Sed error

Sed error

[Aki Hoji]

HI, 

I get the error message (sed: -e expression #1, char 2: invalid usage of line address 0) when I run a following;


parallel -a temp3.fq  -k --block 50M --pipe-part 'sed -e  "$!N;/\nXQ@A00165/!P;D"' > temp4.fq

When I run this one liner in bash, it works just fine but under zsh. Apprently, zsh does not like “!”.   Isn’t  “!” in the script not compatible with zsh and is there any work around for this ? 

I am using my institution’s cluster;

3.10.0-1160.71.1.el7.x86_64
 x86_64 x86_64 x86_64 GNU/Linu

Thanks. 

Re: Sed error

[Bart Schaefer]

On Tue, Aug 2, 2022 at 5:12 PM Aki Hoji <akhst7@gmail.com> wrote:
>
> I get the error message (sed: -e expression #1, char 2: invalid usage of line address 0) when I run a following;
>
> parallel -a temp3.fq  -k --block 50M --pipe-part 'sed -e  "$!N;/\nXQ@A00165/!P;D"' > temp4.fq

$! is the PID of the last background job.  That's being expanded to 0
because the argument to -e is in double quotes, resulting in an
expression starting with "0N;"

In bash, $! expands to the empty string when there are no background
jobs, rather than to zero.  In that case the expression starts with
"N;" which is also probably not what you meant, but isn't a sed error.

You need to quote the expression so that the $! is not expanded.  This
might work:

EXPR='$!N;/\nXQ@A00165/!P;D' parallel -a temp3.fq  -k --block 50M
--pipe-part 'sed -e  "$EXPR"'

Re: Sed error

[Bart Schaefer]

On Tue, Aug 2, 2022 at 5:40 PM Bart Schaefer <schaefer@brasslantern.com> wrote:
>
> This might work:

After removing gmail's annoying line wrapping, that is.

> EXPR='$!N;/\nXQ@A00165/!P;D' parallel -a temp3.fq  -k --block 50M
> --pipe-part 'sed -e  "$EXPR"'