Nextflow 名称冲突

Question

我有名称相同但位于不同文件夹中的文件。Nextflow 将这些文件暂存在同一工作目录中，从而导致名称冲突。我的问题是如何在不重命名文件的情况下处理这个问题。例子：

# Example data
mkdir folder1 folder2
echo 1 > folder1/file.txt
echo 2 > folder2/file.txt

# We read from samplesheet
$ cat samplesheet.csv
sample,file
sample1,/home/atpoint/foo/folder1/file.txt
sample1,/home/atpoint/foo/folder2/file.txt

# Nextflow main.nf
#! /usr/bin/env nextflow

nextflow.enable.dsl=2

// Read samplesheet and group files by sample (first column)
samplesheet = Channel
    .fromPath(params.samplesheet)
    .splitCsv(header:true)
    .map {
            sample = it['sample']
            file   = it['file']
            tuple(sample, file)
}
        
ch_samplesheet = samplesheet.groupTuple(by:0)

// That creates a tuple like:
// [sample1, [/home/atpoint/foo/folder1/file.txt, /home/atpoint/foo/folder2/file.txt]]

// Dummy process that stages both files into the same work directory folder
process PRO {

    input:
    tuple val(samplename), path(files)

    output:
    path("out.txt")

    script:
    """
    echo $samplename with files $files > out.txt
    """

}

workflow { PRO(ch_samplesheet) }

# Run it
NXF_VER=21.10.6 nextflow run main.nf --samplesheet $(realpath samplesheet.csv)

...显然导致：

N E X T F L O W  ~  version 21.10.6
Launching `main.nf` [adoring_jennings] - revision: 87f26fa90b
[-        ] process > PRO -
Error executing process > 'PRO (1)'

Caused by:
  Process `PRO` input file name collision -- There are multiple input files for each of the following file names: file.txt

所以现在怎么办？这里的实际应用程序是对同一 fastq 文件的副本进行排序，这些文件具有相同的名称，但位于不同的文件夹中，我想将它们输入到合并它们的进程中。我知道文档中的这一部分，但不能说其中有任何帮助或我正确理解了它。

Answer 1

stageAs您可以在流程定义中使用选项。

#! /usr/bin/env nextflow
nextflow.enable.dsl=2

samplesheet = Channel
    .fromPath(params.samplesheet)
    .splitCsv(header:true)
    .map {
        sample = it['sample']
        file = it['file']
        tuple(sample, file)
     }
    .groupTuple()
    .set { ch_samplesheet }

// [sample1, [/path/to/folder1/file.txt, /path/to/folder2/file.txt]]

process PRO {
    input:
        tuple val(samplename), path(files, stageAs: "?/*")

    output:
        path("out.txt")

    shell:
        def input_str = files instanceof List ? files.join(" ") : files
        """
        cat ${input_str} > out.txt
        """
}

workflow { PRO(ch_samplesheet) }

请参阅nf-core和输入类型文档中的示例path