If you're looking for an efficient way to compute subsets, import Control.Monad subsets :: [a] -> [[a]] subsets = filterM (pure [True, False]) does the trick nicely. On 05/24/2018 12:36 PM, Mark Wotton wrote:

there are a few things you can do.

I ran it locally and got 3s cpu time and 3gb of allocation, more or less.

first off, your subset implementation computes the subsets twice. taking that off took it down to 2s.

After that it's mostly in the Set implementation. Using Map a () instead gives you access to Data.Map.Merge.Strict.merge, which is a bit more efficient (going over the two data structures once only)

``` module Main where

import Data.List import Data.Map.Strict (Map) import qualified Data.Map.Merge.Strict as Map import qualified Data.Map.Strict as Map

nearest_k :: (Ord a) => Int -> [(Map a (), v)] -> Map a () -> [(Map a (), v)] nearest_k k bs b = take k bs' where bs' = sortOn (hamming b) bs

hamming :: (Ord a)=> Map a () -> (Map a (), v) -> Int hamming x (y, _) = hamming_distance x y

hamming_distance :: (Ord a)=> Map a () -> Map a () -> Int hamming_distance xs ys = Map.size (Map.merge Map.preserveMissing Map.preserveMissing (Map.zipWithMaybeMatched (\_ _ _ -> Nothing)) xs ys)

subsets :: [a] -> [[a]] subsets [] = [[]] subsets (x:xs) = let rs = subsets xs in rs ++ map (x:) rs

int_lists :: [[Int]] int_lists = subsets [1..20]

values :: [(Map Int (), Int)] values = map f (zip [1..] int_lists) where f (i, x) = (Map.fromList(zip x (repeat ())), i)

test = nearest_k 8 values (Map.fromList (zip [1,2,3] (repeat ())))

main = print test ```

that took it to 1s, and now profiling indicates more than half the time is spent in generation of the test values, so I'll leave it there.

I think if you wanted to do better than this you'd have to do some algorithmic changes - for instance, once your worst candidate is n steps away, you can stop calculating the hamming distance for anything else once it's > n, as it can't contribute usefully to the nearest neighbours.

cheers, Mark

On Thu, May 24, 2018 at 12:15 PM, Richard Evans wrote:

...

Dear Haskell Cafe,

Given a set of sets, and a particular target set, I want to find the sets that are nearest (in terms of Hamming distance) to the target set.

I am using the following code:

import Data.List import qualified Data.Set as Set

nearest_k :: Ord a => Int -> [(Set.Set a, v)] -> Set.Set a -> [(Set.Set a, v)] nearest_k k bs b = take k bs' where bs' = sortOn (hamming b) bs

hamming :: Ord a => Set.Set a -> (Set.Set a, v) -> Int hamming x (y, _) = hamming_distance x y

hamming_distance :: Ord a => Set.Set a -> Set.Set a -> Int hamming_distance xs ys = Set.size (Set.difference xs ys) + Set.size (Set.difference ys xs)

subsets :: [a] -> [[a]] subsets [] = [[]] subsets (x:xs) = subsets xs ++ map (x:) (subsets xs)

int_lists :: [[Int]] int_lists = subsets [1..20]

values :: [(Set.Set Int, Int)] values = map f (zip [1..] int_lists) where f (i, x) = (Set.fromList x, i)

test = nearest_k 8 values (Set.fromList [1,2,3])

----

This works ok for the test above (with sets of ints), but is rather slow in my actual application (in which the sets are large sets of ground atoms of first-order logic). Is there some major optimization I should be doing here?

thanks, Richard

_______________________________________________ Haskell-Cafe mailing list To (un)subscribe, modify options or view archives go to: http://mail.haskell.org/cgi-bin/mailman/listinfo/haskell-cafe Only members subscribed via the mailman list are allowed to post.

Yes. Showed up in testing - if you're using a decent chunk of the subsets, you can avoid computing twice. If the pressure is actually memory rather than cpu and you can consume lazily, you are quite right that the original could be better. On Fri, May 25, 2018, 10:00 AM Vanessa McHale wrote:

Also: are you sure

let rs = subsets xs in rs ++ map (x:) rs

is more efficient than

subsets xs ++ map (x:) (subsets xs)

I would have assumed these would be the same due to laziness, etc.

On 05/24/2018 12:36 PM, Mark Wotton wrote:

...

there are a few things you can do.

I ran it locally and got 3s cpu time and 3gb of allocation, more or less.

first off, your subset implementation computes the subsets twice. taking that off took it down to 2s.

After that it's mostly in the Set implementation. Using Map a () instead gives you access to Data.Map.Merge.Strict.merge, which is a bit more efficient (going over the two data structures once only)

``` module Main where

import Data.List import Data.Map.Strict (Map) import qualified Data.Map.Merge.Strict as Map import qualified Data.Map.Strict as Map

nearest_k :: (Ord a) => Int -> [(Map a (), v)] -> Map a () -> [(Map a (), v)] nearest_k k bs b = take k bs' where bs' = sortOn (hamming b) bs

hamming :: (Ord a)=> Map a () -> (Map a (), v) -> Int hamming x (y, _) = hamming_distance x y

hamming_distance :: (Ord a)=> Map a () -> Map a () -> Int hamming_distance xs ys = Map.size (Map.merge Map.preserveMissing Map.preserveMissing (Map.zipWithMaybeMatched (\_ _ _ -> Nothing)) xs ys)

subsets :: [a] -> [[a]] subsets [] = [[]] subsets (x:xs) = let rs = subsets xs in rs ++ map (x:) rs

int_lists :: [[Int]] int_lists = subsets [1..20]

values :: [(Map Int (), Int)] values = map f (zip [1..] int_lists) where f (i, x) = (Map.fromList(zip x (repeat ())), i)

test = nearest_k 8 values (Map.fromList (zip [1,2,3] (repeat ())))

main = print test ```

that took it to 1s, and now profiling indicates more than half the time is spent in generation of the test values, so I'll leave it there.

I think if you wanted to do better than this you'd have to do some algorithmic changes - for instance, once your worst candidate is n steps away, you can stop calculating the hamming distance for anything else once it's > n, as it can't contribute usefully to the nearest neighbours.

cheers, Mark

On Thu, May 24, 2018 at 12:15 PM, Richard Evans wrote:

...

Dear Haskell Cafe,

Given a set of sets, and a particular target set, I want to find the sets that are nearest (in terms of Hamming distance) to the target set.

I am using the following code:

import Data.List import qualified Data.Set as Set

nearest_k :: Ord a => Int -> [(Set.Set a, v)] -> Set.Set a -> [(Set.Set a, v)] nearest_k k bs b = take k bs' where bs' = sortOn (hamming b) bs

hamming :: Ord a => Set.Set a -> (Set.Set a, v) -> Int hamming x (y, _) = hamming_distance x y

hamming_distance :: Ord a => Set.Set a -> Set.Set a -> Int hamming_distance xs ys = Set.size (Set.difference xs ys) + Set.size (Set.difference ys xs)

subsets :: [a] -> [[a]] subsets [] = [[]] subsets (x:xs) = subsets xs ++ map (x:) (subsets xs)

int_lists :: [[Int]] int_lists = subsets [1..20]

values :: [(Set.Set Int, Int)] values = map f (zip [1..] int_lists) where f (i, x) = (Set.fromList x, i)

test = nearest_k 8 values (Set.fromList [1,2,3])

----

This works ok for the test above (with sets of ints), but is rather slow in my actual application (in which the sets are large sets of ground atoms of first-order logic). Is there some major optimization I should be doing here?

thanks, Richard

_______________________________________________ Haskell-Cafe mailing list To (un)subscribe, modify options or view archives go to: http://mail.haskell.org/cgi-bin/mailman/listinfo/haskell-cafe Only members subscribed via the mailman list are allowed to post.

_______________________________________________ Haskell-Cafe mailing list To (un)subscribe, modify options or view archives go to: http://mail.haskell.org/cgi-bin/mailman/listinfo/haskell-cafe Only members subscribed via the mailman list are allowed to post.