Find words containing every vowel

Question

Your program must find all the words in this wordlist that contain all the vowels (a e i o u y). There are easy ways to do this, but I am looking for the shortest answer. I will take any language, but I'd like to see Bash.

Here is an example (could be much improved):

cat wordlist.txt | grep "a" | grep "e" | grep "i" | grep "o" | grep "u" | grep "y"

Your score is the length of the code.

-5 points for counting all occurrences of the word.

Lowest score wins.

Answer 1

GolfScript, 19 characters

n%{'aeiouy'&,6=},n*

Usage:

golfscript vowels.gs < wordlist.txt

Output:

abstemiously
authoritatively
behaviourally
[...]
uninformatively
unintentionally
unquestionably
unrecognisably

If you also want to output the count at the end you can use

n%{'aeiouy'&,6=},.n*n@,

which is four characters longer.

Answer 2

Python - 46 characters

filter(lambda x:set(x)>=set("AEIOUY"),open(f))

Readable version: It's already pretty readable :-)

Answer 3

APL, 21 - 5 = 16

(,≢)w/⍨∧⌿'aeiouy'∘.∊w

Expects to find the list of words as w. Returns a list of the words that contain all the vowels, plus their count. Tested with ngn apl. Here's an example.

Explanation

         'aeiouy'∘.∊w   # generate a truth table for every vowel ∊ every word
       ∧⌿               # reduce with ∧ along the rows (vowels)
    w/⍨                 # select the words that contain all the vowels
(,≢)                    # hook: append to the list its own tally

Answer 4

Ruby 38

Edit 34: Best so far (from @O-I):

a.split.select{|w|'aeiouy'.count(w)>5} 

Edit 1: I just noticed the question asked for 'y' to be included among the vowels, so I've edited my question accordingly. As @Nik pointed out in a comment to his answer, "aeiouy".chars is one character less than %w[a e i o u y], but I'll leave the latter, for diversity, even though I'm risking nightmares over the opportunity foregone.

Edit 2: Thanks to @O-I for suggesting the improvement:

s.split.select{|w|'aeiouy'.delete(w)==''}

which saves 11 characters from what I had before.

Edit 3 and 3a: @O-I has knock off a few more:

s.split.select{|w|'aeiouy'.tr(w,'')==''}

then

s.split.reject{|w|'aeiouy'.tr(w,'')[1]}

and again (3b):

a.split.select{|w|'aeiouy'.count(w)>5} 

I am a mere scribe!

Here are two more uncompettive solutions:

s.split.group_by{|w|'aeiouy'.delete(w)}['']       (43)
s.gsub(/(.+)\n/){|w|'aeiouy'.delete(w)==''?w:''} (48)

Initially I had:

s.split.select{|w|(w.chars&%w[a e i o u y]).size==6}

s is a string containing the words, separated by newlines. An array of words from s that contains all five vowels is returned. For readers unfamiliar with Ruby, %w[a e i o u y] #=> ["a", "e", "i", "o", "u", "y"] and & is array intersection.

Suppose

s = "abbreviations\nabduction\n"
s.split #=> ["abbreviations", "abduction"] 

In the block {...}, initially

w             #=> "abbreviations"
w.chars       #=> ["a", "b", "b", "r", "e", "v", "i", "a", "t", "i", "o", "n", "s"]
w.chars & %w[a e i o u y] #=> ["a", "e", "i", "o"]
(w.chars & %w[a e i o u y]).size == 6 #=> (4 == 6) => false,

so "abbreviations" is not selected.

If the string s may contain duplicates, s.split.select... can be replaced by s.split.uniq.select... to remove duplicates.

Just noticed I could save 1 char by replacing size==6 with size>5.

Answer 5

GolfScript, 20 chars − 5 = 15 points

n%{'aeiouy'\-!},.,+`

Based on Howard's solution, but using a shorter test (\-! saves one char over &,6=), shorter length-appending (.,+ = 3 chars) and shorter output formatting (nobody said the output had to be newline-separated, so ` saves one char over n*).

Here's the output, given the lowercase word list as input (linebreaks inserted for readability):

["abstemiously" "authoritatively" "behaviourally" "consequentially" "counterrevolutionary"
"disadvantageously" "educationally" "encouragingly" "eukaryotic" "evolutionarily"
"evolutionary" "exclusionary" "facetiously" "gregariously" "heterosexuality" "homosexuality"
"importunately" "inconsequentially" "instantaneously" "insurrectionary" "intravenously"
"manoeuvrability" "neurologically" "neurotically" "ostentatiously" "pertinaciously"
"precariously" "precautionary" "questionably" "revolutionary" "simultaneously"
"supersonically" "tenaciously" "uncomplimentary" "uncontroversially" "unconventionally"
"undemocratically" "unemotionally" "unequivocally" "uninformatively" "unintentionally"
"unquestionably" "unrecognisably" 43]

(Ps. Technically, given that the challenge only says the code has to work for this specific input, n%{'aeiouy'\-!},`43 would be one character shorter yet. I consider that cheating, though.)

---

Explanation:

• n% splits the input on newlines into an array.
• { }, is a "grep" operator, executing the code between the braces for each element of the array and selecting those for which it returns a true value.
  • Inside the loop, 'aeiouy'\- takes the literal string aeiouy and removes from it every character found in the candidate word. The ! then logically negates the resulting string, yielding 1 (true) if the string is empty and 0 (false) if it's not.
• .,+ makes a copy of the filtered array, counts the number of words in it, and appends the result to the original array.
• Finally, ` un-evals the array, converting it into a string representation of its contents. (Without it, the words in the array would simply be concatenated in the output, yielding an unreadable mess.)

Answer 6

Haskell - 67

main=interact$unlines.filter(and.flip map "aeiouy".flip elem).lines

Answer 7

Ruby - 28 characters (or 27 if y is excluded from vowels)

("ieaouy".chars-s.chars)==[]

The complete command to run is (48 chars):

ruby -nle 'p $_ if ("ieaouy".chars-$_.chars)==[]'

EDIT: replaced puts with p as suggested by @CarySwoveland

Answer 8

AWK - 29

/a/&&/e/&&/i/&&/o/&&/u/&&/y/

To run: Save the lowercase word list to wordlist.txt. Then, do:

mawk "/a/&&/e/&&/i/&&/o/&&/u/&&/y/" wordlist.txt

If your system does not have mawk, awk can be used as well.

You can also run it from a file by saving the program to program.awk and doing mawk or awk -f program.awk.

Answer 9

Python, 45 characters

[w for w in open(f) if set('aeiouy')<=set(w)]

Answer 10

k [22-5=17 chars]

I have renamed the file "corncob_lowercase.txt" to "w"

Count the words [22 chars]

+/min'"aeiouy"in/:0:`w

Output

43

Find all words [25 chars]

x@&min'"aeiouy"in/:x:0:`w

Overall 43 words containing all the vowels (a e i o u y)

Output

"abstemiously"
"authoritatively"
"behaviourally"
..
..
"unintentionally"
"unquestionably"
"unrecognisably"

Answer 11

Javascript/JScript 147(152-5), 158(163-5) or 184(189-5) bytes:

Here is my Javascript and JScript horribly "ungolfyfied" version (164 152 152-5=147 bytes):

function(s,k,z,x,i,c,r){c='aeiouy'.split('');r=[];for(k in s=(s+'').split(/\b/)){i=0;for(z in c)i+=s[k].indexOf(c[z])>=0;i==6&&(r[r.length]=s[k]);}return r;}

function(s,k,z,x,i,c,r){c='aeiouy'.split('');r=[];for(k in s=(s+'').split(/\b/)){i=6;for(z in c)i-=!!s[k].search(c[z]);i&&(r[r.length]=s[k]);}return r;}

Thank you @GaurangTandon for the search() function, which saved me a byte!

RegExp based with HORRIBLE performance, but support both upper and lowercase (163-5=158 bytes):

function(s,k,z,x,i,c,r){c='aeiouy'.split('');r=[];for(k in s=(s+'').split(/\b/)){i=0;for(z in c)i+=RegExp(c[z],'i').test(s[k]);i==6&&(r[r.length]=s[k]);}return r;}

RegExp based with BETTER performance, BUT takes a lot more bytes (189-5=184 bytes):

function(s,k,z,x,i,c,r,l){l=[];r=[];for(z in c='aeiouy'.split(''))l[z]=RegExp(c[z],'i');for(k in s=(s+'').split(/\b/)){i=0;for(z in c)i+=l[z].test(s[k]);i==6&&(r[r.length]=s[k]);}return r;}

---

This one if just for the fun (175-5 bytes) and won't count as an answer:

function(s,k,z,x,i,c,r){c='aeiouy'.split('');r=[];for(k in s=(s+'').split(/\b/)){i=0;for(z in c)i+=s[k].indexOf(c[z])>=0;i==6&&(r[r[r.length]=s[k]]=1+(r[s[k]]||0));}return r;}

It's based on the 1st answer, but has a 'twist': You can know how many times a word has been found.

You simply do like this:

var b=(function(s,k,z,x,i,c,r){c='aeiouy'.split('');r=[];for(k in s=(s+'').split(/\b/)){i=0;for(z in c)i+=s[k].indexOf(c[z])>=0;i==6&&(r[r[r.length]=s[k]]=1+(r[s[k]]||0));}return r;})('youaie youaie youaie youaie a word');

b.youaie //should be 4

Since that length doesn't have all vowels, it wont be deleted and still would be an answer for the bonus.

---

How do you call it?

"Simple": You wrap the function inside () and then add ('string goes here'); to the end.

Like this: (function(s,k,z,x,i,c,r){c='aeiouy'.split('');r=[];for(k in s=(s+'').split(/\b/)){i=0;for(z in c)i+=s[k].indexOf(c[z])>=0;i==6&&(r[r.length]=s[k]);}return r;})('a sentence youyoy iyiuyoui yoiuae oiue oiuea');

This example will return an array only with 1 string: yoiuae

I know that this is the worst solution, but works!

---

Why am i counting -5?

Well, Javascript/JScript arrays have a property (length) in arrays which tells the number of elements that it have.

After being confirmed in the question, the bonus of -5 is for telling the number of words.

Since the number of words is in that property, automatically I have the score of -5.

Answer 12

Ruby 39 38

Currently the shortest Ruby entry when counting the whole program including input and output.

Saved a char by using map instead of each:

$<.map{|w|w*5=~/a.*e.*i.*o.*u/m&&p(w)}

Another version, 39 characters with prettier output:

puts$<.select{|w|w*5=~/a.*e.*i.*o.*u/m}

Both programs take input from stdin or as a file name passed as a command line argument:
$ ruby wovels.rb wordlist.txt

It costs 3 extra characters to inclyde y as a wovel.

Answer 13

Mathematica (65 or 314)

Two very different approaches, the better one was proposed by Belisarius in the comments to my initial response. First, my brutish effort, which algorithmically generates every possible regular expression that matches a combination of six vowels (including "y"), and then checks every word in the target wordlist against every one of these 720 regular expressions. It works, but it's not very concise and it's slow.

b = Permutations[{"a", "e", "i", "o", "u","y"}]; Table[
 Select[StringSplit[
  URLFetch["http://www.mieliestronk.com/corncob_lowercase.txt"]], 
  StringMatchQ[#, (___ ~~ b[[i]][[1]] ~~ ___ ~~ 
      b[[i]][[2]] ~~ ___ ~~ b[[i]][[3]] ~~ ___ ~~ 
      b[[i]][[4]] ~~ ___ ~~ b[[i]][[5]]~~ ___ ~~ b[[i]][[6]] ~~ ___)] &], {i, 1, 
  Length[b]}]

~320 characters. A few could be saved through using alternate notation, and additional characters are lost preparing the dictionary file as a list of strings (the natural format for the dictionary in Mathematica. Other languages may not need this prep, but Mathematica does). If we omit that step, presuming it to have been handled for us, the same approach can be done in under 250 characters, and if we use Mathematica's built-in dictionary, we get even bigger savings,

Map[DictionaryLookup[(___ ~~ #[[1]] ~~ ___ ~~ #[[2]] ~~ ___ ~~ #[[3]] 
~~ ___ ~~ #[[4]] ~~ ___ ~~ #[[5]]~~ ___ ~~ #[[6]] ~~ ___) .., IgnoreCase -> True] &, 
 Permutations[{"a", "e", "i", "o", "u","y"}]]

Under 200 characters. Counting the number of words found requires only passing the result to Length[Flatten[]], which can be added around either block of code above, or can be done afterwards with, for example, Length@Flatten@%. The wordlist specified for this challenge gives 43 matches, and the Mathematica dictionary gives 64 (and is much quicker). Each dictionary has matching words not in the other. Mathematica finds "unprofessionally," for example, which is not in the shared list, and the shared list finds "eukaryotic," which is not in Mathematica's dictionary.

Belisarius proposed a vastly better solution. Assuming the wordlist has already been prepared and assigned to the variable l, he defines a single test based on Mathematica's StringFreeQ[] function, then applies this test to the word list using the Pick[] function. 65 characters, and it's about 400 times faster than my approach.

f@u_ := And @@ (! StringFreeQ[u, #] & /@ Characters@"aeiouy"); Pick[l,f /@ l]

Answer 14

Perl 6 - 35 characters

Inspired by @CarySwoveland 's Ruby solution:

say grep <a e i o u y>⊆*.comb,lines

This selects (greps) each line that returns True for <a e i o u y> ⊆ *.comb, which is just a fancy way of asking "is the Set ('a','e','i','o','u','y') a subset (⊆) of the Set made up of the letters of the input (*.comb)?"

Actually, both <a e i o u y> and *.comb only create Lists: ⊆ (or (<=) if you're stuck in ASCII) turns them into Sets for you.

To get the number of lines printed, this 42 character - 5 = 37 point script will output that as well:

say +lines.grep(<a e i o u y>⊆*.comb)».say

Answer 15

C - 96 bytes

 char*gets(),*i,j[42];main(p){while(p=0,i=gets(j)){while(*i)p|=1<<*i++-96;~p&35684898||puts(j);}}

I saved several bytes of parentheses thanks to a fortunate coincidence of operator precedence.

Answer 16

Javascript - Score = 124 - 5 = 119 !!!

Edit: 17/02/14

function(l){z=[],l=l.split("\n"),t=0;while(r=l[t++]){o=0;for(i in k="aeiouy")o+=!~r.search(k[i]);if(o==0)z.push(r)}return z}

Big thanks to @Ismael Miguel for helping me cut off ~12 chars!

I removed the fat arrow notation function form because though I have seen it begin used, it doesn't work. No idea why...

---

To make it work:

Pass all the words separated by newline as an argument to the function as shown below.

---

Test:

// string is "facetiously\nabstemiously\nhello\nauthoritatively\nbye"

var answer = (function(l){z=[],l=l.split("\n"),t=0;while(r=l[t++]){o=0;for(i in k="aeiouy")o+=!~r.search(k[i]);if(o==0)z.push(r)}return z})("facetiously\nabstemiously\nhello\nauthoritatively\nbye")

console.log(answer);
console.log(answer.length);

/* output ->    
    ["facetiously", "abstemiously", "authoritatively"]
    3 // count
*/

---

Answer 17

Bash + coreutils, 39

eval cat`printf "|grep %s" a e i o u y`

Takes input from stdin.

Answer 18

sed 29 chars

/y/{/u/{/o/{/i/{/a/{/e/p}}}}}

Order choosed from Letter frequency on wikipedia to speed check.

On my host:

time sed -ne '/a/{/e/{/i/{/o/{/u/{/y/p}}}}}' </usr/share/dict/american-english >/dev/null 
real    0m0.046s

and

time sed -ne '/y/{/u/{/i/{/o/{/a/{/e/p}}}}}'</usr/share/dict/american-english >/dev/null 
real    0m0.030s

Answer 19

Bash (grep) - 36 bytes

g=grep;$g y|$g u|$g o|$g i|$g a|$g e

Note the order of vowels tested, least frequent first. For the test case, this runs about 3 times as fast as testing in order a e i o u y. That way the first test removes a larger number of words so subsequent tests have less work to do. Obviously this has no effect on the length of the code. Many of the other solutions posted here would benefit similarly from doing the tests in this order.

Answer 20

D - 196

import std.regex,std.stream;void main(string[]a){auto f=new File(a[1]);while(!f.eof)if(!(a[0]=f.readLine.idup).match("(?=.*a)(?=.*e)(?=.*i)(?=.*o)(?=.*u)(?=.*y).*").empty)std.stdio.writeln(a[0]);}

Un-golfed:

import std.regex, std.stream;

void main( string[] a )
{
    auto f = new File( a[1] );

    while( !f.eof )
        if( !( a[0] = f.readLine.idup ).match( "(?=.*a)(?=.*e)(?=.*i)(?=.*o)(?=.*u)(?=.*y).*" ).empty )
            std.stdio.writeln( a[0] );
}

Usage: C:\>rdmd vowels.d wordlist.txt

wordlist.txt must contain the lowercase list words.

Answer 21

Rebol (104 chars)

remove-each w d: read/lines %wordlist.txt[6 != length? collect[foreach n"aeiouy"[if find w n[keep n]]]]

Un-golfed:

remove-each w d: read/lines %wordlist.txt [
    6 != length? collect [foreach n "aeiouy" [if find w n [keep n]]]
]

d now contains list of found words. Here is an example from Rebol console:

>> ; paste in golf line.  This (REMOVE-EACH) returns the numbers of words removed from list

>> remove-each w d: read/lines %wordlist.txt[6 != length? collect[foreach n"aeiouy"[if find w n[keep n]]]]
== 58067

>> length? d
== 43

Answer 22

Smalltalk (36/57 chars)

'corncob_lowercase.txt' asFilename contents select:[:w | w includesAll:'aeiouy']

to get the count, send #size to the resulting collection. The result collection contains 43 words ('abstemiously' 'authoritatively' ... 'unquestionably' 'unrecognisably')

The code above has 77 chars, but I could have renamed the wordlist file to 'w', so I count the filename as 1 which gives a score of 57.

Is reading the file part of the problem or not? If not (see other examples), and the list of words is already in a collection c, then the code reduces to:

c select:[:w | w includesAll:'aeiouy']

which is 36 chars (with omittable whitespace removed).

Answer 23

updated: unnecessary spaces removed

Very slow but in bash (81 chars):

while read l;do [ `fold -w1<<<$l|sort -u|tr -dc ieaouy|wc -m` = 6 ]&&echo $l;done

EDIT: echo $l|fold -w1 replaced with fold -w1<<<$l as suggested by @nyuszika7h

Answer 24

JavaScript - 95 bytes

Here's my golf.

function f(s){return[var a=s.match(/^(?=.*a)(?=.*e)(?=.*i)(?=.*o)(?=.*u)(?=.*y).*/),a.length];}

And I would also like to point out that your golf doesn't seem to find all occurrences of vowels.

Ungolfed:

function countVowels(string) {
  var regex   = /^(?=.*a)(?=.*e)(?=.*i)(?=.*o)(?=.*u)(?=.*y).*/;
  var matches = string.match(regex);
  var length  = matches.length;
  return [matches, length];

Answer 25

sort + uniq + sed

This one does not match repeated occurrences of a word. It also does not match the letter 'y' if is occurs at the beginning of a word.

sort wordlist.txt | uniq | sed -n '/a/p' | sed -n '/e/p' | sed -n '/i/p' | sed -n '/o/p' | sed -n '/u/p' | sed -nr '/^[^y]+y/p' 

Answer 26

Bash

Not as short as the OP's , but one line in Bash:

while read p; do if [ $(sed 's/[^aeiouy]//g' <<< $p | fold -w1 | sort | uniq | wc -l) -eq 6 ] ; then echo $p; fi; done < wordlist.txt

Answer 27

C# - 170

using System.Linq;class P{static void Main(string[]a){System.Console.Write(string.Join(",",System.IO.File.ReadAllLines(a[0]).Where(w=>"aeiouy".All(c=>w.Contains(c)))));}}

Formatted:

using System.Linq;
class P
{
    static void Main(string[] a) { 
        System.Console.Write(
            string.Join(",", System.IO.File.ReadAllLines(a[0])
                .Where(w => "aeiouy".All(c => w.Contains(c))))); 
    }
}

Not in the mood right now to implement counting but should be easy. The path to the (lower-case version of the) wordlist should be passed to the program as first argument:

program.exe D:\foo\bar\corncob_lowercase.txt

Output:

abstemiously,authoritatively,behaviourally,consequentially,counterrevolutionary,
disadvantageously,educationally,encouragingly,eukaryotic,evolutionarily,evolutio
nary,exclusionary,facetiously,gregariously,heterosexuality,homosexuality,importu
nately,inconsequentially,instantaneously,insurrectionary,intravenously,manoeuvra
bility,neurologically,neurotically,ostentatiously,pertinaciously,precariously,pr
ecautionary,questionably,revolutionary,simultaneously,supersonically,tenaciously
,uncomplimentary,uncontroversially,unconventionally,undemocratically,unemotional
ly,unequivocally,uninformatively,unintentionally,unquestionably,unrecognisably

I took the liberty of outputting and comma-separating the words; neither of which is specified in the rules (which state "must find all the words", not how (and IF) to output).

Including count (+output): 192 - 5 = 187

using System.Linq;class P{static void Main(string[]a){var r=System.IO.File.ReadAllLines(a[0]).Where(w=>"aeiouy".All(c=>w.Contains(c)));System.Console.Write(string.Join(",",r)+" "+r.Count());}}

Output:

abstemiously,authoritatively,behaviourally,consequentially,counterrevolutionary,
disadvantageously,educationally,encouragingly,eukaryotic,evolutionarily,evolutio
nary,exclusionary,facetiously,gregariously,heterosexuality,homosexuality,importu
nately,inconsequentially,instantaneously,insurrectionary,intravenously,manoeuvra
bility,neurologically,neurotically,ostentatiously,pertinaciously,precariously,pr
ecautionary,questionably,revolutionary,simultaneously,supersonically,tenaciously
,uncomplimentary,uncontroversially,unconventionally,undemocratically,unemotional
ly,unequivocally,uninformatively,unintentionally,unquestionably,unrecognisably 4
3

(Note the count at the end: 43)

No output ("must find all the words"): 137 - 5 = 132

using System.Linq;class P{static void Main(string[]a){var r=System.IO.File.ReadAllLines(a[0]).Where(w=>"aeiouy".All(c=>w.Contains(c)));}}

(Bending the rules a bitm then again: not really) This finds all the words and the count is available by executing r.Count().

Answer 28

C-Sharp

I've never done this before and I'm not exactly sure what the posting procedures are. But this is what i came up with:

185 bytes

Action<string>x=(s=>s.Split('\n').ToList().ForEach(w=>{if("aeiouy".All(v=>w.Contains(v)))Console.Write(w);}));using(var s=new StreamReader(@"C:\corncob_lowercase.txt"))x(s.ReadToEnd());

wordList = a List<string> of all the words.

if you want to display a total:

219 - 5 = 214 bytes

Action<string>x=(s=>{var t=0;s.Split('\n').ToList().ForEach(w=>{if("aeiouy".All(v=>w.Contains(v))){Console.Write(w);t++;}});Console.Write(t);});using(var s=new StreamReader(@"C:\corncob_lowercase.txt"))x(s.ReadToEnd());

---

Expanded

// init
var words = "";
var vowels = "aeiouy";
var total = 0;

using (var stream = new StreamReader(@"C:\corncob_lowercase.txt"))
{
    // read file
    words = stream.ReadToEnd();

    // convert word to List<string>
    var wordList = words.Split('\n').ToList();

    // loop through each word in the list
    foreach (var word in wordList)

        // check if the current word contains all the vowels
        if(vowels.All (w => word.ToCharArray().Contains(w)))
        {
            // Count total
            total += 1;
            // Output word
            Console.Write(word);
        }

    // Display total
    Console.WriteLine(total);
}

Answer 29

vb.net (Score 91 = 96c - 5)_*0

_{*0 +49c min}

This creates an enumeration contain all of the words which contain all of the vowels.

Dim r=IO.File.ReadLines(a(0)).Where(Function(w)"aeiou".Intersect(w).Count=5)
Dim c=r.Count

Answer 30

Mathematica - 136 102

Fold[Flatten@StringCases@## &, 
Flatten@Import@"http://bit.ly/1iZE9kY",
___ ~~ # ~~ ___ & /@ Characters@"aeiouy"]

The shortened link goes to http://www.mieliestronk.com/corncob_lowercase.txt