Well it's possible to code this by passing over the list only once, the question is, is it possible consicely using builtin haskell higher order functions. In my case the list is very short so it's really an academic question. On 18 Nov 2012 10:02, "KC" wrote:

Lists are good if they are short; otherwise, lists are good if you are only traversing them from head to tail or decapitating them.

You want a more complex data structure.

...

well for isSorted, better use the implementation from Data.List.Ordered. That part was poor in performance for sure, but it wasn't my main focus, I was more interested in the rest.

On Sun, Nov 18, 2012 at 8:45 AM, Emmanuel Touzery wrote:

...

Hello,

i wonder what would be the idiomatic way to achieve that algorithm in haskell:

[1,4,56,450,23,46,52] => [1,4,56,450] [1,4,56,450,23,46,52] => [23,46,52]

in other words split the list when one element gets smaller than the previous one. Tge rest of the time the list is sorted. There would be

only

...

two lists, not N. I always need the first or second sublist, I don't need both at once. But of course a more complete algorithm handling the N case and/or returning both sublists would be good.

i could code this by hand, but i'm trying to use as much as possible builtin higher-order functions. However in this case so far I've only come up with this:

import Data.List

isSorted :: Ord a => [a] -> Bool isSorted l = (sort l) == l

secondPart :: Ord a => [a] -> [a] secondPart l = head $ filter isSorted (tails l)

firstPart :: Ord a => [a] -> [a] firstPart l = last $ filter isSorted (inits l)

It is concise alright, but it seems contrived and also in terms of performance I don't think it's OK (for small lists sure but for big

On Sat, Nov 17, 2012 at 11:51 PM, Emmanuel Touzery wrote: lists?).

...

...

Anyway, somehow I think something as simple as this must be doable very concisely and with optimal performance using only builtin higher-order functions. Any idea?

Thanks!

Emmanuel

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-- -- Regards, KC

import Control.Applicative breakup :: Ord a => [a] -> ([a],[a]) breakup [] = ([],[]) breakup xs@(x:_) = unpairs . breakWhenLess . pairs $ x:xs where pairs = zip <$> tail <*> id unpairs (xs,ys) = (fst <$> xs, fst <$> ys) breakWhenLess = break (uncurry (<)) Trick is to duplicate the first element so it doesn't get 'lost' by the tail zip. Applicatives not strictly necessary, but I like them. Peter On 18 November 2012 10:06, Emmanuel Touzery wrote:

That is EXACTLY the kind of answer i was hoping for! Great implementation, I'll be sure to reuse that trick of zipping with the tail of the list.. really really good.

I'm relieved it's not trivial (for me) to write since i could not come up with it, and happy i understand it :-)

Intuitively it's more expensive than what an imperative language would do (new list creations, going several times over the list -zip, spam, map - still O(n) though).

If it was in your project you'd use that, or would you use a more "straightforward" implementation and you came up with this one just because i asked explicitely for such a way? On 18 Nov 2012 10:47, "Tobias Brandt" wrote:

...

Here is a possible solution:

import Data.List

split xs = getSnds $ span (uncurry (<)) $ zip xs (tail xs) where getSnds (as, bs) = (map snd as, map snd bs)

firstPart xs = fst $ split xs

sndPart xs = snd $ split xs

This is a one pass algortihm, it works for infinite lists.

On 18 November 2012 08:45, Emmanuel Touzery wrote:

...

Hello,

i wonder what would be the idiomatic way to achieve that algorithm in haskell:

[1,4,56,450,23,46,52] => [1,4,56,450] [1,4,56,450,23,46,52] => [23,46,52]

in other words split the list when one element gets smaller than the previous one. Tge rest of the time the list is sorted. There would be only two lists, not N. I always need the first or second sublist, I don't need both at once. But of course a more complete algorithm handling the N case and/or returning both sublists would be good.

i could code this by hand, but i'm trying to use as much as possible builtin higher-order functions. However in this case so far I've only come up with this:

import Data.List

isSorted :: Ord a => [a] -> Bool isSorted l = (sort l) == l

secondPart :: Ord a => [a] -> [a] secondPart l = head $ filter isSorted (tails l)

firstPart :: Ord a => [a] -> [a] firstPart l = last $ filter isSorted (inits l)

It is concise alright, but it seems contrived and also in terms of performance I don't think it's OK (for small lists sure but for big lists?).

Anyway, somehow I think something as simple as this must be doable very concisely and with optimal performance using only builtin higher-order functions. Any idea?

Thanks!

Emmanuel

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

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

On Sun, Nov 18, 2012 at 4:47 PM, Tobias Brandt wrote:

split xs = getSnds $ span (uncurry (<)) $ zip xs (tail xs) where getSnds (as, bs) = (map snd as, map snd bs)

You could prepend negative infinity to not lose the first element. -- Kim-Ee On Sun, Nov 18, 2012 at 4:47 PM, Tobias Brandt wrote:

split xs = getSnds $ span (uncurry (<)) $ zip xs (tail xs) where getSnds (as, bs) = (map snd as, map snd bs)

firstPart xs = fst $ split xs

sndPart xs = snd $ split xs

This is a one pass algortihm, it works for infinite lists.

On 18 November 2012 08:45, Emmanuel Touzery wrote:

...

Hello,

i wonder what would be the idiomatic way to achieve that algorithm in haskell:

[1,4,56,450,23,46,52] => [1,4,56,450] [1,4,56,450,23,46,52] => [23,46,52]

in other words split the list when one element gets smaller than the previous one. Tge rest of the time the list is sorted. There would be only two lists, not N. I always need the first or second sublist, I don't need both at once. But of course a more complete algorithm handling the N case and/or returning both sublists would be good.

i could code this by hand, but i'm trying to use as much as possible builtin higher-order functions. However in this case so far I've only come up with this:

import Data.List

isSorted :: Ord a => [a] -> Bool isSorted l = (sort l) == l

secondPart :: Ord a => [a] -> [a] secondPart l = head $ filter isSorted (tails l)

firstPart :: Ord a => [a] -> [a] firstPart l = last $ filter isSorted (inits l)

It is concise alright, but it seems contrived and also in terms of performance I don't think it's OK (for small lists sure but for big lists?).

Anyway, somehow I think something as simple as this must be doable very concisely and with optimal performance using only builtin higher-order functions. Any idea?

Thanks!

Emmanuel

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

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

On Sun, Nov 18, 2012 at 11:53:23AM +0100, Tobias Brandt wrote:

On 18 November 2012 11:33, Kim-Ee Yeoh wrote:

...

On Sun, Nov 18, 2012 at 4:47 PM, Tobias Brandt wrote:

split xs = getSnds $ span (uncurry (<)) $ zip xs (tail xs)

...

where getSnds (as, bs) = (map snd as, map snd bs)

You could prepend negative infinity to not lose the first element.

Oops, didn't noticed that, nice catch. I'd rather do the following, as it works for all types that can be compared with (<), not just for numbers:

split xs = getSnds $ span (uncurry (<)) $ zip xs (tail xs) where getSnds (as, bs) = (head xs : map snd as, map snd bs)

ghci> split [] ([*** Exception: Prelude.head: empty list Better add a special case for split []. Incidentally, this is one splitting pattern (splitting based on a relation between consecutive elements) that the split package [1] does NOT cover. I think I have an idea of how to support it but it would require rewriting a bunch of the internals of the library. -Brent [1] http://hackage.haskell.org/package/split