Re: [Haskell-beginners] Iterating through a list of char...

Ozgur Akgun wrote:

*> bar id (+) [1,2,3] [1,3,5]

Incidentally, scanl1 (+) [1,2,3] == [1,3,5], i.e. scanl1 = bar id. /Jonas On 28 April 2010 17:40, Ozgur Akgun wrote:

Hi!

Since the function to apply depends on the current element, and the previous element, you need to define what to do if an element doesn't have a previous, namely the first element in the list. I guess then it'll be quite easy to implement such a function using explicit recursion.

The function to apply at each step is of type :: a -> a -> a (take the previous element, and this element, and generate an element to replace the current element, right?) The list is of type [a], and the result will be of type [a] again.

foo :: (a -> a -> a) -> [a] -> [a] foo f (x:y:rest) = f x y : foo f (y:rest) foo f _ = []

This function is ready-to-use, except a case to handle the first element. It would simply delete the first element.

*> foo (+) [1,2,3] [3,5]

You can of course have a function like callfoo, which will prepend the first element untouched.

callfoo :: (a -> a -> a) -> [a] -> [a] callfoo f (x:xs) = x : foo f (x:xs)

I would generalise on this a little bit more, and introduce a function to handle the first element differently.

bar :: (a -> b) -> (a -> a -> b) -> [a] -> [b] bar f g (x:xs) = f x : loop g (x:xs)     where loop t (i:j:rest) = t i j : loop t (j:rest)           loop _ _ = [] bar _ _ _ = []

I think, the best thing about this version is that it doesn't limit the input and the output of the function to be of the same type! (Figuring out the meaning of a's and b's is left to the reader) Note that the loop function defined within bar is almost the same with foo defined before.

to test:

*> bar id (+) [1,2,3] [1,3,5]

Hope this will be helpful, and not confusing. If you feel confused, just think about the types one more time :)

Best,

On 28 April 2010 15:56, Jean-Nicolas Jolivet wrote:

...

Hi there!

I'm trying to iterate through each character of a string (that part I can do!) however, I need to apply a transformation to each character...based on the previous character in the string! This is the part I have no clue how to do!

I'm totally new to Haskell so I'm pretty sure I'm missing something obvious... I tried with list comprehensions...map... etc... but I can't figure out how I can access the previous character in my string in each "iteration".... to use simple pseudo code, what i need to do is:

while i < my_string length:        if my_string[i-1] == some_char:                do something with my_string[i]        else                do something else with my_string[i]

I'm using imperative programming here obviously since it's what I am familiar with...but any help as to how I could "translate" this to functional programming would be really appreciated!

Jean-Nicolas Jolivet _______________________________________________ Beginners mailing list Beginners@haskell.org http://www.haskell.org/mailman/listinfo/beginners

-- Ozgur Akgun

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