# Sequence differences Classic List Threaded 13 messages Open this post in threaded view
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## Sequence differences

 I have a Scheme function that calculates sequence differences, i.e., it returns a sequence that is the difference between the 2nd and the 1st element, the 3rd and the 2nd, the 4th and the 3rd, etc.(define s  (lambda (f l)    (cond ((null? (cdr l)) '())          (else (cons (f (cadr l) (car l))                      (s f (cdr l)))))))where(s - '(0,1,3,6,10,15,21,28)) => (1,2,3,4,5,6,7)I'm thinking the same function in Haskell would be something likes :: s f [] = []s f [x] = [x] s f l = [ a f b | (a,b) <- zip (init l) (tail l)]but can't figure out what the function typing would be.Michael

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## Re: Sequence differences

 So, we can walk through it-    >     s f [] = []  >     s f [x] = [x]  >     s f l = [ a f b | (a,b) <- zip (init l) (tail l)] First, we can write some of it to be a little more idiomatic, viz: s _ []  = [] s _ [x] = [x] s f ls  = [f a b | (a,b) <- zip (init ls) (tail ls)] First, we have a function type, we can tell the variable f is a function because it's applied to arguments in the third case, since it's applied to two arguments, it's binary, so `s :: (a -> b -> c) -> ?` however, from the second case, we know that whatever the type of the second argument (a list of some type `a1`) is also the type of the return argument, since the `s` acts as the identity for lists of length less than 2, so     s :: (a -> b -> a1) -> [a1] -> [a1] However, since the arguments for `f` are drawn from the same list, the argument types must _also_ be of type `a1`, leaving us with:     s :: (a -> a -> a) -> [a] -> [a] This is, interestingly enough, precisely the type of foldr1. We can write your original function in another, cleaner way though, too, since zip will "zip" to the smaller of the two lengths, so you don't need to worry about doing the init and the tail, so `s` is really: s _ []  = [] s _ [x] = [x] s f ls  = [f a b | (a,b) <- zip ls (tail ls)] but there is a function which does precisely what the third case does, called "zipWith" which takes a binary function and two lists and -- well -- does what that list comprehension does. In fact, it does what your whole function does... In fact, it _is_ your function, specialized a little, eg: yourZipWith f ls = zipWith f ls (tail ls) Hope that helps /Joe michael rice wrote: > I have a Scheme function that calculates sequence differences, i.e., > it returns a sequence that is the difference between the 2nd and the > 1st element, the 3rd and the 2nd, the 4th and the 3rd, etc. > > (define s >   (lambda (f l) >     (cond ((null? (cdr l)) '()) >           (else (cons (f (cadr l) (car l)) >                       (s f (cdr l))))))) > > where > > (s - '(0,1,3,6,10,15,21,28)) => (1,2,3,4,5,6,7) > > > I'm thinking the same function in Haskell would be something like > > s :: > s f [] = [] > s f [x] = [x] > s f l = [ a f b | (a,b) <- zip (init l) (tail l)] > > > but can't figure out what the function typing would be. > > Michael > > > ------------------------------------------------------------------------ > > _______________________________________________ > Haskell-Cafe mailing list > [hidden email] > http://www.haskell.org/mailman/listinfo/haskell-cafe>   _______________________________________________ Haskell-Cafe mailing list [hidden email] http://www.haskell.org/mailman/listinfo/haskell-cafe jfredett.vcf (308 bytes) Download Attachment
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## Re: Sequence differences

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## Re: Sequence differences

 Hi Michael, michael rice wrote: > Clearly, there's more to the picture than meets the eye. Is there a good > tutorial on types? Have you seen Real World Haskell? http://book.realworldhaskell.org/read/Chapter 2 already dives into types. Hope this helps, Martijn. _______________________________________________ Haskell-Cafe mailing list [hidden email] http://www.haskell.org/mailman/listinfo/haskell-cafe
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## Re: Sequence differences

 In reply to this post by michael rice-3 You might try looking at Real World Haskell, it's not so much a tutorial on types, but a tutorial on Haskell in general. Theres also the Haskell-beginner's list, the backlogs of which are quite nice. Really the best way to learn is to just dive in, you can use Hoogle or Hayoo to search APIs. Just try to come up with a good idea, and start working on it! And of course, you can ask questions here or on Haskell-beginners. The rumors are true, we're awesome people. /Joe michael rice wrote: > All very neat, and it works! Thanks. > > But I copied > > map :: (a -> b) -> [a] -> [b]    <==  I'm assuming this is correct > > from something called "A Tour of the Haskell Prelude" at > > http://undergraduate.csse.uwa.edu.au/units/CITS3211/lectureNotes/tourofprelude.html#all> > which I was looking at to try and roll my own typing. > > > My next question: > > My function > > s f ls > > seems much like > > map f ls > > but instead looks like   > > s :: (a -> a -> a) -> [a] -> [a] > > > Clearly, there's more to the picture than meets the eye. Is there a > good tutorial on types? > > Michael > > > --- On *Fri, 4/10/09, Joe Fredette /<[hidden email]>/* wrote: > > >     From: Joe Fredette <[hidden email]> >     Subject: Re: [Haskell-cafe] Sequence differences >     To: "michael rice" <[hidden email]>, "haskell Cafe mailing >     list" <[hidden email]> >     Date: Friday, April 10, 2009, 2:07 AM > >     So, we can walk through it- > >     >     s f [] = [] >     >     s f [x] = [x] >     >     s f l = [ a f b | (a,b) <- zip (init l) (tail l)] > >     First, we can write some of it to be a little more idiomatic, viz: > >     s _ []  = [] >     s _ [x] = [x] >     s f ls  = [f a b | (a,b) <- zip (init ls) (tail ls)] > >     First, we have a function type, we can tell the variable f is a >     function because it's applied to arguments in the third case, >     since it's applied to two arguments, it's binary, so `s :: (a -> b >     -> c) -> ?` however, from the >     second case, we know that whatever the type of the second argument >     (a list of some type `a1`) is also the type >     of the return argument, since the `s` acts as the identity for >     lists of length less than 2, so > >        s :: (a -> b -> a1) -> [a1] -> [a1] > >     However, since the arguments for `f` are drawn from the same list, >     the argument types must _also_ be of type `a1`, leaving us with: > >        s :: (a -> a -> a) -> [a] -> [a] > >     This is, interestingly enough, precisely the type of foldr1. > >     We can write your original function in another, cleaner way >     though, too, since zip will "zip" to the smaller of the two >     lengths, so you don't need to worry about doing the init and the >     tail, so `s` is really: > >     s _ []  = [] >     s _ [x] = [x] >     s f ls  = [f a b | (a,b) <- zip ls (tail ls)] > >     but there is a function which does precisely what the third case >     does, called "zipWith" which takes a >     binary function and two lists and -- well -- does what that list >     comprehension does. In fact, it does >     what your whole function does... In fact, it _is_ your function, >     specialized a little, eg: > >     yourZipWith f ls = zipWith f ls (tail ls) > > >     Hope that helps > >     /Joe > >     michael rice wrote: >     > I have a Scheme function that calculates sequence differences, >     i.e., it returns a sequence that is the difference between the 2nd >     and the 1st element, the 3rd and the 2nd, the 4th and the 3rd, etc. >     > >     > (define s >     >   (lambda (f l) >     >     (cond ((null? (cdr l)) '()) >     >           (else (cons (f (cadr l) (car l)) >     >                       (s f (cdr l))))))) >     > >     > where >     > >     > (s - '(0,1,3,6,10,15,21,28)) => (1,2,3,4,5,6,7) >     > >     > >     > I'm thinking the same function in Haskell would be something like >     > >     > s :: >     > s f [] = [] >     > s f [x] = [x] >     > s f l = [ a f b | (a,b) <- zip (init l) (tail l)] >     > >     > >     > but can't figure out what the function typing would be. >     > >     > Michael >     > >     > >     > >     ------------------------------------------------------------------------ >     > >     > _______________________________________________ >     > Haskell-Cafe mailing list >     > [hidden email] >     > http://www.haskell.org/mailman/listinfo/haskell-cafe>     >   > > _______________________________________________ Haskell-Cafe mailing list [hidden email] http://www.haskell.org/mailman/listinfo/haskell-cafe jfredett.vcf (308 bytes) Download Attachment
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## Re: Sequence differences

 In reply to this post by jfredett Joe Fredette gmail.com> writes: > We can write your original function in another, cleaner way though, too, > since zip will "zip" to the smaller of the two lengths, so you don't > need to worry about doing the init and the tail, so `s` is really: > > s _ []  = [] > s _ [x] = [x] > s f ls  = [f a b | (a,b) <- zip ls (tail ls)] > > but there is a function which does precisely what the third case does, > called "zipWith" which takes a > binary function and two lists and -- well -- does what that list > comprehension does. In fact, it does > what your whole function does... In fact, it _is_ your function, > specialized a little, eg: > > yourZipWith f ls = zipWith f ls (tail ls) A nice generalization of this that can be really useful is movingWindow :: Int -> [a] -> [[a]] movingWindow 1 xs = map (:[]) xs movingWindow n xs = zipWith (:) xs . tail \$ movingWindow (n-1) xs So for example, > movingWindow 3 [1..10] [[1,2,3],[2,3,4],[3,4,5],[4,5,6],[5,6,7],[6,7,8],[7,8,9],[8,9,10]] Then you can write diff :: (Num a) => [a] -> [a] diff = map (\[x,y] -> y - x) . movingWindow 2 Hopefully the intermediate lists are optimized away, but I haven't done any performance testing. Chad _______________________________________________ Haskell-Cafe mailing list [hidden email] http://www.haskell.org/mailman/listinfo/haskell-cafe
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## Re: Sequence differences

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## Re: Sequence differences

In reply to this post by michael rice-3
 No, I was unaware of this, but it looks good.Thanks.Michael--- On Fri, 4/10/09, Martijn van Steenbergen <[hidden email]> wrote:From: Martijn van Steenbergen <[hidden email]>Subject: Re: [Haskell-cafe] Sequence differencesTo: "michael rice" <[hidden email]>Cc: "haskell Cafe mailing list" <[hidden email]>, "Joe Fredette" <[hidden email]>Date: Friday, April 10, 2009, 11:38 AMHi Michael,michael rice wrote:> Clearly, there's more to the picture than meets the eye. Is there a good tutorial on types?Have you seen Real World Haskell?http://book.realworldhaskell.org/read/Chapter 2 already dives into types.Hope this helps,Martijn.

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## Re: Sequence differences

 In reply to this post by michael rice-3 michael rice <[hidden email]> writes: > map :: (a -> b) -> [a] -> [b]    <==  I'm assuming this is correct This is the type of 'map', yes.  Btw, ou can check types in GHCi with the :i command. > s f ls > > seems much like > > map f ls > > but instead looks like   > > s :: (a -> a -> a) -> [a] -> [a] If you look at the definition: >> s f [] = [] >> s f [x] = [x] >> s f l = [ a f b | (a,b) <- zip (init l) (tail l)] You'll notice that the second clause, namely    s f [x] = [x] produces the second parameter [x] (of type [a]) as its output, and thus the types must be the same as well. Also (assuming it is 'f a b' and not 'a f b' in the list comprehension), f is applied to two parameters, so it'll have to be of type (x -> y -> z), and since the two input parameters come from the originating list, x and y must be the same as a, and since we have seen the result list also has the same type, z must be the same as a, too.  Thus f must have type (a -> a -> a).  Unclear? Clear? Operating thetan? -k -- If I haven't seen further, it is by standing in the footprints of giants _______________________________________________ Haskell-Cafe mailing list [hidden email] http://www.haskell.org/mailman/listinfo/haskell-cafe
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## Re: Re: Sequence differences

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## Re: Sequence differences

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 Clearly, I have some reading to do.Thanks,Michael--- On Fri, 4/10/09, Ketil Malde <[hidden email]> wrote:From: Ketil Malde <[hidden email]>Subject: Re: [Haskell-cafe] Sequence differencesTo: "michael rice" <[hidden email]>Cc: "haskell Cafe mailing list" <[hidden email]>, "Joe Fredette" <[hidden email]>Date: Friday, April 10, 2009, 3:52 PMmichael rice writes:> map :: (a -> b) -> [a] -> [b]    <==  I'm assuming this is correctThis is the type of 'map', yes.  Btw, ou can check types in GHCi with the:i command.> s f ls >> seems much like>> map f ls>> but instead looks like   >> s :: (a -> a -> a) -> [a] -> [a]If you look at the definition:>> s f [] = []>> s f [x] = [x]>> s f l = [ a f b | (a,b) <- zip (init l) (tail l)]You'll notice that the second clause, namely   s f [x] = [x]produces the second parameter [x] (of type [a]) as its output, andthus the types must be the same as well. Also (assuming it is 'f a b' and not 'a f b' in the listcomprehension), f is applied to two parameters, so it'll have to beof type (x -> y -> z), and since the two input parameters come fromthe originating list, x and y must be the same as a, and since we haveseen the result list also has the same type, z must be the same as a,too.  Thus f must have type (a -> a -> a).  Unclear? Clear? Operatingthetan? -k-- If I haven't seen further, it is by standing in the footprints of giants

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