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

[Jonas Almström Duregård]

5 or 6, does it really matter? ;)

Sorry about the confusion...

/J

2010/4/28 Dupont Corentin <corentin.dupont at gmail.com>:
> Hello,
> i have scanl1 (+) [1,2,3] == [1,3,6]
> since scanl apply successive reducing.
> am i missing something?
>
> On 4/28/10, Jonas Almström Duregård <jonas.duregard at gmail.com> wrote:
>> 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 <ozgurakgun at gmail.com> 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 <jeannicolascocoa at gmail.com>
>>> 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 at haskell.org
>>>> http://www.haskell.org/mailman/listinfo/beginners
>>>
>>>
>>>
>>> --
>>> Ozgur Akgun
>>>
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>>>
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