Haskell修复 - 如何减少数组和返回索引？

In GNU Octave this code -

在GNU Octave中这段代码 -

[e, ix] = min(X);

will return minimum element and it's location. How do you this in repa for arbitrary binary function?

将返回最小元素及其位置。你如何在修复任意二进制函数？

This is what I came up with:

这就是我提出的：

min x = z $ foldl' f (e,0,0) es
  where
    (e:es) = toList x
    f (a,ix,r) b = let ix' = ix+1 in if a < b then (a,ix',r) else (b,ix',ix')
    z (a,ix,r) = (a,r)

In above example we convert repa 1D matrix to list and use foldl' (from Data.List) with two accumulators - one for counting iterations (ix) and other to save position of min element (r). But the whole point of using repa is to use arrays, not lists!

在上面的例子中，我们将repa 1D矩阵转换为list并使用foldl'（来自Data.List）和两个累加器 - 一个用于计算迭代（ix），另一个用于保存min元素（r）的位置。但是使用repa的全部意义是使用数组，而不是列表！

In repa there are two folds for Array type (foldS and foldP) - but they can only take function of type (a -> a -> a) - meaning, I cannot pass tuple with accumulators to it. There is also traverse, which can, in principle, reduce 1D array to a scalar array:

在repa中，Array类型有两个折叠（foldS和foldP） - 但它们只能取类型函数（a - > a - > a） - 意思是，我不能将带累加器的元组传递给它。还有遍历，原则上可以将1D数组减少为标量数组：

min x = traverse x to0D min
  where
    to0D (Z:.i) = Z
    min f (Z) = ??? -- how to get elements for comparison?

The first thing that comes to mind is

首先想到的是

[f (Z:.i) | i <- [1..n]], where n = (\(Z:.i) -> i) $ extent x

But this will also convert array to list, instead of doing computation on the array.

但这也会将数组转换为列表，而不是在数组上进行计算。

2 个解决方案

#1

I'm no expert on Repa, but this seems to work for 1-D arrays. It can probably be adapted for other dimensions.

我不是维修专家，但这似乎适用于一维阵列。它可能适用于其他方面。

import Data.Array.Repa

indexed arr = traverse arr id (\src idx@(Z :. i) -> (src idx, i))

minimize arr = foldP f h t
  where
    (Z :. n) = extent arr
    arr' = indexed arr
    h = arr' ! (Z :. 0)
    t = extract (Z :. 1) (Z :. (n-1)) arr'
    f min@(valMin, iMin) x@(val, i) | val < valMin = x
                                    | otherwise = min

#2

At the risk of reviving a zombie post, here is a arbitrary dimension solution I worked out (which, reviewing the comments, is exactly what @hammar suggested). Because of the weird nature of foldAllP, which I use as an identity element when merging tuples, you also need to provide an upper bound for the minimum of the array you are seeking to find.

冒着恢复僵尸帖子的风险，这是我制定的任意维度解决方案（查看评论，正是@hammar建议的那样）。由于foldAllP的奇怪性质，我在合并元组时将其用作标识元素，因此您还需要为要搜索的数组的最小值提供上限。

import Data.Array.Repa
import Data.Array.Repa.Eval
import Data.Array.Repa.Repr.Unboxed

minimize :: (Shape sh, Source r a, Elt a, Unbox a, Monad m, Ord a) => Array r sh a -> a -> m (a,sh)
minimize arr upperBound = do
      -- Zip the array with its (Int representation of) indices
      let zippedWithIndex = Data.Array.Repa.traverse arr id (\f idx -> (f idx, toIndex (extent arr) idx))

      -- Define a function that compares two tuple of (a,Int) on the value of the first entry
      let minimumIndex t@(v,_) t'@(v',_) = if v < v' then t else t'

      -- Do a parallel fold of all elements in the array 
      (min,idx) <- foldAllP minimumIndex (upperBound, error "Empty array") zippedWithIndex

      -- convert the indice back from its Int representation
      return (min, fromIndex (extent arr) idx)

Naturally, if your array contains a type that has an instance of Bounded, you can make the more convenient function

当然，如果您的数组包含具有Bounded实例的类型，则可以使更方便的函数

 minimize' arr = minimize arr maxBound

Some tests you can do at the prompt:

您可以在提示符处执行一些测试：

 λ> let x = fromListUnboxed (ix2 2 2) [20, 30, 10, 40] :: Array U DIM2 Int
 λ> minimize' x
(10,(Z :. 1) :. 0)

#1