Menu
I know how to get a range of lines by using awk and sed. I also do know how to print out every nth line using awk and sed.
However, I don't know how to combined the two.
For example, I have a file with 1780000 lines.
For every 17800th line, I would like to print 17800th line plus the two after that.
So if I have a file with 1780000 lines and it starts from 1 and ends at 1780000, this will print:
1
2
3
17800
17801
17802
35600
35601
35602
# ... and so on.
Does anyone know how to get a range of line every nth interval using awk, sed, or other unix command?
321
Generally I would say grep is the fastest one, sed is the slowest. Of course this depends on what are you doing exactly. I find awk much faster than sed . You can speed up grep if you don't need real regular expressions but only simple fixed strings (option -F).
To print a blank line, use print "" , where "" is the empty string. To print a fixed piece of text, use a string constant, such as "Don't Panic" , as one item. If you forget to use the double-quote characters, your text is taken as an awk expression, and you will probably get an error.
Using GNU sed:
sed -n '0~17800{N;N;p}' input
Meaning,
For every 17800th line: 0~17800
Read two lines: {N;N;
And print these out: p}
We can also add the first three lines:
sed -n -e '1,3p' -e '0~17800{N;N;p}' input
Using Awk, this would be simpler:
awk 'NR%17800<3 || NR==3 {print}' input
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With
