Re: [Haskell-cafe] Re: Ultra-newbie Question
On Sun, Sep 19, 2010 at 5:01 PM, Henrique Becker henriquebecke...@gmail.com wrote: Why not? import Data.Number.Nat as N lastN :: Integral b = b - [a] - [a] lastN n xs = N.drop (N.length xs - n') xs where n' = N.toNat n Wow. That is gorgeous! I think it's basically the same idea as my zipWith implementation, but it is so much clearer here. Thanks :-) Luke ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Re: Ultra-newbie Question
On Mon, Sep 20, 2010 at 5:11 PM, Luke Palmer lrpal...@gmail.com wrote: On Sun, Sep 19, 2010 at 5:01 PM, Henrique Becker henriquebecke...@gmail.com wrote: Why not? import Data.Number.Nat as N lastN :: Integral b = b - [a] - [a] lastN n xs = N.drop (N.length xs - n') xs where n' = N.toNat n Wow. That is gorgeous! I think it's basically the same idea as my zipWith implementation, but it is so much clearer here. Thanks :-) Er forget that zipWith comment. It is quite unlike that. But it is still beautiful and efficient. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Re: Ultra-newbie Question
Thanks, It's my first post. If you not import Prelude is more clear (N. is horrible): import Prelude ((-), Integral) import Data.Number.Nat (drop, length, toNat) lastN :: Integral b = b - [a] - [a] lastN n xs = drop (length xs - n') xs where n' = toNat n P.S.: You benchmarked? I didn't... 2010/9/20, Luke Palmer lrpal...@gmail.com: On Mon, Sep 20, 2010 at 5:11 PM, Luke Palmer lrpal...@gmail.com wrote: On Sun, Sep 19, 2010 at 5:01 PM, Henrique Becker henriquebecke...@gmail.com wrote: Why not? import Data.Number.Nat as N lastN :: Integral b = b - [a] - [a] lastN n xs = N.drop (N.length xs - n') xs where n' = N.toNat n Wow. That is gorgeous! I think it's basically the same idea as my zipWith implementation, but it is so much clearer here. Thanks :-) Er forget that zipWith comment. It is quite unlike that. But it is still beautiful and efficient. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] Re: Ultra-newbie Question
Why not? import Data.Number.Nat as N lastN :: Integral b = b - [a] - [a] lastN n xs = N.drop (N.length xs - n') xs where n' = N.toNat n Not import Prelude maybe make more clear. []'s Henrique Becker ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] Re: Ultra-newbie Question
On Sat, 2010-09-18 at 03:51 -0400, Christopher Tauss wrote: Hello Haskell Community - I am a professional programmer with 11 years experience, yet I just do not seem to be able to get the hang of even simple things in Haskell. I am trying to write a function that takes a list and returns the last n elements. There may be a function which I can just call that does that, but I am trying to roll my own just to understand the concept. Let's call the function n_lastn and, given a list [1,2,3,4,5], I would like n_lastn 3 = [3,4,5] Seems like it would be something like: n_lastn:: [a]-Int-[a] n_lastn 1 (xs) = last(xs) n_lastn n (x:xs) = The issue is I do not see how you can store the last elements of the list. Thanks in advance. ctauss I'll add my $0.03 - unless you are doing it to learn about lists rethink your approach. Taking k elements from end of n-element list will be O(n) operation. For example with appropriate structures (like finger trees) it would look like O(k) operation. Regards signature.asc Description: This is a digitally signed message part ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Re: Ultra-newbie Question
Translation: Look at Data.Sequence sometime. On 9/18/10 11:15 AM, Maciej Piechotka wrote: On Sat, 2010-09-18 at 03:51 -0400, Christopher Tauss wrote: Hello Haskell Community - I am a professional programmer with 11 years experience, yet I just do not seem to be able to get the hang of even simple things in Haskell. I am trying to write a function that takes a list and returns the last n elements. There may be a function which I can just call that does that, but I am trying to roll my own just to understand the concept. Let's call the function n_lastn and, given a list [1,2,3,4,5], I would like n_lastn 3 = [3,4,5] Seems like it would be something like: n_lastn:: [a]-Int-[a] n_lastn 1 (xs) = last(xs) n_lastn n (x:xs) = The issue is I do not see how you can store the last elements of the list. Thanks in advance. ctauss I'll add my $0.03 - unless you are doing it to learn about lists rethink your approach. Taking k elements from end of n-element list will be O(n) operation. For example with appropriate structures (like finger trees) it would look like O(k) operation. Regards ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Re: Ultra-newbie Question
Translation: Look at Data.Sequence sometime. On 9/18/10 11:15 AM, Maciej Piechotka wrote: On Sat, 2010-09-18 at 03:51 -0400, Christopher Tauss wrote: Hello Haskell Community - I am a professional programmer with 11 years experience, yet I just do not seem to be able to get the hang of even simple things in Haskell. I am trying to write a function that takes a list and returns the last n elements. There may be a function which I can just call that does that, but I am trying to roll my own just to understand the concept. Let's call the function n_lastn and, given a list [1,2,3,4,5], I would like n_lastn 3 = [3,4,5] Seems like it would be something like: n_lastn:: [a]-Int-[a] n_lastn 1 (xs) = last(xs) n_lastn n (x:xs) = The issue is I do not see how you can store the last elements of the list. Thanks in advance. ctauss I'll add my $0.03 - unless you are doing it to learn about lists rethink your approach. Taking k elements from end of n-element list will be O(n) operation. For example with appropriate structures (like finger trees) it would look like O(k) operation. Regards ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
