Extract n-th line from file in bash loop - linux

Question

I would like to extract n-th line from file and save it to a new file. For example I have index.txt :
cat index.txt
1 AAAGCGT
2 ACGAAGT
3 ACCTTGT
4 ATAATGT
5 AGGGTGT
6 AGCCAGT
7 AGTTCGT
8 AATGCAG
9 AAAGCGT
10 ACGAAGT
and output should be
cat index.1.txt:
1 AAAGCGT
2 ACGAAGT
cat index.2.txt:
3 ACCTTGT
4 ATAATGT
cat index.3.txt:
5 AGGGTGT
6 AGCCAGT
And so on.. So I would like to extract form input file first 2 rows in cycle and save to new file.

Answer 1

It doesn't give you exactly the names you want, but:
split -l 2 index.txt index.
seems like the easiest solution. It will create files with names beginning with the final argument, so will get names like 'index.aa' and 'index.bb'

Answer 2

This will work for any number of grouped lines just by changing the 2 to a 3 or whatever number you like:
$ awk 'NR%2==1{++i} {print > ("index." i ".txt")}' index.txt
$ ls index.?.txt
index.1.txt index.2.txt index.3.txt index.4.txt index.5.txt
$ tail index.?.txt
==> index.1.txt <==
1 AAAGCGT
2 ACGAAGT
==> index.2.txt <==
3 ACCTTGT
4 ATAATGT
==> index.3.txt <==
5 AGGGTGT
6 AGCCAGT
==> index.4.txt <==
7 AGTTCGT
8 AATGCAG
==> index.5.txt <==
9 AAAGCGT
10 ACGAAGT

Answer 3

awk '{print >"index."(x+=NR%2)".txt"}' file
This increments x every two lines starting from 1 and then prints the line into a file with that name
cat index.1.txt:
1 AAAGCGT
2 ACGAAGT
cat index.2.txt:
3 ACCTTGT
4 ATAATGT
cat index.3.txt:
5 AGGGTGT
6 AGCCAGT
In some awks, extra parens may be required as shown below (As commented by Ed Morton)
awk '{print >("index."(x+=NR%2)".txt")}' file

Answer 4

I would say:
awk '{file=int((NR+1)/2)".txt"; print > file}' file
int((NR+1)/2 maps every line number:
1 --> 1
2 --> 1
3 --> 2
x --> (x+1) / 2
So you get these files:
$ cat 1.txt
1 AAAGCGT
2 ACGAAGT
or
$ cat 3.txt
5 AGGGTGT
6 AGCCAGT