Playing around with repa arrays and got some questions. 1) How I can get list of indexes of array that suffice some predicate? > a1 AUnboxed (Z :. 3) (fromList [False,False,True]) it :: Array U (Z :. Int) Bool Indexes of element that satisfying specific predicate could be obtained like that: > (\a p → Data.List.map (subtract 1 . snd) $ filter (p . fst) $ zip (toList a) [1..]) a1 (== False) [0,1] Looks ugly. How REPA users used to do filtering like that without converting to list? 2) How can I apply some function `f' to each row of 2D array `a' and collect results in single value? f ∷ (Shape sh, Source r Bool) ⇒ Array r sh Bool → Bool f a = (== toList a) $ foldl1 (Prelude.zipWith (||)) $ Prelude.map toList $ foldl (\l k -> filter (\x -> x ! (Z :. k) == False) l) [b1,b2,b3,b4] $ findWhich (== False) a and ∷ [Bool] → Bool [a1,a2] :: [Array U (Z :. Int) Bool] Having all that I could find what I want like that: and $ map f [a1,a2] > True All going on ridiculous and ugly because: - 2D arrays are not 2D arrays but lists of 1D arrays b1,b2,b3,b4,a1,a2 ∷ Array U (Z :. Int) Bool b1 = fromListUnboxed (Z :. (3::Int)) [False, True, False] b2 = fromListUnboxed (Z :. (3::Int)) [False, False, False] b3 = fromListUnboxed (Z :. (3::Int)) [False, False, True] b4 = fromListUnboxed (Z :. (3::Int)) [True, False, False] a1 = fromListUnboxed (Z :. (3::Int)) [False, False, True] a2 = fromListUnboxed (Z :. (3::Int)) [True, True, True] How 2D array could be split to list of 1D arrays? - redundant usage of `toList'; all operations are list-specified. How `f' could be rewritten in REPA terms? -- Best regards, dmitry malikov !
Dmitry Malikov
Playing around with repa arrays and got some questions.
1) How I can get list of indexes of array that suffice some predicate?
> a1 AUnboxed (Z :. 3) (fromList [False,False,True]) it :: Array U (Z :. Int) Bool
Indexes of element that satisfying specific predicate could be obtained like that:
> (\a p → Data.List.map (subtract 1 . snd) $ filter (p . fst) $ zip (toList a) [1..]) a1 (== False) [0,1]
Looks ugly. How REPA users used to do filtering like that without converting to list?
I hope someone will correct me if I am wrong and furthermore I was not entirely clear what you were trying to do but it seems to me that if you want to filter out an unknown number of elements from a collection then repa is the wrong abstraction to use. You can however filter out a known number of elements e.g. xs = Repa.fromListUnboxed (Z :. 3) [1, 2, 3] removeOne ix xs = Repa.fromFunction (Z :. dx - 1) (\(Z :. jx) -> xs ! (Z :. f jx)) where Z :. dx = Repa.extent xs f jx | jx < ix = jx | otherwise = jx + 1 test = Repa.computeP $ removeOne 1 xs :: IO (Array U DIM1 Float) Does that help? Dominic.
On 11/19/2012 01:21 AM, Dominic Steinitz wrote:
Dmitry Malikov
writes: Playing around with repa arrays and got some questions.
1) How I can get list of indexes of array that suffice some predicate?
> a1 AUnboxed (Z :. 3) (fromList [False,False,True]) it :: Array U (Z :. Int) Bool
Indexes of element that satisfying specific predicate could be obtained like that:
> (\a p → Data.List.map (subtract 1 . snd) $ filter (p . fst) $ zip (toList a) [1..]) a1 (== False) [0,1]
Looks ugly. How REPA users used to do filtering like that without converting to list?
I hope someone will correct me if I am wrong and furthermore I was not entirely clear what you were trying to do but it seems to me that if you want to filter out an unknown number of elements from a collection then repa is the wrong abstraction to use.
You can however filter out a known number of elements e.g.
xs = Repa.fromListUnboxed (Z :. 3) [1, 2, 3]
removeOne ix xs = Repa.fromFunction (Z :. dx - 1) (\(Z :. jx) -> xs ! (Z :. f jx)) where Z :. dx = Repa.extent xs f jx | jx < ix = jx | otherwise = jx + 1
test = Repa.computeP $ removeOne 1 xs :: IO (Array U DIM1 Float)
Does that help?
Dominic.
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Does that help? Yep, kinda, this is a nice example of fromFunction usage, but my question was about retrieving indexing of elements that satisfy some predicate.
Thanks for response. -- Best regards, dmitry malikov !
participants (2)
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Dmitry Malikov -
Dominic Steinitz