substitution using sed or awk

[yao_licr]

Hi all,
I just start to use sed and awk. I would like to change the following fastq file

@SRR391606.9.1 1 length=50
AGCACCGCGAGGGCGGAGCTGCGTTCTCCTCTGCACAGCTTTCGGTGGTA
+SRR391606.9.1 1 length=50
@9@?<<AA:3>7A/;@?B57@6;++4=4<,=6=+++17/)1?%%%%%%%%
@SRR391606.10.1 2 length=50
GTTGCGTTCTCCTCAGCACAGACCCGGAGAGCACCGCGAGGGCGGAGCTG
+SRR391606.10.1 2 length=50
?BB?BA?AABAB>->ABB=99<AAB++35)137<>:37=<735(2-9492

into following fastq file:

@SRR391606.9 1 length=50
AGCACCGCGAGGGCGGAGCTGCGTTCTCCTCTGCACAGCTTTCGGTGGTA
+SRR391606.9 1 length=50
@9@?<<AA:3>7A/;@?B57@6;++4=4<,=6=+++17/)1?%%%%%%%%
@SRR391606.10 2 length=50
GTTGCGTTCTCCTCAGCACAGACCCGGAGAGCACCGCGAGGGCGGAGCTG
+SRR391606.10 2 length=50
?BB?BA?AABAB>->ABB=99<AAB++35)137<>:37=<735(2-9492

Basically, it is a task to remove the ".1" on every other line. As I have a huge fastq file around 30 Gb, I only put the 9th and 10th reads out of it.

Does anyone know how to do it using sed or awk? Please also explain the script in detail as I am a beginner.

Thanks a lot!

Yao

[dschika]

This command should do the job:

awk '{if ($1 ~ /SRR/) {split($0, T, ".1 "); print T[1], T[2]}  else print $0}' YOURINPUT.fastq

Since you mentioned sed and awk I assume you know that $1 = first field, $2 = second field (fields separated by whitespaces if not defined otherwise by FS) and $0 = whole line.

The command checks if in the first field SRR is present ("if ($1 ~ /SRR/)"). If yes, it splits the content of the first line by ".1 " and stores the result in T. In this case ".1 " can be found only once in each line with SRR. Therefore, printing of T[0] and T[1] results in the line without ".1 ".

If the line does not contain SRR (i.e., we are at a line with sequence or quality values) print the whole line.

[yao_licr]

Thanks for the detailed explanation. It works well.

[dschika]

No problem, I just realized, that I was thinking way to complicated:

sed 's/\.1 / /g' YOURINPUT.fastq > out.fastq

Just replace ".1 " with " " and escape the "." with a backslash.

[yao_licr]

Sorry, ".1" is the pattern to be replace by " ", so why do you put an empty space after 1? I mean the empty space in "/\.1 /" .

Thanks!

[yao_licr]

I got it now, as I have .1.1 in the first read; if there is no empty space after "\.1", both of them will be replaced. In your script, ".1 " is the pattern but not ".1" .

Thanks!

[dschika]

Correct, you're welcome!