CODEWITHSUNDEEP
awksed
sivakumar V
sivakumar V

Reputation: 135

how to replace the starting with minus value in a file with zero

I am having the text file as below. that file contains some negative data. so I want to replace that data with zero. 

1,2,3
4,5,-6
7,-8,9

I have tried with below command.

sed 's/-*/0/g' 1.txt

but am not getting the reqired output. so can any one please help me on this.

Upvotes: 1

Views: 486

Answers (5)

Claes Wikner
Claes Wikner

Reputation: 1517

awk 'sub(/-./,"0")1' file
1,2,3
4,5,0
7,0,9

Upvotes: 1

James Brown
James Brown

Reputation: 37404

awk:

$ awk 'gsub(/-[0-9]+/,0)1' 1.txt

Identical to @redneb's sed solution but in awk. Replace all - starting number strings ([0-9]+) with a 0 in record $0 and print it nevertheless (1in the end).

Upvotes: 1

glenn jackman
glenn jackman

Reputation: 246774

With awk, you would write

awk -F, -v OFS=, '{for (i=1; i<=NF; i++) if ($i < 0) $i = 0; print}' file

My sed would be

sed -r 's/(^|,)-[^,]+/\10/g' file

Upvotes: 1

rojeeer
rojeeer

Reputation: 2011

You have to better understand regular expression for sed. The expression -* in s /-*/0/g will match zero or more '-' characters. * in regular expression means zero or more duplicates, not wild character match, . does all character set match for a single character. See http://www.grymoire.com/Unix/Regular.html

To your question, you have to build a RE to match any negative number (maybe you need also consider float numbers). I would suggest:

-[0-9]*\.?[0-9]+

This is an extended regular expression for sed program, meaning that starts with -, can with or without pre-dot numbers or dot, and must contain at least one number.

So 

sed -r s/-[0-9]*\.?[0-9]+/0/g 1.txt >2.txt

is what you need. -r option means use extended regular expression.

Some tests:

echo '12, 0.12, -2' | sed -r s/-[0-9]*\.?[0-9]+/0/g
      12, 0.12, 0
echo '12, -234, -2' | sed -r s/-[0-9]*\.?[0-9]+/0/g
      12, 0, 0
echo '-0.98, 2.34, -2' | sed -r s/-[0-9]*\.?[0-9]+/0/g
          0, 2.34, 0
echo '-9.8, 0.34, -2' | sed -r s/-[0-9]*\.?[0-9]+/0/g
         0, 0.34, 0
echo '-9.8, 1.34, -2' | sed -r s/-[0-9]*\.?[0-9]+/0/g
         0, 1.34, 0

Hope this would be helpful, check tutors on http://www.grymoire.com/Unix/Sed.html.

Thanks

Upvotes: 1

redneb
redneb

Reputation: 23850

Try this:

sed 's/-[0-9]\+/0/g' 1.txt

The way this works is, sed will find all sequences of characters that match the pattern -[0-9]\+ and replace them with a 0. Now, they meaning of the -[0-9]\+ pattern is as follows: it will find a - character followed by a nonempty sequence of characters (this is the meaning of \+) that consists entirely of characters in the set between 0 and 9.

Upvotes: 2

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