How to select certain lines (n, n+4, n+8, n+12...) from the file?

Question

Input:

1
hgh
h2b
h4h
2
ok
koko
lkopk
3
uh
ju
nfjvn
4

Expected output:

1
2
3
4

So, I need to have only 1st, 5th, 9th, 13th value of the file in the output file. How to do this?

Answer 1

Using AWK:

awk '!((NR - 1) % 4)' input > output

Figuring out how this works is left as an exercise for the reader.

Answer 2

Using split (GNU coreutils):

split -nr/1/4 input > output

• -n generate CHUNKS output files

and CHUNKS as

• r/K/N use round robin distribution and only output Kth of N to stdout without splitting lines/records

Answer 3

With GNU sed:

sed '1~4!d' < input > output

With standard sed:

sed -n 'p;n;n;n' < input > output

With 1 and 4 in $n and $i variables:

sed "$n~$i!d" # GNU only

awk -v n="$n" -v i="$i" 'NR >= n && (NR % i) == (n % i)'

Answer 4

Adding the obligatory perl solution:

perl -ne 'print if $. % 4 == 1' input > output

Answer 5

Python version, just for fun:

with open('input.txt') as f:
    for i, line in enumerate(f.readlines()):
        if i%4 == 0:
            print(line.strip())

Answer 6

POSIX sed: this method uses the posixly sed and so can be run everywhere, or atleast those seds that respect posix.

 $ sed -ne '
   /\n/!{
    H;s/.*//;x
   }

   :loop
       $bdone
       N;s/\n/&/4
       tdone
   bloop

   :done
   s/.//;P
 ' input.file

Another is a programmatic sed code generation for scalability purposes:

$ code=$(yes n | head -n 4 | paste -sd\; | sed s/n/p/)
$ sed -ne "$code" input.file

Perl: we fill-up array A till it is 4 in size. Then we print its first element and also clear out the array.

$ perl -pe '
   $A[@A] = @A ? <> : $_ while @A < 4;
   $_ = (splice @A)[0];
' input.file

Answer 7

Call with scriptname filename skip (4 in your case) It works by pulling iter lines from the top of the file and then only outputting the last. It then increments iter by skips and repeats as long as the value of iter hasn't exceeded the lines in file.

#!/bin/bash
file="$1"
lines=`wc -l < "$file"`
skips="$2" || "4"
iter=1
while [ "$iter" -le "$lines" ]; do
 head "$file" -n $iter | tail -n 1
 iter=$(( $iter + $skips ))
done

Answer 8

Pure Bash:

mapfile -t lines < input
for (( i=0; i < ${#lines[@]}; i+=4 ))
do printf "%s\n" "${lines[$i]}"
done

mapfile is a builtin added in Bash 4 which reads standard input into an array, here named lines, with one line per entry. The -t option strips the final newlines.

If you want to print every fourth line starting from line 4, then you can do that in one command using mapfile's callback option -C, which runs the provided code every so many lines, with the interval given by -c. The current array index and the next line to be assigned are given to the code as arguments.

mapfile -t -c4 -C 'printf "%.0s%s\n"' < input

This uses the printf builtin; the format code %.0s suppresses the first argument (the index), so only the line is printed.

You could use the same command to print every fourth line starting from line 1, 2, or 3, but you'd have to prepend 3, 2, or 1 lines to input before feeding it to mapfile, which I think is more trouble than it's worth.

This also works:

mapfile -t lines < input
printf "%s%.0s%.0s%.0s\n" "${lines[@]}"

Here, printf consumes four entries of the array lines at a time, only printing the first and skipping the other three with %.0s. I don't like this since you have to manually fiddle with the format string for different intervals or starting points.