Grep beginning of line

Solution 1:

The symbol for the beginning of a line is ^. So, to print all lines whose first character is a (, you would want to match ^(:

1. grep

   grep '^(' file
   

2. sed

   sed -n '/^(/p' file
   

Solution 2:

Using perl

perl -ne '/^\(/ && print' foo

Output:

(((jfojfojeojfow 
(((jfojfojeojfow

Explanation (regex part)

• /^\(/

  • ^ assert position at start of the string
  • \( matches the character ( literally

Solution 3:

Here is a bash one liner:

while IFS= read -r line; do [[ $line =~ ^\( ]] && echo "$line"; done <file.txt

Here we are reading each line of input and if the line starts with (, the line is printed. The main test is done by [[ $i =~ ^\( ]].

Using python:

#!/usr/bin/env python2
with open('file.txt') as f:
    for line in f:
        if line.startswith('('):
            print line.rstrip()

Here line.startswith('(') checks if the line starts with (, if so then the line is printed.

Solution 4:

awk

awk '/^\(/' testfile.txt

Result

$ awk '/^\(/' testfile.txt                   
(((jfojfojeojfow 
(((jfojfojeojfow

Python

As python one-liner:

$ python -c 'import sys;print "\n".join([x.strip() for x in sys.stdin.readlines() if x.startswith("(")])' < input.txt    
(((jfojfojeojfow
(((jfojfojeojfow

Or alternatively:

$ python -c 'import sys,re;[sys.stdout.write(x) for x in open(sys.argv[1]) if re.search("^\(",x)]' input.txt

BSD look

look is one of the classic but little known Unix utilities, which appeared way back in AT&T Unix version 7. From man look:

The look utility displays any lines in file which contain string as a prefix

The result:

$ look "(" input.txt
(((jfojfojeojfow 
(((jfojfojeojfow