Get 20% of lines in File randomly


 
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# 15  
Old 10-22-2012
Quote:
Originally Posted by chercheur857
where X and Y are digits? That means: do you have -XY. once and only once in each line?

Yes X and Y are digits, XY no redendant in other line (once and only once in each line)
Ok. So we have a sequence of chars (a -, then some digits you may want to use for exclusion, then a .) once and only once in each line. So here it is:

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 red.
--
Bye
This User Gave Thanks to Lem For This Post:
# 16  
Old 10-22-2012
Thank you so much for help Smilie
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