获得计算列表haskell的平均值 [英] Getting average of calculated list haskell

问题描述

我有

  avgRatingsForDirector :: String  - > [电影]  - > [Int] 
 avgRatingsForDirector _ [] = [] 
 avgRatingsForDirector requestedDirector（（Film _ director _（（_，rating）：ratings））：restOfFilms）
 | requestedDirector == director = [rating] ++ avgRatingsForDirector requestedDirector restOfFilms 
 |否则= avgRatingsForDirector requestedDirector restOfFilms 
  

这会输出一个数字列表，我想让结果成为平均值这些数字。是可以在结果上使用foldr函数吗？

结果带有一个函数来计算任何数字列表的平均值。

编码它的一种方法是只对列表进行一次遍历

  { - ＃LANGUAGE BangPatterns＃ - } 
 import Data.List 
 
 avg ::（Integral a，Fractional b）=> [a]  - > b 
 avg xs = g $ foldl'c（0,0）xs 
其中
c（！a，！n）x =（a + x，n + 1）
g （a，n）= fromIntegral a / fromIntegral n 
  

Bang模式使计算更有效率。

另见：最富有你可以有：美丽的折叠

I have

avgRatingsForDirector :: String -> [Film] -> [Int]
avgRatingsForDirector _ [] = []
avgRatingsForDirector requestedDirector ((Film _ director _ ((_, rating):ratings)):restOfFilms)
    | requestedDirector == director = [rating] ++ avgRatingsForDirector requestedDirector restOfFilms
    | otherwise = avgRatingsForDirector requestedDirector restOfFilms

This outputs a list of numbers, I want to have the result be the average of these numbers. is is possible to use the foldr function on the result somehow?

解决方案

You can keep your function as is and post-process its result with a function to calculate an average of any list of numbers.

One way to code it so it does only one traversal of the list is

{-# LANGUAGE BangPatterns #-}
import Data.List

avg :: (Integral a, Fractional b) => [a] -> b
avg xs = g $ foldl' c (0,0) xs
 where
   c (!a,!n) x = (a+x,n+1)
   g (a,n) = fromIntegral a / fromIntegral n

Bang patterns make the calculation efficient.

see also: The Most Fuun You Can Have: Beautiful Folding

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