CODEWITHSUNDEEP
pythonbashawksed
jtan
jtan

Reputation: 9

deleting lines after a specific string is greater than a value (bash or python)

I have a data file with a list of numbers like that:

5 5
6 6
4 4
3 3
4 4
3 3
2 2

I wish to pick out the first row that has the two columns with numbers <=3 and delete all the rows after that. Hence my result file will just be:

5 5 
6 6
4 4 
3 3

Does anyone know how to use awk/sed for this?

Upvotes: 0

Views: 289

Answers (3)

Raymond Hettinger
Raymond Hettinger

Reputation: 226516

For Python, print every line and just break out of the loop when your criteria is met:

>>> while open('somefile.txt') as f:
        for line in f:
            print(line, end='')
            p, q = map(int, line.split())
            if p <= 3 or q <= 3:
                break


5 5
6 6
4 4
3 3

Here is an AWK solution:

$ awk 'BEGIN {flag=0} flag==0 {print ; flag=$1<=3 && $1<=3}' somefile.txt
5 5
6 6
4 4
3 3

Upvotes: 0

James Brown
James Brown

Reputation: 37424

Here's an awk solution:

$ awk '1; $1<=3&&$2<=3{exit}' file
5 5
6 6
4 4
3 3

Explained:

awk '
1;                           # output record
$1 <= 3 && $2 <= 3 { exit }  # if it was as requested, exit
' file

Upvotes: 1

RomanPerekhrest
RomanPerekhrest

Reputation: 92874

Consider the following sed approach:

sed -n '1,/[0-3] [0-3]/p' file

The output:

5 5    
6 6    
4 4    
3 3

1,/[0-3] [0-3] - address range starting from the first line and captures lines till encountering 2 digits which are less or equal to 3(inclusively).

An address range is specified with two addresses separated by a comma (,). Addresses can be numeric, regular expressions, or a mix of both.

Upvotes: 1

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