using Grep

Poison Ivy · #1 (**permalink**) 07-30-2006

I'm trying to figure out what command I would enter in to get all 4 letter words (only a-z characters, punctuations should not be counted as a letter)

I have this so far:

Code:

cat /usr/share/dict/words | egrep '^....$'

But that command counts punctuations such as ' as letters

xinusys · #2 (**permalink**) 07-30-2006

use regular expression man.it's very easy.

Poison Ivy · #3 (**permalink**) 07-30-2006

umm sorry i dont know what u mean lol im very new to unix

nathan · #4 (**permalink**) 07-30-2006

A google search for "regular expressions" will give you all the information you need to know. Please note - they're not just Unix-specific.

Code:

$ cat /usr/share/dict/words | egrep '^[a-z]{4}$' | head -3
aahs
aals
abac

Poison Ivy · #5 (**permalink**) 07-30-2006

that's good actually but what i meant was to include words such as she's as 4 letter word etc..

nathan · #6 (**permalink**) 07-31-2006

Use a regular expression to match the pattern:

letter + [not a letter] 0 or more times +
letter + [not a letter] 0 or more times +
letter + [not a letter] 0 or more times +
letter + [not a letter] 0 or more times

It's quite easy, I used it to find these words that start with "she" in my /usr/share/dict/words file.

shea
she'd
shed
shee
shem
sher
shes
shew

vino · #7 (**permalink**) 07-31-2006

Code:

grep -wE  "[[:lower:]]{3}[[:punct:]]{0,}[[:lower:]]{1}"

3 letter combination + any number of punctuation marks + 1 letter

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