running nanofilt with snakemake - bioinformatics

Question

I am new to using snakemake. I want to run my fastq files that are in one folder with nanofilt. I want to run this with snakemake because I need it to create a pipeline. This is my snake make script:
rule NanoFilt:
input:
"data/samples/{sample}.fastq"
output:
"nanofilt_out.gz"
shell:
"gunzip -c {input} | NanoFilt -q 8 | gzip > {output}"
When I run it I get the following error message:
WildcardError in line 2:
Wildcards in input files cannot be determined from output files:
'sample'
EDIT
I tried searching the error message but still couldnt figure out how to make it work. Can anyone help me?
So I tried what people on here told me so the new script is this:
samples = ['fastqrunid4d89b52e7b9734bd797205037ef201a30be415c8293','fastqrunid4d89b52e7b9734bd797205037ef201a30be415c8292','fastqrunid4d89b52e7b9734bd797205037ef201a30be415c8291','fastqrunid4d89b52e7b9734bd797205037ef201a30be415c8290']
rule all:
input:
[f"nanofilt_out_{sample}.gz" for sample in samples]
rule NanoFilt:
input:
"zipped/zipped{sample}.gz"
output:
"nanofilt_out_{sample}.gz"
shell:
"gunzip -c {input} | NanoFilt -q 8 | gzip > {output}"
but when I run this I get the following error message:
Error in rule NanoFilt:
Removing output files of failed job NanoFilt since they might be corrupted:
nanofilt_out_fastqrunid4d89b52e7b9734bd797205037ef201a30be415c8292.gz
jobid: 4
output: nanofilt_out_fastqrunid4d89b52e7b9734bd797205037ef201a30be415c8290.gz
shell:
gunzip -c zipped/zippedfastqrunid4d89b52e7b9734bd797205037ef201a30be415c8290.gz | NanoFilt -q 8 | gzip > nanofilt_out_fastqrunid4d89b52e7b9734bd797205037ef201a30be415c8290.gz
(one of the commands exited with non-zero exit code; note that snakemake uses bash strict mode!)
Does anyone know how to fix this?

Answer 1

The 'idea' of SnakeMake is that you specify what output you want (through for instance rule all), and that SnakeMake takes a look at all the rules defined and knows how to get to the desired output.
When you tell SnakeMake you want as output nanofilt_out.gz, how does it know what sample to take? It doesn't.. if it would just take any of the possible sample files, then we would lose the information about which file it belongs to. To solve this we need in the output also the same wildcard as in the input:
rule NanoFilt:
input:
"data/samples/{sample}.fastq"
output:
"nanofilt_out_{sample}.gz"
shell:
"gunzip -c {input} | NanoFilt -q 8 | gzip > {output}"
This way SnakeMake can make an output for every sample. You do somehow need to adjust the pipeline that you specify which output you want, maybe something like this:
samples = [1,2,3]
rule all:
input:
[f"nanofilt_out_{sample}.gz" for sample in samples]

Answer 2

I made it work. The working code looks like this.
rule NanoFilt:
input:
expand("zipped/zipped.gz", sample=samples)
output:
"filtered/nanofilt_out.gz"
conda:
"envs/nanoFilt.yaml"
shell:
"gunzip -c {input} | NanoFilt -q 6 | gzip > {output}"