Snakemake 中的互连变量

Question

假设我有示例 SAMPLE_A，分为两个文件 SAMPLE_A_1、SAMPLE_A_2 并与条形码 AATT、TTAA 和 SAMPLE_B 关联code>与条形码CCGG、GGCC、GCGC关联，分为4个文件SAMPLE_B_1...SAMPLE_B_4。

我可以创建 getSampleNames() 来获取 [SAMPLE_A,SAMPLE_A,SAMPLE_B,SAMPLE_B,SAMPLE_B,SAMPLE_B] 和 [1,2,1,2,3, 4]，然后压缩它们以获得组合{sample}_{id}。然后我可以对条形码执行相同的操作：[SAMPLE_A,SAMPLE_A,SAMPLE_B,SAMPLE_B,SAMPLE_B] 和 [AATT, TTAA,CCGG, GGCC, GCGC]。

SAMPLES_ID,IDs = getSampleNames()
SAMPLES_BC,BCs = getBCs(set(SAMPLES_ID))

rule refine:
input:
    '{sample}/demultiplex/{sample}_{id}.demultiplex.bam'
output:
    bam = '{sample}/polyA_trimming/{sample}_{id}.fltnc.bam',
shell:
    "isoseq3 refine {input} "


rule split:
input:
    expand('{sample}/polyA_trimming/{sample}_{id}.fltnc.bam', zip, sample = SAMPLES_ID, id = IDs),
output:
    expand("{sample}/cells/{barcode}_{sample}/fltnc.bam", zip, sample = SAMPLES_BC, barcode = BCs),
shell:
    "python {params.script_dir}/split_cells_bam.py"


rule dedup_split:
input:
    "{sample}/cells/{barcode}_{sample}/fltnc.bam"
output:
    bam = "{sample}/cells/{barcode}_{sample}/dedup/dedup.bam",
shell:
    "isoseq3 dedup {input} {output.bam} "

rule merge:
input:
    expand("{sample}/cells/{barcode}_{sample}/dedup/dedup.bam",
        zip, sample = SAMPLES_BC, barcode = BCs),

如何防止规则拆分成为管道中的瓶颈？目前，它等待对所有样本完成细化规则，虽然没有必要，每个样本都应该独立运行，但我不能，因为每个样本的条形码集都不同。有没有办法获得类似

expand("{sample}/cells/{barcode}_{sample}/fltnc.bam", zip, Sample = SAMPLES_BC, Barcode = BCs[SAMPLES_BC]) ，其中 SAMPLES_BC 的每个 {sample} 都是 BCs 字典中的键？ ID 也一样吗？我知道我可以使用函数，但我不确定如何通过规则传播 {barcode}

Answer 1

根据您的评论，有几种方法可以采用，其中涉及更改保存样本、条形码和 ID 的数据结构。现在，您可以为每个样本创建一条规则：

for sample in set(SAMPLES_ID):  # get uniq samples
    # get ids and barcodes for this sample
    ids = [tup[1] for tup in zip(SAMPLES_ID, IDs) if tup[0] == sample]
    bcs = [tup[1] for tup in zip(SAMPLES_BC, BCs) if tup[0] == sample]

    rule:
        name: f'{sample}_split'
        input:
            expand('{sample}/polyA_trimming/{sample}_{id}.fltnc.bam', 
                   sample = sample, id = ids),
        output:
            expand("{sample}/cells/{barcode}_{sample}/fltnc.bam", 
                   sample = sample, barcode = bcs),
        shell:
            "python {params.script_dir}/split_cells_bam.py"

您不需要在扩展中使用 zip，因为 ids 和 bcs 用于单个样本。我认为这通常不是最好的方法，但对于您当前的工作流程来说是最简单的。

只需注意您的 shell 命令，您如何将输入/输出传递给脚本？

Answer 2

我找到了如何通过函数使用字典，这解决了我的问题！

此解决方案的主要默认设置是您必须创建一个虚拟文件作为分割规则的输出，而不是检查是否创建了每个“{sample}/cells/{barcode}_{sample}/fltnc.bam”文件，所以我仍在寻找更优雅的东西......

IDs = getSampleNames() #{SAMPLE_A:[1,2], SAMPLE_B:[1,2,3,4]}
SAMPLES = list(IDs.keys()) 
BCs = getBCs(SAMPLES) #{SAMPLE_A:[AATT, TTAA], SAMPLE_B:[CCGG,GGCC,GCGC]}
    
# function linking IDs and SAMPLE
def sample2ids(wildcards):
    return expand('{{sample}}/polyA_trimming/{{sample}}_{id}.fltnc.bam', 
               id = IDs[wildcards.sample])

# function linking BCs and SAMPLE
def sample2ids(wildcards):
    return expand('{{sample}}/cells/{barcode}_{{sample}}/dedup/dedup.bam',
               barcode = BCs[wildcards.sample])

rule refine:
input:
    '{sample}/demultiplex/{sample}_{id}.demultiplex.bam'
output:
    bam = '{sample}/polyA_trimming/{sample}_{id}.fltnc.bam',

rule split:
input:
    sample2ids
output:
    # cannot use a function here, so I create a dummy file to pipe
    'dummy_file.txt'

rule dedup_split:
input:
    'dummy_file.txt'
output:
    bam = "{sample}/cells/{barcode}_{sample}/dedup/dedup.bam",


rule merge:
input:
    sample2bc