I'm writing a function like this:
testing :: [Int] -> [Int] -> [Int]
testing lst1 lst2 =
let t = [ r | (x,y) <- zip lst1 lst2, let r = if y == 0 && x == 2 then 2 else y ]
let t1 = [ w | (u,v) <- zip t (tail t), let w = if (u == 2) && (v == 0) then 2 else v]
head t : t1
What the first let does is: return a list like this: [2,0,0,0,1,0], from the second let and the following line, I want the output to be like this: [2,2,2,2,1,0]. But, it's not working and giving parse error!!
What am I doing wrong?
There are two kinds of lets: the "let/in" kind, which can appear anywhere an expression can, and the "let with no in" kind, which must appear in a comprehension or do block. Since your function definition isn't in either, its let's must use an in, for example:
testing :: [Int] -> [Int] -> [Int]
testing lst1 lst2 =
let t = [ r | (x,y) <- zip lst1 lst2, let r = if y == 0 && x == 2 then 2 else y ] in
let t1 = [ w | (u,v) <- zip t (tail t), let w = if (x == 2) && (y == 0) then 2 else y] in
return (head t : t1)
Alternately, since you can define multiple things in each let, you could consider:
testing :: [Int] -> [Int] -> [Int]
testing lst1 lst2 =
let t = [ r | (x,y) <- zip lst1 lst2, let r = if y == 0 && x == 2 then 2 else y ]
t1 = [ w | (u,v) <- zip t (tail t), let w = if (x == 2) && (y == 0) then 2 else y]
in return (head t : t1)
The code has other problems, but this should get you to the point where it parses, at least.
With an expression formed by a let-binding, you generally need
let bindings
in
expressions
(there are exceptions when monads are involved).
So, your code can be rewritten as follows (with simplification of r and w, which were not really necessary):
testing :: [Int] -> [Int] -> [Int]
testing lst1 lst2 =
let t = [ if y == 0 && x == 2 then 2 else y | (x,y) <- zip lst1 lst2]
t1 = [ if (v == 0) && (u == 2) then 2 else v | (u,v) <- zip t (tail t)]
in
head t : t1
(Note, I also switched u and v so that t1 and t has similar forms.
Now given a list like [2,0,0,0,1,0], it appears that your code is trying to replace 0 with 2 if the previous element is 2 (from the pattern of your code), so that eventually, the desired output is [2,2,2,2,1,0].
To achieve this, it is not enough to use two list comprehensions or any fixed number of comprehensions. You need to somehow apply this process recursively (again and again). So instead of only doing 2 steps, we can write out one step, (and apply it repeatedly). Taking your t1 = ... line, the one step function can be:
testing' lst =
let
t1 = [ if (u == 2) && (v == 0) then 2 else v | (u,v) <- zip lst (tail lst)]
in
head lst : t1
Now this gives:
*Main> testing' [2,0,0,0,1,0]
[2,2,0,0,1,0]
, as expected.
The rest of the job is to apply testing' as many times as necessary. Here applying it (length lst) times should suffice. So, we can first write a helper function to apply another function n times on a parameter, as follows:
apply_n 0 f x = x
apply_n n f x = f $ apply_n (n - 1) f x
This gives you what you expected:
*Main> apply_n (length [2,0,0,0,1,0]) testing' [2,0,0,0,1,0]
[2,2,2,2,1,0]
Of course, you can wrap the above in one function like:
testing'' lst = apply_n (length lst) testing' lst
and in the end:
*Main> testing'' [2,0,0,0,1,0]
[2,2,2,2,1,0]
NOTE: this is not the only way to do the filling, see the fill2 function in my answer to another question for an example of achieving the same thing using a finite state machine.
I'm still learning Haskell, and I was wondering if there is a less verbose way to express the below statement using 1 line of code:
map (\x -> (x, (if mod x 3 == 0 then "fizz" else "") ++
if mod x 5 == 0 then "buzz" else "")) [1..100]
Produces:
[(1,""),(2,""),(3,"fizz"),(4,""),(5,"buzz"),(6,"fizz"),(7,""),(8,""),(9,"fizz"),(10,"buzz"),(11,""),(12,"fizz"),(13,""),(14,""),(15,"fizzbuzz"),(16,""),(17,""),(18,"fizz"),(19,""),(20,"buzz"),(21,"fizz"),(22,""),(23,""),(24,"fizz"),(25,"buzz"),(26,""),(27,"fizz"),(28,""),(29,""),(30,"fizzbuzz"), etc
It just feels like I'm fighting the syntax more than I should. I've seen other questions for this in Haskell, but I'm looking for the most optimal way to express this in a single statement (trying to understand how to work the syntax better).
We need no stinkin' mod...
zip [1..100] $ zipWith (++) (cycle ["","","fizz"]) (cycle ["","","","","buzz"])
or slightly shorter
import Data.Function(on)
zip [1..100] $ (zipWith (++) `on` cycle) ["","","fizz"] ["","","","","buzz"]
Or the brute force way:
zip [1..100] $ cycle ["","","fizz","","buzz","fizz","","","fizz","buzz","","fizz","","","fizzbuzz"]
If you insist on a one-liner:
[(x, concat $ ["fizz" | mod x 3 == 0] ++ ["buzz" | mod x 5 == 0]) | x <- [1..100]]
How's about...
fizzBuzz = [(x, fizz x ++ buzz x) | x <- [1..100]]
where fizz n | n `mod` 3 == 0 = "fizz"
| otherwise = ""
buzz n | n `mod` 5 == 0 = "buzz"
| otherwise = ""
Couldn't resist going in the other direction and making it more complicated. Look, no mod...
merge as#(a#(ia,sa):as') bs#(b#(ib,sb):bs') =
case compare ia ib of
LT -> a : merge as' bs
GT -> b : merge as bs'
EQ -> (ia, sa++sb) : merge as' bs'
merge as bs = as ++ bs
zz (n,s) = [(i, s) | i <- [n,2*n..]]
fizzBuzz = foldr merge [] $ map zz [(1,""), (3,"fizz"), (5,"buzz")]
Along the same lines as larsmans' answer:
fizzBuzz = [(x, f 3 "fizz" x ++ f 5 "buzz" x) | x <- [1..100]]
where f k s n | n `mod` k == 0 = s
| otherwise = ""
I think the reason why you feel like you are fighting the syntax is because you are mixing too many types.
Instead of trying to print:
[(1, ""), (2,""), (3,"Fizz")...]
Just think of printing strings:
["1","2","Fizz"...]
My attempt:
Prelude> let fizzBuzz x | x `mod` 15 == 0 = "FizzBuzz" | x `mod` 5 == 0 = "Buzz" | x `mod` 3 == 0 = "Fizz" | otherwise = show x
Prelude> [fizzBuzz x | x <-[1..100]]
["1","2","Fizz","4","Buzz","Fizz","7","8","Fizz","Buzz","11","Fizz","13","14","FizzBuzz"...]
In order to convert an Int to String you use the:
show x
Just for studying
zipWith (\a b -> b a) (map show [1..100]) $ cycle [id,id,const "fizz",id,const "buzz",const "fizz",id,id,const "fizz",const "buzz",id,const "fizz",id,id,const "fizzbuzz"]
produces
["1","2","fizz","4","buzz","fizz","7","8","fizz","buzz","11","fizz","13","14","fizzbuzz","16","17","fizz","19","buzz","fizz","22","23","fizz","buzz","26","fizz","28","29","fizzbuzz","31","32","fizz","34","buzz","fizz","37","38","fizz","buzz","41","fizz","43","44","fizzbuzz","46","47","fizz","49","buzz","fizz","52","53","fizz","buzz","56","fizz","58","59","fizzbuzz","61","62","fizz","64","buzz","fizz","67","68","fizz","buzz","71","fizz","73","74","fizzbuzz","76","77","fizz","79","buzz","fizz","82","83","fizz","buzz","86","fizz","88","89","fizzbuzz","91","92","fizz","94","buzz","fizz","97","98","fizz","buzz"]
Writer monad may look nice (if you don't like concat):
fizzBuzz = [(x, execWriter $ when (x `mod` 3 == 0) (tell "fizz") >> when (x `mod` 5 == 0) (tell "buzz")) | x <- [1..100]]
It's not particularly succinct though.
I am doing yet another projecteuler question in Haskell, where I must find if the sum of the factorials of each digit in a number is equal to the original number. If not repeat the process until the original number is reached. The next part is to find the number of starting numbers below 1 million that have 60 non-repeating units. I got this far:
prob74 = length [ x | x <- [1..999999], 60 == ((length $ chain74 x)-1)]
factorial n = product [1..n]
factC x = sum $ map factorial (decToList x)
chain74 x | x == 0 = []
| x == 1 = [1]
| x /= factC x = x : chain74 (factC x)
But what I don't know how to do is to get it to stop once the value for x has become cyclic. How would I go about stopping chain74 when it gets back to the original number?
When you walk through the list that might contain a cycle your function needs to keep track of the already seen elements to be able to check for repetitions. Every new element is compared against the already seen elements. If the new element has already been seen, the cycle is complete, if it hasn't been seen the next element is inspected.
So this calculates the length of the non-cyclic part of a list:
uniqlength :: (Eq a) => [a] -> Int
uniqlength l = uniqlength_ l []
where uniqlength_ [] ls = length ls
uniqlength_ (x:xs) ls
| x `elem` ls = length ls
| otherwise = uniqlength_ xs (x:ls)
(Performance might be better when using a set instead of a list, but I haven't tried that.)
What about passing another argument (y for example) to the chain74 in the list comprehension.
Morning fail so EDIT:
[.. ((length $ chain74 x x False)-1)]
chain74 x y not_first | x == y && not_first = replace_with_stop_value_:-)
| x == 0 = []
| x == 1 = [1]
| x == 2 = [2]
| x /= factC x = x : chain74 (factC x) y True
I implemented a cycle-detection algorithm in Haskell on my blog. It should work for you, but there might be a more clever approach for this particular problem:
http://coder.bsimmons.name/blog/2009/04/cycle-detection/
Just change the return type from String to Bool.
EDIT: Here is a modified version of the algorithm I posted about:
cycling :: (Show a, Eq a) => Int -> [a] -> Bool
cycling k [] = False --not cycling
cycling k (a:as) = find 0 a 1 2 as
where find _ _ c _ [] = False
find i x c p (x':xs)
| c > k = False -- no cycles after k elements
| x == x' = True -- found a cycle
| c == p = find c x' (c+1) (p*2) xs
| otherwise = find i x (c+1) p xs
You can remove the 'k' if you know your list will either cycle or terminate soon.
EDIT2: You could change the following function to look something like:
prob74 = length [ x | x <- [1..999999], let chain = chain74 x, not$ cycling 999 chain, 60 == ((length chain)-1)]
Quite a fun problem. I've come up with a corecursive function that returns the list of the "factorial chains" for every number, stopping as soon as they would repeat themselves:
chains = [] : let f x = x : takeWhile (x /=) (chains !! factC x) in (map f [1..])
Giving:
take 4 chains == [[],[1],[2],[3,6,720,5043,151,122,5,120,4,24,26,722,5044,169,363601,1454]]
map head $ filter ((== 60) . length) (take 10000 chains)
is
[1479,1497,1749,1794,1947,1974,4079,4097,4179,4197,4709,4719,4790,4791,4907,4917
,4970,4971,7049,7094,7149,7194,7409,7419,7490,7491,7904,7914,7940,7941,9047,9074
,9147,9174,9407,9417,9470,9471,9704,9714,9740,9741]
It works by calculating the "factC" of its position in the list, then references that position in itself. This would generate an infinite list of infinite lists (using lazy evaluation), but using takeWhile the inner lists only continue until the element occurs again or the list ends (meaning a deeper element in the corecursion has repeated itself).
If you just want to remove cycles from a list you can use:
decycle :: Eq a => [a] -> [a]
decycle = dc []
where
dc _ [] = []
dc xh (x : xs) = if elem x xh then [] else x : dc (x : xh) xs
decycle [1, 2, 3, 4, 5, 3, 2] == [1, 2, 3, 4, 5]