I have a very large file 100Mb+ where all the content is on one line. I wish to find a pattern in that 开发者_JAVA技巧file and a number of characters around that pattern.
For example I would like to call a command like the one below but where -A and -B are number of bytes not lines:
cat very_large_file | grep -A 100 -B 100 somepattern
So for a file containing content like this:
1234567890abcdefghijklmnopqrstuvwxyz
With a pattern of
890abc
and a before size of -B 3
and an after size of -A 3
I want it to return:
567890abcdef
Any tips would be great. Many thanks.
You could try the -o option:
-o, --only-matching
Show only the part of a matching line that matches PATTERN.
and use a regular expression to match your pattern and the 3 preceding/following characters i.e.
grep -o -P ".{3}pattern.{3}" very_large_file
In the example you gave, it would be
echo "1234567890abcdefghijklmnopqrstuvwxyz" > tmp.txt
grep -o -P ".{3}890abc.{3}" tmp.txt
Another one with sed (you may need it on systems where GNU grep is not available):
sed -n '
s/.*\(...890abc...\).*/\1/p
' infile
Best way I can think of doing this is with a tiny Perl script.
#!/usr/bin/perl
$pattern = $ARGV[0];
$before = $ARGV[1];
$after = $ARGV[2];
while(<>) {
print $& if( /.{$before}$pattern.{$after}/ );
}
You would then execute it thusly:
cat very_large_file | ./myPerlScript.pl 890abc 3 3
EDIT: Dang, Paolo's solution is much easier. Oh well, viva la Perl!
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