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

Ozgur Akgun ozgurakgun at gmail.com
Wed Apr 28 12:02:47 EDT 2010

Exactly:

*> bar id (+) [1..5]
[1,3,5,7,9]
*> scanl1 (+) [1..5]
[1,3,6,10,15]

But they are close enough, I guess :)

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
> >>
> >> _______________________________________________
> >> Beginners mailing list
> >> Beginners at haskell.org
> >> http://www.haskell.org/mailman/listinfo/beginners
> >>
> >>
> > _______________________________________________
> > Beginners mailing list
> > Beginners at haskell.org
> > http://www.haskell.org/mailman/listinfo/beginners
> >
>

--
Ozgur Akgun
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