Cheapest way to get line X from file

treesloth · #1 (**permalink**) 3 Weeks Ago

So, I have a couple of scripts that go through lots and lots of data-- sometimes thousands of smaller files, sometimes a few very large ones. It's very common that I need a particular line, by line number X. There are lots of ways to do this:

Code:

awk 'NR == X {print ; exit}' file
sed -n 'Xp' file
head -n X file | tail -n 1

There are others, too... My question is, what's the most economical way you know to do this? In my primitive tests (involving a large file and the time command), head/tail seem to be the winner. What is surprising is how badly sed did. On a file with 4237905 lines, here are the results:

Code:

awk                   sed                     head/tail
real    0m1.258s      real    0m22.469s       real    0m1.011s
user    0m1.010s      user    0m22.170s       user    0m0.550s
sys     0m0.250s      sys     0m0.280s        sys     0m0.730s

I actually expected sed to win. Strange... Anyway, are there any other methods I should consider? Are there improvements to these methods? Thanks.

vgersh99 · #2 (**permalink**) 3 Weeks Ago

try these from sed1liners:

Code:

 # print line number 52
 sed -n '52p'                 # method 1
 sed '52!d'                   # method 2
 sed '52q;d'                  # method 3, efficient on large files

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