haskell - 压缩和计算

Question

我正在寻找不使用递归的解决方案。我有3个数据，例如：

names = ["Alice", "Bob", "Charlie", "David"]

-- their grades:
grades :: [[Int]]
grades = [[3, 2], [4, 6], [2, 3], [1, 4]]

-- And the weights for a total note
weights :: [Float]
weights = [0.75, 0.25]

这应该理解为：

Alice's grades:   3 and 2 -- The grade 3 weighted with 75% and the grade 2 with 25%
Bob's grades:     4 and 6 -- The grade 4 weighted with 75% and the grade 6 with 25%
Charlie's grades: 2 and 3 -- The grade 2 weighted with 75% and the grade 3 with 25%
David's grades:   1 and 4 -- The grade 1 weighted with 75% and the grade 4 with 25%

我写了一个计算averageGrade一个学生的函数：

averageGrade :: [Float] -> [Int] -> Float

我必须根据这种类型完成一个功能：

allStudents :: [String] -> [[Int]] -> [Float] -> [(String, Float)]

这个函数应该给出每个人的平均成绩的元组，例如：

[("Alice", 2.75), ("Bob", 4.5), etc....]

计算由 进行averageGrade (weights) ([Specific Grades])，例如

averageGrade (weights) ([3,2])

如果我想为爱丽丝计算它。

如何在不通过headand遍历列表的情况下迭代列表tail（因为不允许递归）？

标签： haskellhigher-order-functions

Answer 1

所以你要

allStudents  :: [String] -> [[Int]] -> [Float]            -> [(String, Float)]

这与

allStudents2 :: [String] -> [Float] -> [[Int]]            -> [(String, Float)]

你有

averageGrade ::             [Float] ->  [Int]  ->  Float

这样

averageGrade (weights ::    [Float]) :: [Int]  ->  Float

并且因为

map (foo ::                               a    ->   b      ) 
                           ::          [  a  ] -> [ b   ]

我们有

map (averageGrade weights) ::          [[Int]] -> [Float]

和

map (averageGrade weights) (grades :: [[Int]]) :: [Float]

最后一个缺失的部分是

_ ::            [ c    ]  ->                      [ b   ] -> [( c    , b    )]

这样

_ (names ::     [String]) ::                      [ b   ] -> [(String, b    )]