median is defined as the "center" element for datasets of odd magnitude, or the arithmetic mean of the two center elements for datasets of even magnitude.

    If you made it use an array, you'd also get the vast benefit of not needing the len traverse the full list, nor to traverse it when accessing a specific index, which should cut execution time for that function roughly in half.

    According to the profiling, that will also vastly reduce overall execution time.

  • The argument to convert f in findRedDot is repeated unnecessarily a lot, it should be possible to do the following:

    convert f a@(Z:. i :. j) = let r = f (a :. 0)
                                   g = f (a :. 1)
                                   b = f (a :. 2)  in
                               if r > 237 && (g+b) < 50
                               then (i,j) 
                               else (-1, -1)
    

    This makes it a tad easier to understand what exactly you do here, IMO

  • You speak of not being able to figure out a way to filter the repa array. A bit of hoogling reveals selectP, which should help with that. They explicitly mention:

    Produce an array by applying a predicate to a range of integers. If the predicate matches, then use the second function to generate the element.

    • This is a low-level function helpful for writing filtering operations on arrays.
    • Use the integer as the index into the array you're filtering.
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2
  • \$\begingroup\$ selectP does only work for integers, but I have to filter the array by looking at the last dimension in a whole. What is your suggestion concerning the median problem, I don't really see, what you are trying to say. \$\endgroup\$
    – Erich
    Oct 10 '17 at 16:31
  • \$\begingroup\$ @Erich your median implementation iterates the list thrice. If you had an array, that'd more likely be around 1.7 times for sorting and then a constant-time access to the median element(s). Re: selectP: Read the quoted section again. This is what they use for filtering arrays. Reread if you don't understand it yet... \$\endgroup\$
    – Vogel612
    Oct 10 '17 at 16:43

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