Sed command to print out every 1st and 14th line of a file every 46 lines [duplicate]

Question

So parse a text file and print out line 1 and line 14, and then do nothing with lines 15-46, and then print out line 47 and line 60, etc until the end of the file. So basically every 46 lines, print out the 1st and 14th line, repeatedly for every 46 lines until EOF

Answer 1

Using GNU sed's first~step address range extension:

sed -n '1~46p; 14~46p;' file

Answer 2

Since you have awk in your tags, I will provide a solution with awk:

awk '(NR%46==1||NR%46==14){print}' file

Answer 3

No job for POSIX sed:

sed 'H;1h;$!d;x;y/\n#/#\n/;s/\(#[^#]*\)\{12\}#\([^#]*\)\([^#]*#\)\{33\}/#\2##/g;s/\(.*\)##.*/\1/;s/##/#/g;y/\n#/#\n/'

Although I consider it to be nonsense using plain POSIX sed for such a task when other tools are better suited, I'll add an explanation, because there are useful elements here you may need in actual tasks:

• H;1h;$!d;x is a pattern to collect the whole file in the pattern space, which is often useful (with GNU sed, you can replace it by using the -z option). You can figure out how it works.
• y/\n#/#\n/ exchanges the newlines with another char (in this case #). Do this as a workaround before and after your processing if you need expressions like "every char except newline". Again, you don't need it with GNU sed, as [^\n] is allowed there.
• That weird pattern in the s command matches 12+1+33=46 # (formerly newlines) and 45 [^#]*, which are the line contents. The every first line is untouched, 12 are removed, 1 preserved as \2 and 33 removed. This is globally done. The ## stuff is done to remove trailing lines.