Haskell +师

Question

我能够对所有元组的第z个元素求和，但由于一些奇怪的原因，我将结果除以2后得到1。

averageYear :: [(String, String, Int)] -> Int
averageYear [] = 0
averageYear ((x,y,z):xs) = (z + (averageYear xs)) `div` (length xs)

谢谢！

Answer 1

数学是完全错误的。

averageYear [(_, _, 2010)] = (2010 + averageYear []) `div` 2
                           = (2010 + 0) `div` 2
                           = 2010 `div` 2
                           = 1005

要计算平均值，您必须将总和除以项目数。

average :: [Int] -> Int
average xs = sum xs `div` length xs

averageYear :: [(String, String, Int)] -> Int
averageYear = average . map (\(_, _, year) -> year)

Answer 2

@Dietrich Epp  在您的示例中，长度xs为零而不是两个。

   averageYear ((x,y,z):xs) = (z + (averageYear xs)) `div` (length xs)

如果xs为空，则长度xs等于零，并除以零。