3

我正在尝试在不同的行上打印自相关结果的不同分量：

import Data.Vector as V
import Statistics.Autocorrelation
import Data.Typeable

sampleA = [1.0, 2.0, 3.0, 4.0, 1.0, 2.0, 3.0, 4]

main = do
    let res = autocorrelation $ V.fromList sampleA
    putStr "Type of result of autocorrelation test: "
    print $ typeOf res
    print res
    -- Prelude.mapM_ print res      -- not working;

输出是：

Type of result of autocorrelation test: ((Vector Double),(Vector Double),(Vector Double))

([1.0,2.5e-2,-0.45,-0.325,0.5,0.125,-0.15],[1.0,-1.3255000000000001,-1.4039375473415425,-1.442999810318651,-1.5311377955236107,-1.5364636906417393,-1.544097864842309],[1.0,1.0755000000000001,1.1539375473415425,1.192999810318651,1.2811377955236107,1.2864636906417393,1.294097864842309])

但是，如果我取消注释最后一行，我会收到错误消息：

• No instance for (Foldable ((,,) (Vector Double) (Vector Double)))
    arising from a use of ‘Prelude.mapM_’
• In a stmt of a 'do' block: Prelude.mapM_ print res
  In the expression:
    do { let res = autocorrelation $ fromList sampleA;
         putStr "Type of result of autocorrelation test: ";
         print $ typeOf res;
         print res;
         .... }
  In an equation for ‘main’:
      main
        = do { let res = ...;
               putStr "Type of result of autocorrelation test: ";
               print $ typeOf res;
               .... }

如何在单独的行上打印结果的所有部分？谢谢你的帮助。

4

3 回答 3

5

最简单的事情就是模式匹配。

main = do
    let (a,b,c) = autocorrelation $ V.fromList sampleA
    print a
    print b
    print c
于 2019-09-14T01:51:14.270 回答
4

如果您真的不想自己进行模式匹配，可以使用固定向量包来帮助您。请注意，它有自己的mapM_，您必须使用：

import Data.Vector as V
import qualified Data.Vector.Fixed as F
import Statistics.Autocorrelation
import Data.Typeable

sampleA = [1.0, 2.0, 3.0, 4.0, 1.0, 2.0, 3.0, 4]

main = do
    let res = autocorrelation $ V.fromList sampleA
    putStr "Type of result of autocorrelation test: "
    print $ typeOf res
    print res
    F.mapM_ print res

或者，您可以制作一个newtype并将您的元组粘贴在其中：

{-# LANGUAGE DeriveFoldable #-}

import Data.Vector as V
import Statistics.Autocorrelation
import Data.Typeable

newtype Triple a = Triple (a, a, a) deriving(Foldable)

sampleA = [1.0, 2.0, 3.0, 4.0, 1.0, 2.0, 3.0, 4]

main = do
    let res = autocorrelation $ V.fromList sampleA
    putStr "Type of result of autocorrelation test: "
    print $ typeOf res
    print res
    Prelude.mapM_ print $ Triple res
于 2019-09-14T03:55:42.897 回答
2

我们可以使用lens[Hackage]包将元组转换为列表：

import Control.Lens(toListOf, each)
import qualified Data.Vector as V
import Statistics.Autocorrelation

sampleA = [1.0, 2.0, 3.0, 4.0, 1.0, 2.0, 3.0, 4]

main :: IO ()
main = mapM_ print (toListOf each (autocorrelation (V.fromList sampleA)))

但是像@DanielWagner所说的那样进行模式匹配可能会更好，因为这样你就可以更明确地说明正在发生的事情。

于 2019-09-14T08:56:18.230 回答