使用 Snakemake 的分组样本每组单个输出文件

Question

我正在尝试编写一个 snakemake 管道来处理分成多个组的样本，每个组需要一个摘要文件。例如样本划分如下：

Group 1:
    Sample 1
    Sample 2
Group 2: 
    Sample 3 
    Sample 4

每个样本都使用 bedtools 进行处理，为每个样本生成一个输出文件。接下来，我需要对每个Group收集的样本进行分组层面的总结。缩写的 snakemake 文件如下所示：

 rule intersect:
     input:
         bam = join('{group}','{sample}.bam'),
         reg_bed = join('{group}', 'region.bed')
     output:
         reg_intersect = join('{group}', '{sample}.intersect.bed')
    shell:
        'bedtools intersect -wa -wb -a {input.reg_bed} -b {input.bam} > {output.reg_intersect}'

rule aggregate:
    input:
        rules.interesect.output
    output:
        join('{group}','summary_stats.csv')
    shell:
        touch(join('{group}','summary_stats.csv'))
        #this would be a call to a python function that operates on all the output files to generate a single file

我收到关于通配符在输入和输出之间不一致的投诉（输入包含 {group} 和 {sample} 而输出只有 {group}。我试过使用 expand() 但我必须包括两者的值组和样本，样本是不可能的，因为它依赖于组。

欢迎提出任何建议。

Answer 1

正如@Maarten-vd-Sande 所说，最好的方法是使用输入函数获取 aggregate 规则中某个组的所有样本。

samplesInGroups = {"group1":["sample1","sample2"],"group2":["sample3","sample4"]}

def getSamplesInGroup(wildcards):
    return [join(wildcards.group,s+".intersect.bed") for s in samplesInGroups[wildcards.group]]

rule all:
     input:  expand("{group}/summary_stats.csv",group=list(samplesInGroups.keys()))

rule intersect:
     input:
         bam = join('{group}','{sample}.bam'),
         reg_bed = join('{group}', 'region.bed')
     output:
         reg_intersect = join('{group}', '{sample}.intersect.bed')
     shell:
        'bedtools intersect -wa -wb -a {input.reg_bed} -b {input.bam} > {output.reg_intersect}'

rule aggregate:
     input:
        getSamplesInGroup
     output:
        join('{group}','summary_stats.csv')
     shell:
        "python ... {input}"

使用 Snakemake 的分组样本每组单个输出文件

Single output file per group for grouped samples using Snakemake

python

wildcard

snakemake