Delete lines from a text file except the first and every nth - bash

Question

I have a long text file comprised of numbers, such as:
1
2
9.252
9.252
9.272
1
1
6.11
6.11
6.129
I would like to keep the first line, delete the subsequent three and then keep the next one. I would like to do this process for the whole file. Following that logic, considered the input above, I would like to have the following output:
1
9.272
1
6.129

Answer 1

Using GNU sed (needed for the ~ extension):
sed -n '1~5p;5~5p' file

Answer 2

Saving your numbers in a "textfile.txt" I can use the following with sed:
sed -n 'p;n;n;n;n;p;' textfile.txt
Sed prints the first line, reads the next 4 and prints the last line.
Or the following using while read in bash:
while read -r firstline && read -r nextone1 && read -r nextone2 && read -r nextone3 && read -r lastone; do
printf "%s\n" "$firstline" "$lastone";
done < textfile.txt
This just reads 5 lines at a time and prints only the first and 5th lines.

Answer 3

You can simply say:
awk 'NR%5<2' input.txt
Explanation: Considering the entire pattern repeats every five lines, let's start with applying modulo operation to the line number NR by five. Then we'll see the 1st line of the five-line block yields "1" and the 5th line of the block yields "0". Now they can be separated from other lines by comparing it to two.

Answer 4

To print the 1st and 5th line of every block of 5 lines (remember that 5%5 = 0):
$ awk '(NR%5) ~ /[10]/' file
1
9.272
1
6.129
If you want to print the 2nd, 3rd, and 4th line of every block of 5 lines instead of the 1st and 5th:
$ awk '(NR%5) ~ /[234]/' file
2
9.252
9.252
1
6.11
6.11
If you wanted to print the 27th and 53rd line of every block of 100:
awk '(NR%100) ~ /^(27|53)$/' file
We couldn't use a bracket expression there as we're now beyond single char numbers.

Answer 5

This might work for you (GNU sed):
sed '2~5,+2d' file
Starting from line 2, delete the next three lines using modulo 5.
An alternative:
sed -n '1p;5~5,+1p' file

Answer 6

Considering your groups are packed as 5 lines, you could use awk with a mod 5 operation.
awk '{i=(NR-1)%5;if(i==0||i==4)print $0}' input.txt
With indentation it looks like this:
{
i=(NR-1)%5;
if (i==0||i==4)
print $0;
}
i=(NR-1)%5 gets the line number and computes the modulo with 5, but since the line numbers start at 1 (instead of 0), you need to subtract 1 to it before computing the modulo.
This leaves you with an integer i that ranges from 0 to 4. You want to print the first line (index 0), skip the next three lines (indexes 1-3) and print the last line (index 4), which is exactly what does if (i==0||i==4) print $0
Alternately you can do the same thing with a shorter (and probably slightly more optimized version):
awk '((NR-1)%5==0||(NR-1)%5==4)' input.txt
This tells awk to do something for every 1st out of 5 lines and every 5th out of 5 lines. Since the "something" is not defined, by default it outputs the current line. If it helps, this is strictly equivalent to:
awk '((NR-1)%5==0||(NR-1)%5==4){print $0}' input.txt