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linuxbashshellcommand-linecommand

Print 3, 4 and 5 line of the file in Linux

I have a problem with a Linux command. I have to put on the output 3, 4 and 5 line of the file /etc/passwd, but I have no idea how to do it. I can print first five line: 

head -n 5 /etc/passwd

but I don't know how to remove first two lines or do all from scratch.

Solution

  • Using sed:

    sed -n '3,5p' /etc/passwd

    or

    sed -n '3,5p;6q' /etc/passwd

    (The second version would quit upon encountering line 6 so it would be efficient for huge files.)

    Using awk:

    awk 'NR==3,NR==5' /etc/passwd

    or

    awk 'NR>=3{print}NR==5{exit}' /etc/passwd

    (The second variant quit after printing line 5 so it's more efficient.)

    Using perl:

    perl -ne 'print if $.>=3 and $.<=5;' /etc/passwd

    or

    perl -ne 'print if $.>=3; last if $.>5' /etc/passwd

    (The second variant is, again, more efficient.)

    For fun, lets time these different approaches on an input of 10 million lines:

    $ time seq 10000000 | sed -n '3,5p'
3
4
5

real    0m10.086s
user    0m9.173s
sys     0m0.101s

$ time seq 10000000 | sed -n '3,5p;6q'
3
4
5

real    0m0.012s
user    0m0.010s
sys     0m0.001s

$ time seq 10000000 | awk 'NR==3,NR==5'
3
4
5

real    0m12.906s
user    0m11.475s
sys     0m0.134s

$ time seq 10000000 | awk 'NR>=3{print}NR==5{exit}'
3
4
5

real    0m0.013s
user    0m0.001s
sys     0m0.010s

$ time seq 10000000 | perl -ne 'print if $.>=3 and $.<=5;'
3
4
5

real    0m15.982s
user    0m14.217s
sys     0m0.179s

$ time seq 10000000 | perl -ne 'print if $.>=3; last if $.>5'
3
4
5
6

real    0m0.013s
user    0m0.000s
sys     0m0.011s

    It's evident that quitting in case of large inputs once the desired lines are obtained is much more efficient. On small inputs, the difference would be negligible, though.