在Haskell中有效地计算列表的均值 [英] Computing the mean of a list efficiently in Haskell

问题描述

我设计了一个函数来计算列表的均值。虽然它工作正常，但我认为它可能不是最好的解决方案，因为它需要两个功能而不是一个。

I've designed a function to compute the mean of a list. Although it works fine, but I think it may not be the best solution due to it takes two functions rather than one. Is it possible to do this job done with only one recursive function ?

calcMeanList (x:xs) = doCalcMeanList (x:xs) 0 0

doCalcMeanList (x:xs) sum length =  doCalcMeanList xs (sum+x) (length+1)
doCalcMeanList [] sum length = sum/length

推荐答案

您的解决方案很好，使用两个函数并不比一个更差。尽管如此，你可能会把尾递归函数放在其中子句中。

Your solution is good, using two functions is not worse than one. Still, you might put the tail recursive function in a where clause.

但是，如果你想在一行：

But if you want to do it in one line:

calcMeanList = uncurry (/) . foldr (\e (s,c) -> (e+s,c+1)) (0,0)

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