how can i easily (quick and dirty) change, say 10, random lines of a file with a simple shellscript?
i though about abusing ed and generating random commands and line ranges, but i'd like to know if there was a better way
+1
A:
awk 'BEGIN{srand()}
{ lines[++c]=$0 }
END{
while(d<10){
RANDOM = int(1 + rand() * c)
if( !( RANDOM in r) ) {
r[RANDOM]
print "do something with " lines[RANDOM]
++d
}
}
}' file
or if you have the shuf command
shuf -n 10 $file | while read -r line
do
sed -i "s/$line/replacement/" $file
done
ghostdog74
2010-09-06 10:31:00
You probably want to do this: `BEGIN { srand() }`
Dennis Williamson
2010-09-06 15:38:30
+2
A:
This seems to be quite a bit faster:
file=/your/input/file
c=$(wc -l < "$file")
awk -v c=$c 'BEGIN {
srand();
for (i=0;i<10;i++) lines[i] = int(1 + rand() * c);
asort(lines);
p = 1
}
{
if (NR == lines[p]) {
++p
print "do something with " $0
}
else print
}' "$file"
I
Dennis Williamson
2010-09-06 16:08:17
can you explain why the shuf command doesn't keep order? shuf just brings out 10 random lines, then uses sed to change each of them. Not efficient though i can say.
ghostdog74
2010-09-06 23:52:36
@ghostdog74: I'm not sure I understand. `shuf` is short for "shuffle" and that's what it does.
Dennis Williamson
2010-09-07 02:13:28
@ghostdog74: I'm sorry, I overlooked a tiny little `-i` which caused me to misunderstand what yours is doing. My apologies.
Dennis Williamson
2010-09-07 06:10:35
This solution assumes lines[] will be unique. Run the above code against a file with 11 lines. You will rarely have 10 lines modified; usually there will be fewer.
themis
2010-09-10 23:53:30
+1
A:
Playing off @Dennis' version, this will always output 10. Doing random numbers in a separate array could create duplicates and, consequently, fewer than 10 modifications.
file=~/testfile
c=$(wc -l < "$file")
awk -v c=$c '
BEGIN {
srand();
count = 10;
}
{
if (c*rand() < count) {
--count;
print "do something with " $0;
} else
print;
--c;
}
' "$file"
themis
2010-09-07 02:38:17