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Top Forums Shell Programming and Scripting Get 20% of lines in File randomly Post 302719111 by Lem on Monday 22nd of October 2012 05:34:32 AM
Old 10-22-2012
Quote:
Originally Posted by chercheur857
My question please how can i eliminate 50 an 51 , i mean i'd like 25% of this list other than this two lines 50 and 51, and i'd like to specify it in the command like that
Code:
./program Myfile 10 50 51

Have you an idea please ?
Thank you so much for help Smilie
Code:
#!/bin/bash

inputfile="/tmp/buddyfile"
cp "$1" $inputfile
(( $# < 3 )) || {
        string="-${3}$"
        for ((a=4;a<=$#;a++)); do
                string+="\|-${!a}$"
        done
        sed -i "/$string/d" $inputfile; }

range=$(wc -l $inputfile | cut -d " " -f1)
(( $range > 32768 )) && exit              ### max number of lines for this script: 32768
percent=$2                                ### set this  as you like, [1-100]
(( 0 < $2 )) && (( $2 < 101 )) || exit
lim=$(( $range * $percent / 100 ))
for ((i=0;i<lim;i++)); do
        num=$(( $RANDOM % $range + 1 ));
        arr[$i]=$num;
        for ((j=0;j<i;j++)); do
                (( ${arr[$j]} == $num )) && {
                        let i--
                        break; }
        done
done

for linenum in "${arr[@]}"; do
        line=$(sed -n "$linenum p" $inputfile)
        ### your stuff here, for example: ###
        echo "$linenum"$'\t'"### $line"
done

exit 0

Changes are in bold. I didn't test it, but it should work.

Of course if 50, 51 and friends are true line numbers counting from 1 (first line) onwards, we can simplify the new part of the script.

Beware: if you have a file of 100 lines, and you exclude 40 lines, and you want 30% of the lines, you'll get 30% of the remaining 60 lines, so you'll get 18 lines. Is this right for you, or do you still want 30 lines among the 60 remaining lines?
--
Bye

Last edited by Lem; 10-22-2012 at 06:43 AM..
This User Gave Thanks to Lem For This Post:
 

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