具有不同文件名的多个目录上的glob_wildcards

时间:2018-05-18 14:19:00

标签: snakemake

我正在尝试编写一个规则,它从不同的目录中获取两个文件,并将规则的输出放在与下面文件结构相同的目录中:

DIR_A 
    dir1    
        file1.clean.vcf  
    dir2  
        file2.clean.vcf  
    dir3
        file1.output.vcf
        file2.output.vcf

到目前为止,我已尝试使用glob_wildcards:

(DIR,NAME) = glob_wildcards("DIR_A/{dir}/{name}.clean.vcf") 
input: expand("DIR_A/{dir}/{name}.clean.vcf", dir=DIR, name=NAME)
output: "DIR_A/dir3/{name}.output.vcf

但它会引发错误:

MissingInputException in line 80 of DIR_A:
Missing input files for rule convert_output:

DIR_A/dir1/file2.clean.vcf
DIR_A/dir2/file1.clean.vcf

将zip添加到输入:

input: expand("DIR_A/{dir}/{name}.clean.vcf", zip, dir=DIR, name=NAME)

如果$ snakemake -s snakefile -n(干运行):

rule conv_output:
input: DIR_A/dir1/file1.clean.vcf, DIR_A/file2/file2.clean.vcf

这可以被snakemake接受并防止出现上述错误,但现在file1.clean.vcf和file2.clean.vcf都是规则的输入,但{name}通配符使规则每个文件运行一次。这最终成为一个文件到一个文件,而不是我正在寻找的一对一文件。

有没有办法设置它,所以我可以得到规则conv_output的输出作用于每个文件然后将输出放在dir3?任何帮助将不胜感激!!

1 个答案:

答案 0 :(得分:2)

使用python,将输入vcf sample / filename配对到其路径,然后使用它在snakemake规则中指定输入路径。下面的示例适用于问题中给出的目录结构。

from pathlib import Path   

def pair_name_to_infiles():
    # get all *.clean.vcf files recursively under DIR_A
    vcf_path = Path('DIR_A').glob('**/*.clean.vcf')

    # pair vcf name to infile path using a dictionary
    vcf_infiles_dict = {}
    for f in vcf_path:
        vcf_name = f.name.replace('.clean.vcf', '')
        vcf_infiles_dict[vcf_name] = str(f)

    return vcf_infiles_dict


# using function written in python code, map vcf name to their infile path
vcf_infiles_dict = pair_name_to_infiles()


rule all:
    input:
        expand('DIR_A/dir3/{vcf_name}.output.vcf', vcf_name=vcf_infiles_dict.keys())


rule foo:
    input:
        lambda wildcards: vcf_infiles_dict[wildcards.vcf_name]
    output:
        'DIR_A/dir3/{vcf_name}.output.vcf'
    shell:
        'touch {output}'