7 ms·

Just for fun, a solution in language j to the problem of selecting the longest word with less than 3 vowels. solution =: >@:{.@:(\: #&>)@:(((+/@:(e.&'aeiou') <

Just for fun, a solution in language j to the problem of selecting the longest word with less than 3 vowels. solution =: >@:{.@:(\: #&>)@:(((+/@:(e.&'aeiou') <: 2:) # ])&.>)@:;: Example solution 'yes, today you are reading something that is not so easy to grasp' the result is: today Ruby : frase.split.select{|x| x.count("aeiou")<3}.sort_by(&:length).last => "today" Edited: Added a comparison with Ruby. It seems Ruby here is easier to read and to compose.

mr_toad 3y ago
:(
elcaro 3y ago
FWIW, I solved this in J as follows F =: >@{.@(\: #@>)@(#~ (2 >: +/@e.&'aeoiu')@>) which I think maybe composes a little better This doesn't include the splitting. I would imagine I would be feeding this function from a boxed list from a word list, like: F 'b' fread 'dict.txt' Also `;:` is unreliable on hyphenated words, etc. Better to use `cut`.
llmzero 3y ago
Shorter ruby version: phrase.split.select{|x| x.count("aeiou")<=2}.max_by(&:length) The enumerable module of Ruby provides many methods that can be easily implemented in J. Just to show one of them max_by =: 1 : '{~ (i. >./)@:(u&>)' For example the list.max_by(&:length) is (# max_by)
antonvs 3y ago
Haskell: maximumBy (on compare length) . filter ((< 3) . length . filter (`elem` "aeiou")) $ words phrase That returns "grasp", though, because it doesn't sort the list.
- antonvs 3y ago
  Plus, defining a few aliases for the punctuation-happy terseness-lovers among us, we can reduce the above to: maxBy(on cmp (#)).(((<3).(#).(el"aeiou"|=))|=)$words phrase
- leephillips 3y ago
  The problem as stated does not have a unique solution in general, as you’ve found. One Julia program for this is `sort(split(p)[count.(r"[aeiou]", split(p)) .< 3]; by=r -> length(r))[end]` which also returns "grasp". But this one appeals to me more, because it doesn’t split twice: `sort(filter(w -> count(r"[aeiou]", w) < 3, split(p)); by=r -> length(r))[end]`

mr_toad 3y ago

elcaro 3y ago

FWIW, I solved this in J as follows F =: >@{.@(\: #@>)@(#~ (2 >: +/@e.&'aeoiu')@>) which I think maybe composes a little better This doesn't include the splitting. I would imagine I would be feeding this function from a boxed list from a word list, like: F 'b' fread 'dict.txt' Also `;:` is unreliable on hyphenated words, etc. Better to use `cut`.

llmzero 3y ago

Shorter ruby version: phrase.split.select{|x| x.count("aeiou")<=2}.max_by(&:length) The enumerable module of Ruby provides many methods that can be easily implemented in J. Just to show one of them max_by =: 1 : '{~ (i. >./)@:(u&>)' For example the list.max_by(&:length) is (# max_by)

antonvs 3y ago

Haskell: maximumBy (on compare length) . filter ((< 3) . length . filter (`elem` "aeiou")) $ words phrase That returns "grasp", though, because it doesn't sort the list.

antonvs 3y ago
Plus, defining a few aliases for the punctuation-happy terseness-lovers among us, we can reduce the above to: maxBy(on cmp (#)).(((<3).(#).(el"aeiou"|=))|=)$words phrase
leephillips 3y ago
The problem as stated does not have a unique solution in general, as you’ve found. One Julia program for this is `sort(split(p)[count.(r"[aeiou]", split(p)) .< 3]; by=r -> length(r))[end]` which also returns "grasp". But this one appeals to me more, because it doesn’t split twice: `sort(filter(w -> count(r"[aeiou]", w) < 3, split(p)); by=r -> length(r))[end]`