Sed. How to remove line match with pattern and strings arround it?

Question

I have a file where you want to delete line matching by pattern and remove strings above and below.

By example:

FFFFIFIBBFFFFFFFFFFFFFBBBBFBBBBFBBBB77<<BBBBBB7B<BBBBBB<B< @HISEQ:102:h9u5badxx:1:1101:13002:2147 1:N:0:CTGT GATCCCCGTCTATCAGATACACGTTACTCAGCTAGTGCGAATGCGAACGCGAAATTTT + FFFFFFFFBBFFFFFFFFFFFFFBFBFFFFFFFFFBFFFBFFFFFBFFFFFFFFFBFB @HISEQ:102:h9u5badxx:1:1101:15368:2194 1:N:0:CTGT + FFIFBFFIFFBBBFFFFFFFBBFFBFFBBBFFFBB7BBBBBBFFFBB700<7770<BBB0<0<BFFBFBFFFFF @HISEQ:102:h9u5badxx:1:1101:19167:2169 1:N:0:CTGT GATCTCATATAGGGCAGCGTGGTCGCGGC

I want to remove second block which does not contain the nucleotide sequence.

The end result:

`FFFFIFIBBFFFFFFFFFFFFFBBBBFBBBBFBBBB77<<BBBBBB7B<BBBBBB<B<
@HISEQ:102:h9u5badxx:1:1101:13002:2147 1:N:0:CTGT
GATCCCCGTCTATCAGATACACGTTACTCAGCTAGTGCGAATGCGAACGCGAAATTTT
+
FFIFBFFIFFBBBFFFFFFFBBFFBFFBBBFFFBB7BBBBBBFFFBB700<7770<BBB0<0<BFFBFBFFFFF
@HISEQ:102:h9u5badxx:1:1101:19167:2169 1:N:0:CTGT
GATCTCATATAGGGCAGCGTGGTCGCGGC
`

Pattern which matched this block

'^.+$(\n)^(@HISEQ).*$(\n)^\+'

works in perl and javascript, but not sed.

Because sed does not work with line break.

I found the solution

sed -e ':a;N;$!ba;s/\n/ /' test

But this code replace line break to space. If insert to this code my regexp:

sed -e ':a;N;$!ba;/^.+$(\n)^(@HISEQ).*$(\n)^\+/d' test

this does not work. Can you help me find the solution of this problem?

---

I'm just stupid. I misunderstood the file format. Input:

@HWI-ST383:199:D1L73ACXX:3:1101:1309:1956 1:N:0:ACAGTGA 
+ 
JJJHIIJFIJJJJ=BFFFFFEEEEEEDDDDDDDDDDBD 
@HWI-ST383:199:D1L73ACXX:3:1101:3437:1952 1:N:0:ACAGTGA
GATCTCGAAGCAAGAGTACGACGAGTCGGGCCCCTCCA 
+ 
IIIIFFF<?6?FAFEC@=C@1AE############### 

How to edit the regular exp to get what you want

output:

@HWI-ST383:199:D1L73ACXX:3:1101:3437:1952 1:N:0:ACAGTGA
GATCTCGAAGCAAGAGTACGACGAGTCGGGCCCCTCCA 
+ 
IIIIFFF<?6?FAFEC@=C@1AE###############

Answer 1

If I understand you correctly, then

sed ':loop; N; /\n+/ ! { $ ! b loop }; /\n@HISEQ[^\n]\+\n+/ d' foo.txt

will work. This is as follows:

:loop                    # in a loop
N                        # fetch more lines
/\n+/ ! { $ ! b loop }   # until one starts with + or is the last line
/\n@HISEQ[^\n]\+\n+/ d   # if the penultimate line of all that begins with @HISEQ,
                         # discard the lot.

That last pattern is using the fact that it is checked right after the first line that begins with + is found, so the \n+ at the end of it uniquely matches the start of the last line in the block.

Answer 2

To remove the second block, you can just do:

awk 'NR!=2' RS=+ ORS=+ input

But I would suspect you want something more like:

awk '/[GATC]{5,}\n/' RS=+ ORS=+ input

or

awk '/\n[GATC]*\n/' RS=+ ORS=+ input

Answer 3

If I understand you correctly, then

sed ':loop; N; /\n+/ ! { $ ! b loop }; /\n@HISEQ[^\n]\+\n+/ d' foo.txt

will work. This is as follows:

:loop                    # in a loop
N                        # fetch more lines
/\n+/ ! { $ ! b loop }   # until one starts with + or is the last line
/\n@HISEQ[^\n]\+\n+/ d   # if the penultimate line of all that begins with @HISEQ,
                         # discard the lot.

That last pattern is using the fact that it is checked right after the first line that begins with + is found, so the \n+ at the end of it uniquely matches the start of the last line in the block.