Average of user input - function

Question

I want to take the user input of a list of numbers and find the average. However, after looking for examples I have not found any that seems to match what I am doing because I can have from 2 - 100 numbers in a list. Any help/advice is appreciated.
Below is my working code as is.
main = do
putStrLn "Enter how many numbers:"
listlen <- getLine
if ((read listlen) <= 100) -- read converts a string to a number
then do
putStrLn "Enter a String of numbers:"
--numberString <- getLine
numberString<- getLine
let ints = map read (words numberString) :: [Int]
putStrLn("The List: " ++(numberString))
putStrLn("The average: ")
putStrLn("Number of values greater than average: ")
else do
putStrLn " Error: listlen must be less than or = to 100"
main

Answer 1

Ok, this is homework, but homework can be really tough when you have to do it in Haskell. I'll try to explain step by step how you can do.
Good to know
First, Haskell is functional. You can find different defintions of "functional", but basically you can think of it as a property of the language: everything is constant (no side effect).
Second, you can start a REPL by typing ghci in a terminal:
jferard#jferard-Z170XP-SLI:~$ ghci
GHCi, version 8.0.2: http://www.haskell.org/ghc/ :? for help
Prelude> :set +m -- enable parsing of multiline commands
Prelude> -- type the expression you want to evaluate here
Recursion
How do you compute the sum of the elements of a list? In an imperative language, you will do something like that (Python like):
s = 0
for every x in xs:
s <- s + x
But s is not constant. It's updated on every iteration until we get the sum. Can we reformulate the algorithm to avoid this mutation? Fortunately yes. Here's the key idea:
sum [x] = x
sum [x1, ..., xn] = x1 + sum [x2, ..., xn]
With a little imagination, you can say that sum [] = 0. So we can write it in Haskell:
sum [] = 0
sum (x:xs) = x + sum xs
-- sum [1, 2, 3, 5] == 11
(x:xs) means: x (the head) followed by the list xs (the tail). If you have understood that, you know how we can avoid side effects in many situations: just call another function to do the rest of the job. (Note: if you know about the stack, you can imagine what happens under the hood.)
Now, how do you compute the length of a list? In a Python-like language, you'd do something like (forget len(..)):
l = 0
for every x in xs:
l <- l + 1
Again, with a recursive definition, you have:
length [] = 0
length (x:xs) = 1 + length xs
-- len [1, 2, 3, 5] == 4
Folds
Computing sum and length is so common that they are built-in functions in Haskell. But there is something more important: if you examine the two functions carefully, you'll notice this pattern:
f [] = <initial value>
f (x:xs) = g(f xs, x)
For sum, initial value is 0 and g(f xs, x) = x + f xs. For length, initial value is 0 and g(f xs, x) = 1 + f xs. This pattern is so common that Haskell has a built-in function (actually several built-in functions) for it: foldl. foldl takes a function, an initial value and a list and returns the function repeatedly applied to the previous result and the current element, until the list is consumed. You can think of the function as the the body of the loop:
sum xs = foldl (+) 0 xs
(Note on curryfication: 1. You will maybe learn some day that Haskell functions always take one argument, but that's not the point here. 2. You can remove xs on both sides: sum = foldl (+) 0)
Prelude> foldl (+) 0 [1,2,3,5]
11 -- (((0+1)+2)+3)+5
With scanl, you can in some way "debug" the foldl:
Prelude> scanl (+) 0 [1,2,3,5]
[0,1,3,6,11]
length is more tricky, since you don't have a built-in function g(f xs, x) = 1 + f xs. You can use a lambda function: \acc x -> 1 + acc where acc is the current value:
length xs = foldl (\acc x -> 1 + acc) 0 xs
Your question
Average
Let's try to write average with the built-in sum and length functions:
Prelude> average xs = sum xs / length xs
<interactive>:1:14: error:
• Could not deduce (Fractional Int) arising from a use of ‘/’
...
What does that mean? I won't get into details, but you have to know that Haskell is very strict with numbers. You can't divide two integers and expect a float result without a little work.
Prelude> :t (/)
(/) :: Fractional a => a -> a -> a
This means that / will take Fractionals. Thus, the work is: cast integers into Fractionals.
average xs = fromIntegral (sum xs) / fromIntegral (length xs)
-- average [1, 2, 3, 5] == 2.75
Number of values greater than average
Now, the number of values greater than the mean. In a Python-like langage, you'll write:
c = 0
for every x in xs:
if x > avg:
c <- c + 1
Let's try the recursive method:
gtAvg [] = 0
gtAvg (x:xs) = (if x>avg then 1) + sum xs -- WRONG
You see that there is something missing. In the imperative version, if x <= avg, we simply do nothing (and thus do not update the value). Here, we must return something:
gtAvg [] = 0
gtAvg (x:xs) = (if x>avg then 1 else 0) + gtAvg xs
But where does the avg value come from? We need to precompute it. Let's define a function that takes avg as an argument:
gtAvg' [] _ = 0
gtAvg' (x:xs) avg = (if fromIntegral x>avg then 1 else 0) + gtAvg' xs avg
-- gtAvg' [1, 2, 3, 5] (average [1, 2, 3, 5]) == 2
And then:
gtAvg xs = gtAvg' xs (average xs)
-- gtAvg [1, 2, 3, 5] == 2
Obviously, this is more simple with a foldl:
gtAvg xs = foldl (\acc x -> if fromIntegral x>average xs then acc+1 else acc) 0 xs
More (map, filter and list comprehension)
When we are on the basics of Haskell, you may need three more constructs.
Filter
First, a filter:
Prelude> filter (>2.75) [1, 2, 3, 5]
[3.0,5.0]
If you take the length of that list, you get the number of elements greater than the average:
gtAvg xs = length $ filter (\x -> fromIntegral x >average xs) xs
(Or with a composition of functions: length $ filter ((> average xs).fromIntegral) xs) Don't be disturbed by the $ sign: it means that the right side of the expression (filter...) is one block, like if it were in parenthesis.
Map
Second, map applies a function to every element of a list and returns the list of mapped elements. For instance, if you want to some squares of elements of a list:
Prelude> sum $ map (**2) [1, 2, 3, 5]
39.0
You can use it like that:
gtAvg xs = length $ filter (>average xs) $ map fromIntegral xs
It converts elements to Fractional, then it applies the filter.
List comprehension
Third, you can have filter and a map with a list comprehension:
gtAvg xs = length [x | x<-xs, fromIntegral x>average xs]
I left a lot of things aside and made probably approximations, but now you should have the basic knowledge to answer your question.

Answer 2

listlen :: [a] -> Int
listlen xs = go 0 xs -- 0 = initial value of accumulator
where go s [] = s -- return accumulator
go s (a:as) = go (s+1) as -- compute the next value of the accumulator
sumx :: [a] -> Int
sumx xs = go 0 xs
where go s [] = s
go s (a:as) = go ... as -- flll in the blank ... -- and recurse
lenAndSum :: [a] -> (Int,Int)
lenAndSum xs = go (0,0) xs -- (0,0) = initial values of both accumulators
where go (s1,s2) [] = (s1,s2) -- return both accumulators at the end
go (s1,s2) (a:as) = go ... as -- left as an exercise
main = do
putStrLn "Enter how many numbers:"
listlen <- getLine
if ((read listlen) <= 100) -- read converts a string to a number
then do
putStrLn "Enter a String of numbers:"
--numberString <- getLine
numberString<- getLine
let ints = map read (words numberString) :: [Int]
putStrLn("The List: " ++(numberString))
putStrLn("The average: ")
putStrLn("Number of values greater than average: ")
else do
putStrLn " Error: listlen must be less than or = to 100"
main`enter code here`