How do I remove all lines in a file that are less than 6 characters?

Solution 1:

There are many ways to do this.

Using grep:

grep -E '^.{6,}$' file.txt >out.txt

Now out.txt will contain lines having six or more characters.

Reverse way:

grep -vE '^.{,5}$' file.txt >out.txt

Using sed, removing lines of length 5 or less:

sed -r '/^.{,5}$/d' file.txt

Reverse way, printing lines of length six or more:

sed -nr '/^.{6,}$/p' file.txt

You can save the output in a different file using > operator like grep or edit the file in-place using -i option of sed:

sed -ri.bak '/^.{6,}$/' file.txt

The original file will be backed up as file.txt.bak and the modified file will be file.txt.

If you do not want to keep a backup:

sed -ri '/^.{6,}$/' file.txt

Using shell, Slower, Don't do this, this is just for the sake of showing another method:

while IFS= read -r line; do [ "${#line}" -ge 6 ] && echo "$line"; done <file.txt

Using python,even slower than grep, sed:

#!/usr/bin/env python2
with open('file.txt') as f:
    for line in f:
        if len(line.rstrip('\n')) >= 6:
            print line.rstrip('\n')

Better use list comprehension to be more Pythonic:

#!/usr/bin/env python2
with open('file.txt') as f:
     strip = str.rstrip
     print '\n'.join([line for line in f if len(strip(line, '\n')) >= 6]).rstrip('\n')

Solution 2:

It's very simple:

grep ...... inputfile > resultfile   #There are 6 dots

This is extremely efficient, as grep will not try to parse more than it needs, nor to interpret the chars in any way: it simply send a (whole) line to stdout (which the shell then redirects to resultfile) as soon as it saw 6 chars on that line (. in a regexp context matches any 1 character).

So grep will only output lines having 6 (or more) chars, and the other ones are not outputted by grep so they don't make it to resultfile.

Solution 3:

Solution #1: using C

Fastest way: compile and run this C program:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define MAX_BUFFER_SIZE 1000000

int main(int argc, char *argv[]) {
    int length;

    if(argc == 3)
        length = atoi(argv[2]);
    else
        return 1;

    FILE *file = fopen(argv[1], "r");

    if(file != NULL) {
        char line[MAX_BUFFER_SIZE];

        while(fgets(line, sizeof line, file) != NULL) {
            char *pos;

            if((pos = strchr(line, '\n')) != NULL)
                *pos = '\0';
            if(strlen(line) >= length)
                printf("%s\n", line);
        }

        fclose(file);
    }
    else {
        perror(argv[1]);
        return 1;
    }
    
    return 0;
}

Compile with gcc program.c -o program, run with ./program file line_length (where file = path to the file and line_length = minimum line length, in your case 6; the maximum line length is limited to 1000000 characters per line; you can change this by changing the value of MAX_BUFFER_SIZE).

(Trick to substitute \n with \0 found here.)

Comparison with all the other solutions proposed to this question except the shell solution (test run on a ~91MB file with 10M lines with an average lenght of 8 characters):

time ./foo file 6

real    0m1.592s
user    0m0.712s
sys 0m0.160s

time grep ...... file

real    0m1.945s
user    0m0.912s
sys 0m0.176s

time grep -E '^.{6,}$'

real    0m2.178s
user    0m1.124s
sys 0m0.152s

time awk 'length>=6' file

real    0m2.261s
user    0m1.228s
sys 0m0.160s

time perl -lne 'length>=6&&print' file

real    0m4.252s
user    0m3.220s
sys 0m0.164s

sed -r '/^.{,5}$/d' file >out

real    0m7.947s
user    0m7.064s
sys 0m0.120s

./script.py >out
real    0m8.154s
user    0m7.184s
sys 0m0.164s

Solution #2: using AWK:

awk 'length>=6' file

length>=6: if length>=6 returns TRUE, prints the current record.

Solution #3: using Perl:

perl -lne 'length>=6&&print' file

If lenght>=6 returns TRUE, prints the current record.

% cat file
a
bb
ccc
dddd
eeeee
ffffff
ggggggg
% ./foo file 6
ffffff
ggggggg
% awk 'length>=6' file   
ffffff
ggggggg
% perl -lne 'length>=6&&print' file
ffffff
ggggggg