CODEWITHSUNDEEP
grep
manolius
manolius

Reputation: 395

remove word using "grep -v" exact match only

How can I do an exact match using grep -v?

For example: the following command

for i in 0 0.3 a; do echo $i | grep -v "0"; done

returns a. But I want it to return 0.3 a.

Using 

grep -v "0\b"

is not working

Upvotes: 0

Views: 1033

Answers (2)

kvantour
kvantour

Reputation: 26481

The safest way for single-column entries is using awk. Normally, I would use grep with the -w flag, but since you want to exactly match an integer that could be part of a float, it is a bit more tricky. The <dot>-character makes it hard to use any of 

grep -vw 0
grep -v '\b0\b'
grep -v '\<0\>'

The proposed solution also will only work on perfect lines, what if you have a lost space in front or after your zero. The line will fail. So the safest would be:

single column file:

awk '($1!="0")' file

multi-word file: (adopt the variable FS to fit your needs)

awk '{for(i=1;i<=NF;++i) if($i == "0") next}1' file

Upvotes: 1

mikeb
mikeb

Reputation: 11267

for i in 0 0.3 a; do echo $i | grep -v "^0$"; done

You need to match the start and end of the string with ^ and $

So, we say "match the beginning of a line, the char 0 and then the end of the line.

$ for i in 0 0.3 a; do echo $i | grep -v "^0$"; done

0.3

a

Upvotes: 2

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