Question

我需要对数字列表进行平方并将其除以列表的平均值。到目前为止我有：

square :: Float -> Float
square x = x * x

s2 :: [Float] -> Float
s2 xs = map square  (map (\c -> c - mean) xs)

任何帮助都会非常准确。

Answer 1

问题尚不清楚，但我假设您的意思是将平方的和除以平均值。简单版本：

sumOfSquaresOverMean :: Fractional a => [a] -> a
sumOfSquaresOverMean as = sum (map (^2) as) / mean as

mean :: Fractional a => [a] -> a
mean as = sum as / fromIntegral (length as)

这确实存在一个缺点，即列表将被遍历三次。 Gabriel Gonzalez的"Composable Streaming Folds"讨论了一个可能的解决方案，但它更先进，所以你现在可能想跳过它。

Answer 2

s2 :: [Float] -> Float
s2 xs = sum $ map ((/mean).(^2)) xs
  where mean= sum xs / fromIntegral (length xs)

测试：

λ: s2 [1.0, 2.0, 3.0]
7.0

将数字列表平方并将其除以列表的平均值

2 个答案: