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Grep -v lines starting with pattern 1 and not matching pattern 2


 
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# 1  
Grep -v lines starting with pattern 1 and not matching pattern 2

Hi all!

Thanks for taking the time to view this!

I want to grep out all lines of a file that starts with pattern 1 but also does not match with the second pattern.

Example:

Code:
Drink a soda
Eat a banana
Eat multiple bananas
Drink an apple juice
Eat an apple
Eat multiple apples

I want to remove all lines that starts with Eat but does not contain apple. Other lines that dont start with Eat should be ignored. Result:

Code:
Drink a soda
Drink an apple juice
Eat an apple
Eat multiple apples

I have looked for a solution but so far no luck... Is it possible to grep it?

Thanks a lot!

Last edited by Don Cragun; 07-14-2015 at 08:43 PM.. Reason: Add CODE and ICODE tags.
# 2  
I believe awk would be easier. Your logic would be ! ( /^Eat/ && ! /apple/), or, following De Morgan's law:

Code:
$ awk '! /^Eat/ || /apple/' input
Drink a soda
Drink an apple juice
Eat an apple
Eat multiple apples

or if it really has to be grep:

Code:
mute@tiny:~$ grep -e '^[^E][^a][^t]' -e apple input
Drink a soda
Drink an apple juice
Eat an apple
Eat multiple apples

These 2 Users Gave Thanks to neutronscott For This Post:
# 3  
Thanks a lot!

The awk worked out just nice, however I could not get the grep one to do the same... anyways, thank you very much!
# 4  
grep only solution, but NOT elegant nor efficient:
Code:
{ grep -iv "^eat" file; grep "apple" file; } | grep -f- file
Drink a soda
Drink an apple juice
Eat an apple
Eat multiple apples

# 5  
Code:
sed '/^Eat/{/apple/!d;}' input

No easy solution with grep, because it reports all positives or all negatives (-v option).

---------- Post updated at 06:37 AM ---------- Previous update was at 03:53 AM ----------

neutronscott's grep solution does not print
Code:
Fat fish

This User Gave Thanks to MadeInGermany For This Post:
# 6  
Another inelegant grep, but one with a single pass :
Code:
grep -E '^[^E]|^E[^a]|^Ea[^t]|apple' file

Like MadeInGermany says, grep isn't really well suited for this..
This User Gave Thanks to Scrutinizer For This Post:
# 7  
Hi.

An alternative to grep, a Ruby code, glark, can do this:
Code:
#!/usr/bin/env bash

# @(#) s1	Demonstrate multiple matches, inverted, glark.
# For glark, see repository (Debian, Mac OS X "port"), otherwise:
# https://github.com/jpace/glark

# Utility functions: print-as-echo, print-line-with-visual-space, debug.
# export PATH="/usr/local/bin:/usr/bin:/bin"
LC_ALL=C ; LANG=C ; export LC_ALL LANG
pe() { for _i;do printf "%s" "$_i";done; printf "\n"; }
pl() { pe;pe "-----" ;pe "$*"; }
db() { ( printf " db, ";for _i;do printf "%s" "$_i";done;printf "\n" ) >&2 ; }
db() { : ; }
C=$HOME/bin/context && [ -f $C ] && $C ruby glark

FILE=${1-data1}

pl " Input data file $FILE:"
cat $FILE

pl " Expected output:"
cat expected-output.txt

# Remove lines that start with Eat  and does not contain apple
pl " Results (no color, no line numbers):"
glark -U -N -v --and 0 '/^Eat/' '!/apple/' $FILE

exit 0

producing:
Code:
$ ./s1

Environment: LC_ALL = C, LANG = C
(Versions displayed with local utility "version")
OS, ker|rel, machine: Linux, 2.6.26-2-amd64, x86_64
Distribution        : Debian 5.0.8 (lenny, workstation) 
bash GNU bash 3.2.39
ruby 1.8.7 (2008-08-11 patchlevel 72) [x86_64-linux]
glark version 1.8.0

-----
 Input data file data1:
Drink a soda
Eat a banana
Eat multiple bananas
Drink an apple juice
Eat an apple
Eat multiple apples
Fat fish

-----
 Expected output:
Drink a soda
Drink an apple juice
Eat an apple
Eat multiple apples
Fat fish

-----
 Results (no color, no line numbers):
Drink a soda
Drink an apple juice
Eat an apple
Eat multiple apples
Fat fish

Best wishes ... cheers, drl
 

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