- Beginners - Haskell.org

Saving intermediate calculations
by Abhinav Kalawatia 11 Mar '15

11 Mar '15

Hello, I am facing a slight difficulty trying to do something while saving intermediate calculations week actual_ sales(a) forecast(f) forecast(f) forecast(f) 1 20 f1 = a1 f1 = 20 f1 = 20 2 27 f2 = f1 + 0.5*(a1 - f1) f2 = 20 +0.5(20-20) f2 = 20 3 25 f3 = f2 + 0.5*(a2-f2) f3 = 20+0.5*(27-20) f3=23.5 4 22 When I execute a function to achieve this in Haskell, I get the forecast for the fourth period. Can I save the values for intermediate period(1..3) also? Please find my Haskell code below: a = 0.5 ipt = [20,27,25,22] avg :: [Double] -> Double avg (x:xs) = (a*x) + (1-a)*(avg xs) avg [] = 0 Could you please help me with this, thanking you in anticipation Regards,Abhinav

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Avoiding stack space overflow
by Hans Georg Schaathun 10 Mar '15

10 Mar '15

Hi, can someone give some hints on how to get around a stack space overflow? My problem is with the training function for a neural network: trainNetwork :: Double -> Samples -> Int -> Network -> Network trainNetwork _ _ 0 n = n trainNetwork eta samples c n = trainNetwork eta samples (c-1) $! epoch eta n samples epoch :: Double -> Network -> Samples -> Network So trainNetwork runs epoch c times, each time taking a Network in and modifying the Network as output. Clearly, space complexity can be made constant in c, but I get stack overflow if and only if c is too large. As you can see, I have tried to make the epoch evaluation strict ($!). I have also tried bang patterns on the input parameter n, and I have tried rewriting with foldr/foldl/foldl', and I have tried switchin the inner and outer calls (epoch vs. trainNetwork), all to no avail. I reckon this loop like pattern should be fairly common ... does it have a common solution too? TIA -- :-- Hans Georg

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GHC not buying what I am offering this afternoon
by Geoffrey Bays 10 Mar '15

10 Mar '15

My main function looks like this: main :: [IO()] main = do let stud1 = Student {name = "Geoff", average = -99.0, grades = [66,77,88]} let stud2 = Student {name = "Doug", average = -99.0, grades = [77,88,99]} let stud3 = Student {name = "Ron", average = -99.0, grades = [55,66,77]} let studList = [stud1,stud2] let newList = calcAvg studList [putStrLn $ show s | s <- newList] --putStrLn $ show (newList !! 0) --putStrLn $ show (newList !! 1) With this final line, putStrLn $ show (newList !! 0), the type IO () in the function declaration compiles fine. But with [putStrLn $ show s | s <- newList] as the final line, [IO ()] in the function declaration will not compile, I get this error: Couldn't match expected type `IO t0' with actual type `[IO ()]' What does the declared type need to be for a final line of: [putStrLn $ show s | s <- newList] ??? Thanks, Geoffrey

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Laziness to efficiently get one element from a set
by Jeffrey Brown 08 Mar '15

08 Mar '15

Hi, Suppose S is a big set, and I'd like to access just a single element from it. The only natural way I've come up with to do that is this: head $ Data.Set.toList S. If I do that, am I correct that Haskell will not try to convert all of S to a list; instead it will only convert one element, and then return it, and leave the rest of the list unevaluated? Thanks, Jeff

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Strange behavior of program
by m00nlight 08 Mar '15

08 Mar '15

Hi Haskellers, I encounter an strange behavior of haskell program recently. The following is my program ```haskell main = do _ <- getLine arr1 <- getLine _ <- getLine arr2 <- getLine let a = map (read :: String -> Int) (words arr1) b = map (read :: String -> Int) (words arr2) putStrLn $ show $ (foldl (*) 1 a) putStrLn $ show $ a == [1,2,4,8,32,64,128,256,512,1024,2048,4906,8192] putStrLn $ show $ (foldl (*) 1 [1,2,4,8,32,64,128,256,512,1024,2048,4906,8192]) ``` With the input test file as following: ```test.in 13 1 2 4 8 32 64 128 256 512 1024 2048 4906 8192 9 1 3 9 27 81 243 729 2187 6561 ``` The output is as: ```output 0 True 185343439719667835347140608 ``` In fact, from the program, we know that a is equal to list [1,2,4,8,32,64,128,256,512,1024,2048,4906,8192] , but the product of a and the literal list is different. Can anyone tell me why? Thanks --m00nlight

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Re: [Haskell-beginners] State Monad stack example
by Animesh Saxena 08 Mar '15

08 Mar '15

Thanks for the links, I am able to get my head around this a bit. Here's my simple question 1. In the stack example what is the state. As per my understanding the state is the new stack. If above statement is correct than by definition of >>= (>>=) :: State s a -> (a -> State s b) -> State s b push 3 stack This returns ((), newstack1) Now in this case the state or context is newstack1 which has to be passed to the next function if i apply >>=. push 3 stack >>= pop newstack1 This might make sense coz I am shoving (or binding) the state to the pop function. State in this case is the stack which is being passed around or shoved around. Problem is pop doesn't need "a" argument like the definition of (>>=) indicates. It can very well produce a new state or return a new state. So if state = stack then this makes sense to me, where am i wrong...?? push 3 stack >>= pop newstack1 On Mar 06, 2015, at 09:09 PM, "Sumit Sahrawat, Maths & Computing, IIT (BHU)" <sumit.sahrawat.apm13(a)iitbhu.ac.in> wrote: I won't comment on what state exactly is, but you can read up on that and gain some intuition here: https://en.wikibooks.org/wiki/Haskell/Understanding_monads/State It's helpful to implement it using a pen and paper, and consider how the state flows and gets transformed. According to the below example, stackManip stack = let ((), newStack1) = push 3 stack (a, newStack2) = pop newStack1 in pop newStack2 We get, push :: a -> Stack a -> ((), Stack a) -- Assuming 'Stack a' is a defined datatype pop :: Stack a -> (a, Stack a) -- Representing a stack with elements of type 'a' Thus, push 3 >>= pop ~~ (Stack a -> ((), Stack a)) >>= (Stack a -> (a, Stack a)) { Replacing by types } Bind (>>=) has the type, (for "State s") (>>=) :: State s a -> (a -> State s b) -> State s b This is a type mismatch. The conversion to do syntax is at fault here. First, you must write the computation using bind (>>=), and then convert to do-notation. On 7 March 2015 at 10:12, Animesh Saxena <animeshsaxena(a)icloud.com> wrote: I am trying to relate the state monad to a stack example and somehow found it easy to get recursively confused! instance Monad (State s) where return x = State $ \s -> (x,s) (State h) >>= f = State $ \s -> let (a, newState) = h s (State g) = f a in g newState Considering the stack computation stackManip stack = let ((), newStack1) = push 3 stack (a, newStack2) = pop newStack1 in pop newStack2 in do notation this would become do push 3 a <- pop pop If I consider the first computation push 3 >>= pop and try to translate it to the definition there are problems.... Copy paste again, I have (State h) >>= f = State $ \s -> let (a, newState) = h s (State g) = f a in g newState f is the push function to which we are shoving the old state. I can't exactly get around to what exactly is the state computation h? Ok assuming it's something which gives me a new state, but then thats push which is function f. Then push is applied to a which is assuming 3 in this case. This gives me a new state, which I would say newStack1 from the stockManip above. Then somehow I apply g to newState?? All the more confusion. Back to the question what exactly is state computation and how is it different from f? It seems to be the same function? -Animesh _______________________________________________ Beginners mailing list Beginners(a)haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners -- Regards Sumit Sahrawat _______________________________________________ Beginners mailing list Beginners(a)haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners

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State Monad stack example
by Animesh Saxena 07 Mar '15

07 Mar '15

I am trying to relate the state monad to a stack example and somehow found it easy to get recursively confused! instance Monad (State s) where return x = State $ \s -> (x,s) (State h) >>= f = State $ \s -> let (a, newState) = h s (State g) = f a in g newState Considering the stack computation stackManip stack = let ((), newStack1) = push 3 stack (a, newStack2) = pop newStack1 in pop newStack2 in do notation this would become do push 3 a <- pop pop If I consider the first computation push 3 >>= pop and try to translate it to the definition there are problems.... Copy paste again, I have (State h) >>= f = State $ \s -> let (a, newState) = h s (State g) = f a in g newState f is the push function to which we are shoving the old state. I can't exactly get around to what exactly is the state computation h? Ok assuming it's something which gives me a new state, but then thats push which is function f. Then push is applied to a which is assuming 3 in this case. This gives me a new state, which I would say newStack1 from the stockManip above. Then somehow I apply g to newState?? All the more confusion. Back to the question what exactly is state computation and how is it different from f? It seems to be the same function? -Animesh

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Using LTS et. al, a small thank you.
by emacstheviking 06 Mar '15

06 Mar '15

First, find yourself a Pharrel Williams youtube video of Happy! ;) It might seem crazy what I'm about to say Functional is here, you can take a break No more OO confusion that takes up too much space Punch the air like you don't care it's cool as I say... Because I code Haskell Code along if you feel like a project without a boss Because I code Haskell Code along if you know that OO is just a bunch of dross Because I code Haskell Code along if you know what productivity is to you Because I code Haskell Clap along if you that's what you wanna do I am sure I could write more but that floated in out of nowhere and stuck as the song came on the radio!! LMAO In all honesty, I have now got OpenGL, http-client and a bunch of stuff all playing nicely. Thanks very much everybody for your help yesterday, it's noce to be hacking Haskell once more. I'm happy..........

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Function Composition/Map
by Animesh Saxena 06 Mar '15

06 Mar '15

I have a simple function fmove::Char->(Char,Char)->Char fmove s (x,y) | s == x = y | otherwise = s If I say fmove 'a' ('a','b') it will move to 'b' If I say fmove 'a' ('d','c') it will stay at 'a' Now I am trying to use this for graph, coz one move can result in multiple result nodes. For example in graphs you can have move from a to b and a to c. So I was trying something like this, :t (map fmove s) where s is the set of possible start nodes "abcd" (map fmove s) :: [(Char, Char) -> Char] Now I want to be able to pass the graph to this function and get all possible list of endings. For example my graph can be like [('a','b'),('b','c'),('c','a'),('c','d'),('e','a')]. You can see that both 'a' and 'd' are possible from node 'c'. I can see that above type signature is wrong for passing the whole graph coz it accepts [(Char, Char) -> Char]. I initially started with foldl but problem is recursion on top of foldl. The actual objective of the problem is to figure out if a node is reachable from a start node or not. Hope the problem is clear, can anyone suggest how to convert the map function to accept the graph? For each of the possible start nodes (accumulating at each step) it should apply it to the same graph. -Animesh

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<$> and <*> pronunciation
by Max Schneider 06 Mar '15

06 Mar '15

(I really hope 'no stupid questions' holds here) I'm not even going to try framing this into a weighty or practical question; how do you pronounce the <$> and <*> Applicative functions?

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