Print all words in a file sorted by length

for w in $(tr 'A-Z ,."()?!;:' 'a-z\n' < sample.txt); do echo ${#w} $w; done | sort -u | sort -n
Take a file and ,."()?!;: give a list of all the words in order of increasing length. First of all use tr to map all alphabetic characters to lower case and also strip out any puntuation. A-Z become a-z ,."()?!;: all become \n (newline) I've ignored - (hyphen) and ' (apostrophe) because they occur in words. Next use bash to print the length ${#w} and the word Finally sort the list numerically (sort -n) and remove any duplicates (sort -u). Note: sort -nu performs strangely on this list. It outputs one word per length.
Sample Output
9 suggested
9 unsightly
10 advantages
10 categorise
10 considered
10 everything
12 experimental
12 grandparents
13 insignificant
13 self-assembly
14 apologetically

By: flatcap
2012-03-15 14:14:11

3 Alternatives + Submit Alt

What Others Think

This yields the same: egrep -Eo "[0-9a-Z\'\-]*" sample.txt|tr A-Z a-z|sort -u|awk '{ print length(), $0 | "sort -n" }'
knoppix5 · 543 weeks and 1 day ago
you dont need 2 sorts you can just do sort -un
chrismccoy · 543 weeks and 1 day ago

What do you think?

Any thoughts on this command? Does it work on your machine? Can you do the same thing with only 14 characters?

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