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danielbmartin 09-14-2022 09:39 AM
Seven-letter words with only one vowel
 
This is my solution.
Code:
echo; echo 'Find seven-letter words which have only one vowel.'
    Path=${0%%.*}
  OutFile=$Path"out.txt"
 WordList='/usr/share/dict/words'
  Sevens=$Path"sevens.txt"
 egrep "^.{7}$" $WordList      \
|grep -v '[^a-z]' >$Sevens
 sed 's/[^aeéiouy]//' $Sevens  \
|sed 's/[^aeéiouy]//'          \
|sed 's/[^aeéiouy]//'          \
|sed 's/[^aeéiouy]//'          \
|sed 's/[^aeéiouy]//'          \
|sed 's/[^aeéiouy]//'          \
|paste -d" " $Sevens -          \
|egrep "^.{9}$"                \
>$OutFile
It works but seems like brute force. Is there a better, faster, cleaner method? Please advise.

Daniel B. Martin

.

danielbmartin 09-14-2022 09:55 AM
I searched my personal library and found this one.
Code:
egrep "^.{7}$" $WordList      \
|grep -v '[^a-z]' >$Sevens

tr -dc '[aeiouéèy\n]' <$Sevens  \
|paste $Sevens -                \
|sed -nr '/^.{9}$/p'            \
>$OutFile
Daniel B. Martin

.

Turbocapitalist 09-14-2022 11:18 AM
There don't seem to be any modules or built-in character sets which include a comprehensive group of vowels, at least not externally. However, CPAN's Unicode::Normalize can convert to ASCII and then you can check with a pattern similar to one of the ones which you have used above.

Code:
#!/usr/bin/perl                                                               
                                                                   
use Unicode::Normalize;                                                                               
use strict;                                                                   
use warnings;

while (my $s = <>) {
    if ($s =~ m/^\w{7}$/) {
        my $d = NFKD($s);
        if ($d =~ m/[aeiou]{1}/ && $d !~ m/[aeiou](?=.*[aeiou])/i) {
            print $s;
        }
    }
}

exit(0);
That reads in a line and checks if it contains a seven-letter string. Then it normalizes it to ASCII and checks for the presence of a single vowel, except for y in English and w in Welsh etc.

I think that's about as short as it can be with the Unicode constraint. If you are sticking with ASCII then sed or grep would be enough.

Edit: The old version allowed up to one vowel but did not require one. The modification requires exactly one vowel.

grail 09-14-2022 10:56 PM
Code:
awk -F'[aeiou]' '/^\w{7}$/ && NF == 2' $WordList

pan64 09-15-2022 12:18 AM
you can also try to
1. filter 7 letter words,
2. remove all the non-wovels (like your sed)
3. check length again, should be 1
but the post #3 and #4 are definitely much better. I would avoid using pipe chains if possible.

astrogeek 09-15-2022 01:28 AM
"Holy Grail, Batman!"

I think I'll keep my inelegant solution to myself!

MadeInGermany 09-15-2022 07:12 AM
Code:
awk '
  { u0=toupper($0) }
  length==7 && gsub(/[BCDFGHJKLMNPQRSTVWXZ]/, "", u0)==1
' "$WordList"
If you want to keep+print the vowel:
Code:
awk '
  { u0=toupper($0) }
  length==7 && gsub(/[^BCDFGHJKLMNPQRSTVWXZ]/, "", u0)==6 {
    print u0,$0
  }
' "$WordList"

sundialsvcs 09-15-2022 11:59 AM
Hint:

Filter for words which do match [aeéiouy] in a suitable words-file, then pipe the results to keep only those which do not match [aeéiouy].*[aeéiouy]. Finally, filter out all which are not seven-letter words. (Rearrange the order of piped operations as you like.)

A "shell one-liner," given a suitable "words file," can solve this. No programming is required.

Q.E.D.


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