{"id":1263,"date":"2015-03-30T06:00:19","date_gmt":"2015-03-30T04:00:19","guid":{"rendered":"http:\/\/www.glc.us.es\/~jalonso\/exercitium\/?p=1263"},"modified":"2016-05-01T20:12:48","modified_gmt":"2016-05-01T18:12:48","slug":"agrupamiento-de-consecutivos-iguales","status":"publish","type":"post","link":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/agrupamiento-de-consecutivos-iguales\/","title":{"rendered":"Agrupamiento de consecutivos iguales"},"content":{"rendered":"<p>Definir las funciones<\/p>\n<pre lang=\"text\">\n   agrupa  :: Eq a => [a] -> [(a,Int)]\n   expande :: [(a,Int)] -> [a]\n<\/pre>\n<p>tales que<\/p>\n<ul>\n<li>(agrupa xs) es la lista obtenida agrupando las ocurrencias consecutivas de elementos de xs junto con el n\u00famero de dichas ocurrencias. Por ejemplo: <\/li>\n<\/ul>\n<pre lang=\"text\">\n     agrupa \"aaabzzaa\" == [('a',3),('b',1),('z',2),('a',2)] \n<\/pre>\n<ul>\n<li>(expande xs) es la lista expandida correspondiente a ps (es decir, es la lista xs tal que la comprimida de xs es ps. Por ejemplo,  <\/li>\n<\/ul>\n<pre lang=\"text\">\n     expande [('a',2),('b',3),('a',1)] == \"aabbba\"\n<\/pre>\n<p>Comprobar con QuickCheck que dada una lista de enteros, si se la agrupa y despu\u00e9s se expande se obtiene la lista inicial.<\/p>\n<h4>Soluciones<\/h4>\n<pre lang=\"haskell\">\nimport Data.List (group)\nimport Test.QuickCheck\n\n-- Definiciones de agrupa\n-- ======================\n\n-- 1\u00aa definici\u00f3n (por recursi\u00f3n)\nagrupa :: Eq a => [a] -> [(a,Int)]\nagrupa xs = aux xs 1\n    where aux (x:y:zs) n | x == y    = aux (y:zs) (n+1)\n                         | otherwise = (x,n) : aux (y:zs) 1\n          aux [x]      n             = [(x,n)]\n          aux []       _             = []\n\n-- 2\u00aa definici\u00f3n (por recursi\u00f3n usando takeWhile):\nagrupa2 :: Eq a => [a] -> [(a,Int)]\nagrupa2 [] = []\nagrupa2 (x:xs) = \n    (x,1 + length (takeWhile (==x) xs)) : agrupa2 (dropWhile (==x) xs)\n\n-- 3\u00aa definici\u00f3n (por comprensi\u00f3n usando group):\nagrupa3 :: Eq a => [a] -> [(a,Int)]\nagrupa3 xs = [(head ys,length ys) | ys <- group xs]\n\n-- 4\u00aa definici\u00f3n (usando map y group):\nagrupa4 :: Eq a => [a] -> [(a,Int)]\nagrupa4 = map (\\xs -> (head xs, length xs)) . group\n\n-- Definiciones de expande\n-- =======================\n\n-- 1\u00aa definici\u00f3n (por comprensi\u00f3n)\nexpande :: [(a,Int)] -> [a]\nexpande ps = concat [replicate k x | (x,k) <- ps]\n\n-- 2\u00aa definici\u00f3n (por concatMap)\nexpande2 :: [(a,Int)] -> [a]\nexpande2 = concatMap (\\(x,k) -> replicate k x) \n\n-- 3\u00aa definici\u00f3n (por recursi\u00f3n)\nexpande3 :: [(a,Int)] -> [a]\nexpande3 [] = []\nexpande3 ((x,n):ps) = replicate n x ++ expande3 ps\n\n-- Comprobaci\u00f3n\n-- ============\n\n-- La propiedad es\nprop_expande_agrupa :: [Int] -> Bool \nprop_expande_agrupa xs = expande (agrupa xs) == xs\n\n-- La comprobaci\u00f3n es\n--    ghci> quickCheck prop_expande_agrupa\n--    +++ OK, passed 100 tests.\n<\/pre>\n","protected":false},"excerpt":{"rendered":"<p>Definir las funciones agrupa :: Eq a => [a] -> [(a,Int)] expande :: [(a,Int)] -> [a] tales que (agrupa xs) es la lista obtenida agrupando las ocurrencias consecutivas de elementos de xs junto con el n\u00famero de dichas ocurrencias. Por ejemplo: agrupa \u00abaaabzzaa\u00bb == [(&#8216;a&#8217;,3),(&#8216;b&#8217;,1),(&#8216;z&#8217;,2),(&#8216;a&#8217;,2)] (expande xs) es la lista expandida correspondiente a ps (es&#8230;<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"jetpack_post_was_ever_published":false,"_kad_post_transparent":"","_kad_post_title":"","_kad_post_layout":"","_kad_post_sidebar_id":"","_kad_post_content_style":"","_kad_post_vertical_padding":"","_kad_post_feature":"","_kad_post_feature_position":"","_kad_post_header":false,"_kad_post_footer":false,"_jetpack_newsletter_access":"","_jetpack_dont_email_post_to_subs":false,"_jetpack_newsletter_tier_id":0,"_jetpack_memberships_contains_paywalled_content":false,"footnotes":"","_jetpack_memberships_contains_paid_content":false},"categories":[4],"tags":[12,59,13,71,28,10,11,6,19,34,146],"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/1263"}],"collection":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/comments?post=1263"}],"version-history":[{"count":3,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/1263\/revisions"}],"predecessor-version":[{"id":1304,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/1263\/revisions\/1304"}],"wp:attachment":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/media?parent=1263"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/categories?post=1263"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/tags?post=1263"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}