How do I extract lines from a file using their line number on unix?

Question

Using sed or similar how would you extract lines from a file? If I wanted lines 1, 5, 1010, 20503 from a file, how would I get these 4 lines?

What if I have a fairly large number of lines I need to extract? If I had a file with 100 lines, each representing a line number that I wanted to extract from another file, how would I do that?

Answer 1

Something like "sed -n '1p;5p;1010p;20503p'. Execute the command "man sed" for details.

For your second question, I'd transform the input file into a bunch of sed(1) commands to print the lines I wanted.

Answer 2

with awk it's as simple as:

awk 'NR==1 || NR==5 || NR==1010' "file"

Answer 3

@OP, you can do this easier and more efficiently with awk. so for your first question

awk 'NR~/^(1|2|5|1010)$/{print}' file

for 2nd question

awk 'FNR==NR{a[$1];next}(FNR in a){print}' file_with_linenr file

Answer 4

I'd investigate Perl, since it has the regexp facilities of sed plus the programming model surrounding it to allow you to read a file line by line, count the lines and extract according to what you want (including from a file of line numbers).

my $row = 1
while (<STDIN>) {
   # capture the line in $_ and check $row against a suitable list.
   $row++;
}

Answer 5

This ain't pretty and it could exceed command length limits under some circumstances^*:

sed -n "$(while read a; do echo "${a}p;"; done < line_num_file)" data_file

Or its much slower but more attractive, and possibly more well-behaved, sibling:

while read a; do echo "${a}p;"; done < line_num_file | xargs -I{} sed -n \{\} data_file

A variation:

xargs -a line_num_file -I{} sed -n \{\}p\; data_file

You can speed up the xarg versions a little bit by adding the -P option with some large argument like, say, 83 or maybe 419 or even 1177, but 10 seems as good as any.

_{*xargs --show-limits </dev/null can be instructive}

Answer 6

In Perl:

perl -ne 'print if $. =~ m/^(1|5|1010|20503)$/' file