Hi,
I am trying to do the exercise which asks to define built-in functions
'last' and 'init' using 'foldr' function, such as last Greggery Peccary =
'y'
the type for my function is:
myLast :: [Char] - Char
I am not generalizing type so that make it less complicated. But what ever
i am
Well, i have tried cons (:) operator but when it passed to foldr doesn't work
because cons operator operates first character and then the list but the
foldr argument takes a function (a-a-a). Maybe i am missing the point
here?
Aaron Denney wrote:
On 2007-08-14, Alexteslin [EMAIL PROTECTED
Aaron Denney wrote:
(Quoting reformatted. Try to have your responses below what you are
responding to. It makes it easier to read as a conversation.)
On 2007-08-14, Alexteslin [EMAIL PROTECTED] wrote:
Aaron Denney wrote:
Folds replace the cons operator (:) with the function you pass
Hello,
I just came across with this question on the exam and can not think of
implementing it.
mapPair :: (a - a - a) - [a] - [a]
such that mapPairs f [x1, x2, x3, x4...] = [f x1 x2, f x3 x4,...]
and if the list contains an odd number of elements, the last one is kept
unchanged, for example
Alexteslin wrote:
Hello,
I just came across with this question on the exam and can not think of
implementing it.
mapPair :: (a - a - a) - [a] - [a]
such that mapPairs f [x1, x2, x3, x4...] = [f x1 x2, f x3 x4,...]
and if the list contains an odd number of elements, the last one
Brent Yorgey wrote:
On 8/29/07, Alexteslin [EMAIL PROTECTED] wrote:
Hello,
I just came across with this question on the exam and can not think of
implementing it.
Wait, is this an exam for a class you're taking? Or just a problem from
an
exam that you're trying to solve for fun
Hi, i am a beginner to Haskell and i have a beginner's question to ask.
An exercise asks to define function unique :: [Int] - [Int], which outputs
a list with only elements that are unique to the input list (that appears no
more than once). I defined a function with list comprehension which
of it from the list,
so
it won't get included later.
unique2 (x:xs)
|elemNum2 x xs == 1 = x:unique2 xs
|otherwise = unique2 (deleteElt x xs)
I'll let you figure out how to implement the deleteElt function.
hope this is helpful!
-Brent
On 7/10/07, Alexteslin [EMAIL
ys
|otherwise = y:deleteElem x ys
And I did think about for every call it will go through an entire list - now
i know where the solution lies.
Thanks again
Dan Weston wrote:
Alexteslin wrote:
I'v got it - it produces the right output.
Thank you.
Now that you've done
[1,4,5,3] but you
can get this with head and tail). I'm just learning about the fold family
myself. - Greg
Prelude foldl (\a b - if (any (\x - x == b) a) then a else b:a) [1]
[4,5,3,3,4]
[3,5,4,1]
- Original Message
From: Alexteslin [EMAIL PROTECTED]
To: haskell-cafe
Oh, I am lost now - for now anyway.
I am attempting to do next exercise in the book to define reverse function
using primitive recursion and pattern matching on lists. But getting stack
because when i con in front of xs (xs:x) i get en error, which i thought i
would be getting anyway. I tried
Hello,
I am trying to define a function as part of the exercises that gives a
result of a minimum value of the input list of ints. Such as this:
minimumValue :: [Int] - Int
minimumValue ns ...
using either filter or map functions but Not foldr1, because the exercise
precedes the section on
they are not interested ones
but I am just trying to learn the thinking behind the function definitions.
Thank you
Joachim Breitner-2 wrote:
Hi,
Am Mittwoch, den 18.07.2007, 13:42 -0700 schrieb Alexteslin:
I am trying to define a function as part of the exercises that gives a
result
I have defined the first line it seems right to me but second line not sure.
I have True or False and whatever value i give it produces that value.
allEqual :: [Int] - Bool
allEqual (x1:x2:xs) = (x1 == x2) allEqual xs
allEqual _ = ???
Steve Schafer wrote:
On Thu, 19 Jul 2007
I don't
forget it.
Dan
Alexteslin wrote:
I have defined the first line it seems right to me but second line not
sure.
I have True or False and whatever value i give it produces that value.
allEqual :: [Int] - Bool
allEqual (x1:x2:xs) = (x1 == x2) allEqual xs
allEqual
What wrong with my original solution?
allEqual2 :: [Int] - Bool
allEqual2 xs = length xs == length (filter isEqual xs)
where
isEqual n = (head xs) == n
It looks simpler to me
Dan Weston wrote:
The real lesson here is that
1) no problem is too easy to cheat good software
Hi,
first I like to thank all of you guys - it really helps!
I am still on a same chapter with higher order functions and this function
is also confusing.
But before i even define this function i am getting the type error - i don't
know why? So i wrote the simpler one like:
filterAlpha :: (a -
apfelmus wrote:
Alexteslin wrote:
filterAlpha :: (a - Bool) - [a] - [a]
filterAlpha f [] = []
filterAlpha f (x:xs)
|f x= x : filterAlpha xs
|otherwise = filterAlpha xs
and i am getting this error message:
Type error in application
Expression :filterAlpha xs
Thank you guys, i didn't understand the concept of the chapter.
Brent Yorgey wrote:
On 7/25/07, Alexteslin [EMAIL PROTECTED] wrote:
Hi,
I am going through examples from the textbook and trying them out but
some
don't work.
For example:
addNum :: Int - (Int - Int)
addNum n = addN
the technique is even more valuable.
Dan Weston
Alexteslin wrote:
filterAlpha :: (a - Bool) - [a] - [a]
filterAlpha f [] = []
filterAlpha f (x:xs)
|f x = x : filterAlpha f xs-- corrected
|otherwise = filterAlpha f xs-- corrected
Alexteslin wrote:
Hi, I am doing some simple exercises about recursive algebraic types and
this particular exercise asks to define a function which counts the number
of operators in an expression. I defined the function below, but i am not
sure if on the second line changing from
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