Print first, second, every nth, and last record

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# 1  
Old 09-02-2011
Print first, second, every nth, and last record

does anyone have an awk one-liner to:

print the first line, the second line, then every Nth line, and the last line of a file.

# 2  
Old 09-02-2011
I think this will work. Set n= to the desired value (9th line in the example):

    awk -v n=9 ' NR < 3  ||  !(NR % n)  { print; }'

You can always hard code the variable (doing it this way allows inclusion in a script with flexibility. You can also make it much more cryptic:

awk 'NR < 3 || !(NR % 9)'

# 3  
Old 09-03-2011
The following builds on agama's suggestion so that it handles the last line:
awk '{p=0; r=$0} NR<3 || !(NR%n) {print r; p=1} END {if (!p) print r}' n=9

p: flag to track whether a record has been printed.
r: most recent record

This User Gave Thanks to alister For This Post:
# 4  
Old 09-03-2011
Originally Posted by alister
The following builds on agama's suggestion so that it handles the last line:

Very embarrassing -- I completely missed that requirement!! Thanks.
# 5  
Old 09-03-2011
Nah. Embarassing is what happened to me a couple days ago. I overlooked a requirement and incorrectly critiqued two other suggestions in the thread. Fortunately (I think), I was able to edit/delete my post before it was seen. Smilie
# 6  
Old 09-03-2011
But this doesn't print the last line. With GNU sed:
sed -n '${p;d};1,2{p;d};9~9p'

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