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在这种情况下如何使用mapM

haskell dictionary

rnso · 技术社区 · 6 年前

我正在尝试打印的结果的不同组成部分 autocorrelation

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;

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,如果取消最后一行的注释,则会出现错误:

â¢ No instance for (Foldable ((,,) (Vector Double) (Vector Double)))
    arising from a use of âPrelude.mapM_â
â¢ 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 âmainâ:
      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。

1 回复 | 直到 6 年前

Daniel Wagner 6 年前

最简单的就是模式匹配。

main = do
    let (a,b,c) = autocorrelation $ V.fromList sampleA
    print a
    print b
    print c

Joseph Sible-Reinstate Monica 6 年前

不想模式匹配你自己,你可以用 the fixed-vector package 来帮助你。请注意,它有自己的 mapM_ 你必须使用:

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,你可以做一个 newtype

{-# 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

willeM_ Van Onsem 6 年前

我们可以利用 lens [Hackage]

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)))

但最好是像这样进行模式匹配 @DanielWagner 也就是说,从那以后你就更清楚地知道发生了什么。