strelka variant calling

Question

[fra.r.silvestro]

Conda installed strelka. Then I've created a configuration file.

configureStrelkaGermlineWorkflow.py \
    --bam 03.align/$FQ.align.sort.marked.bam \
    --referenceFasta  $REF \
    --runDir 04.varcalling \
    --exome \
    --callRegions Homo_sapiens_assembly38-3.sorted.bed.gz

After executing:

/mountpoint/fastQ/04.varcalling/runWorkflow.py -m local -j 2
/mountpoint/fastQ/04.varcalling/runWorkflow.py: PERMISSION denied.

Do you know how to resolve?

Answer 1

[DBScan]

Make sure the file exists and give it permissions.

[fra.r.silvestro]

I have the run script file

Indeed, I get this error:

configureStrelkaGermlineWorkflow.py: error: Run directory already contains workflow script file '/mountpoint/fastQ/04.varcalling/runWorkflow.py'. Each analysis must be configured in a separate directory.

Then mountpoint/fastQ/04.varcalling/runWorkflow.py and the appropriate flags, but it displays: PERMISSION DENIED.

[fra.r.silvestro]

There is a shebang in my script:

#!/usr/bin/env python

Then:

chmod +x myscript.py    

But it doens't change permission (executable).

I'm using a Windows machine and mount a Linux Virtual machine. Hence, my script is in a directory where I have my script

[GenoMax]

Please do not add additional information as answers. You should use ADD REPLY/ADD COMMENT for that purpose.

Is your script called myscript.py or runWorkflow.py?

[fra.r.silvestro]

chmod +x runWorkflow.py

Strelka variant caller doesn't work on Windows OS, but I'm using an Ubuntu VM, hence I don't think this is the issue

By looking at file Properties, it display for system and user: Read and execute= ALLOW