List 如何更新多个列表中的数据Haskell

。更新最新里程数据列表中的数据， 删除每个位置最早的英里数

data People= People{ person:: String, age:: Int, weigth:: Float, miles:: [Float]} 

testData :: [People]
testData = [ person "sara"       19  50   [ 5, 8, 8],
          person "ed"      50  60   [0 , 1, 2],
          person "norman"      25   75     [ 2, 3 ,5]]

我正在尝试删除所有最后的数字，并将新的数字添加到给定的每个列表中[2,1,5]

---

因此，我将有[[2,5,8]，[1,0,1]，[5,2,3]]作为每个人的新里程

你可以使用

zipWith:：（a->b->c）->[a]->[b]->[c]

。它使用助手函数将两个列表压缩在一起

首先，您需要定义此帮助器，它接受

人员

和

浮点

，并返回更新的

人员

，例如

update :: Float -> People -> People
update n people = people { miles = newMiles }
   where newMiles = n : init (miles people)

这可以与

zipWith

一起使用，如下所示：

zipWith update [2, 1, 5] testData

---

请注意，

testData

的定义不正确。您必须使用

人员

数据构造函数，如下所示：

testData :: [People]
testData = [ People "sara" 19 50 [5, 8, 8]
           -- etc.
           ]

---

所以基本上你可以这样做

data People = People { person :: String
                     , age    :: Int
                     , weight :: Float
                     , miles  :: [Float]
                     } deriving (Show)

testData :: [People]
testData = [ People "sara"   19 50 [5,8,8]
           , People "ed"     50 60 [0,1,2]
           , People "norman" 25 75 [2,3,5]
           ]

addMiles :: [People] -> [Float] -> [People]
addMiles ps ms = zipWith f ps ms
                 where
                 f p m = p {miles = m : init (miles p)}

或者你也可以喜欢

addMiles :: [People] -> [Float] -> [People]
addMiles = zipWith (\p m -> p {miles = m : init (miles p)})

然后,

---

首先格式化代码，以便我们能够理解：/T%hanks对testData定义的更正

addMiles testData [2,1,5]
[ People {person = "sara", age = 19, weight = 50.0, miles = [2.0,5.0,8.0]}
, People {person = "ed", age = 50, weight = 60.0, miles = [1.0,0.0,1.0]}
, People {person = "norman", age = 25, weight = 75.0, miles = [5.0,2.0,3.0]}
]