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Yasmine Mustafa
Yasmine Mustafa

Reputation: 11

Haskell Programming

I need to implement a function that takes a list of Dists and returns a list of Dists. I need the function to return only the Dists with the label "pass", but somehow this does not work. Any help?

data Ex = Ex Float Float String String deriving Show 
data NewSt = NewSt Float Float String deriving Show 
data Dist = Dist Float NewSt Ex deriving Show 

helper1 [] = []
helper1 (Dist x (NewSt midterm1 quiz1 name1) (Ex midterm quiz name label) : xs) = if (label == "pass")
  then Dist x (NewSt midterm1 quiz1 name1) (Ex midterm quiz name label) : (helper1 xs) 
  else helper1 xs

Upvotes: 0

Views: 75

Answers (2)

chepner
chepner

Reputation: 532418

This is a little simpler to write with more pattern matching than with an if expression.

helper1 :: [Dist] -> [Dist]
helper1 [] = []
helper1 (Dist x newst (Ex midterm quiz name "pass") : xs) = Dist x newst (Ex midterm quiz name "pass") : (helper1 xs)
helper (_:xs) = helper1 xs

However, it is even simpler once you recognize that your recursion is already implemented by the filter function.

helper1 :: [Dist] -> [Dist]
helper1 = filter passed
          where passed (Dist _ _ (Ex _ _ _ "pass")) = True
                passed _ = False

Upvotes: 2

Redu
Redu

Reputation: 26201

I guess you may do as follows;

data Ex    = Ex Float Float String String deriving Show 
data NewSt = NewSt Float Float String deriving Show 
data Dist  = Dist Float NewSt Ex deriving Show 

myData :: [Dist]
myData = [Dist 1 (NewSt 66 100 "John") (Ex 55 90 "John" "pass"),
          Dist 2 (NewSt 20 45 "Tom") (Ex 33 50 "Tom" "fail"),
          Dist 3 (NewSt 75 75 "Mary") (Ex 90 100 "Mary" "pass")]

helper1 [] = []
helper1 d@(Dist _ _ (Ex _ _ _ label):ds) | label == "pass" = (head d):(helper1 ds)
                                         | otherwise       = helper1 ds

Upvotes: 0

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