我很难解决我在snakemake中遇到的一个问题。我的样本目前命名为“1-2-Brain_187_006_S77_L002_R1_001.fastq.gz”。我希望最终将它们重命名为一个较短的名称,如“1_2_Brain_S77_L002_R1”,然后使用扩展名“_trim.fastq.gz”作为我的规则。我正在用bbduk修剪。对于我的输入,我想调用我的字典列表allSamples。然后我想访问每个字典中的值。具体而言,“shortName1”和“shortName2”值。我的问题是在我的dry中,它将整个列表显示为一次运行的输入。我不知道如何使它注册为每个元素都是它自己的运行。我以3个文件名为例,实际上我有114个文件名。因此,我希望我的试运行有114个计数用于修剪工作
config.json
{
"allSamples" : ['1_2_Brain_S77_L002', '10_4_Kidney_S82_L002', '11_4_BB_S105_L002' ......],
"1_2_Brain_S77_L002":{
"sampleName1": "1-2-Brain_187_006_S77_L002_R1_001.fastq.gz",
"sampleName2": "1-2-Brain_187_006_S77_L002_R2_001.fastq.gz",
"shortName1": "1_2_Brain_S77_L002_R1",
"shortName2": "1_2_Brain_S77_L002_R2",
"stemName": "1_2_Brain_S77_L002"
}, ....
}
我正在获取位于rawReads/中的文件,并将新修剪的文件存储在trimmedReads/中
蛇锉
configfile: "refs/config.json"
# variables
sampleDict = config["allSamples"]
sampleNames1 = [config[i]["sampleName1"] for i in sampleDict]
sampleNames2 = [config[i]["sampleName2"] for i in sampleDict]
shortNames1 = [config[i]["shortName1"] for i in sampleDict]
shortNames2 = [config[i]["shortName2"] for i in sampleDict]
rule all:
input:
expand("trimmedReads/{trim1}_trim.fastq.gz", trim1 = shortNames1),
expand("trimmedReads/{trim2}_trim.fastq.gz", trim2 = shortNames2)
rule trim:
input:
R1 = expand("rawReads/{sample1}", sample1 = sampleNames1),
R2 = expand("rawReads/{sample2}", sample2 = sampleNames2)
output:
trim1 = expand("trimmedReads/{trim1}_trim.fastq.gz", trim1 = shortNames1),
trim2 = expand("trimmedReads/{trim2}_trim.fastq.gz", trim2 = shortNames2)
shell:
"""
bbduk.sh in1={input.R1} in2={input.R2} out1={output.trim1} out2={output.trim2} ref=ref/adapters.fa ktrim=r k=23 mink=11 hdist=1 tpe tbo
"""
当我做一次试跑时,我得到了这个
Building DAG of jobs...
Job counts:
count jobs
1 all
1 trim
2
[Mon May 24 22:42:36 2021]
rule trim:
input: rawReads/1-2-Brain_187_006_S77_L002_R1_001.fastq.gz, rawReads/10-4-Kidney_127_066_S82_L002_R1_001.fastq.gz, rawReads/11-4_BB_041_152_S105_L002_R1_001.fastq.gz, ...
output: trimmedReads/1_2_Brain_S77_L002_R1_trim.fastq.gz, trimmedReads/10_4_Kidney_S82_L002_R1_trim.fastq.gz, trimmedReads/11_4_BB_S105_L002_R1_trim.fastq.gz, ...
jobid: 1
bbduk.sh in1=rawReads/1-2-Brain_187_006_S77_L002_R1_001.fastq.gz rawReads/10-4-Kidney_127_066_S82_L002_R1_001.fastq.gz rawReads/11-4_BB_041_152_S105_L002_R1_001.fastq.gz ... out1=trimmedReads/1_2_Brain_S77_L002_R1_trim.fastq.gz trimmedReads/10_4_Kidney_S82_L002_R1_trim.fastq.gz trimmedReads/11_4_BB_S105_L002_R1_trim.fastq.gz ... ref=ref/adapters.fa ktrim=r k=23 mink=11 hdist=1 tpe tbo
[Mon May 24 22:42:36 2021]
localrule all:
input: trimmedReads/1_2_Brain_S77_L002_R1_trim.fastq.gz, trimmedReads/10_4_Kidney_S82_L002_R1_trim.fastq.gz, trimmedReads/11_4_BB_S105_L002_R1_trim.fastq.gz, ...
jobid: 0
Job counts:
count jobs
1 all
1 trim
2
This was a dry-run (flag -n). The order of jobs does not reflect the order of execution.
rule test:
的通配符是空的dict。在此规则中没有指定通配符值wildcards.sample
。每个通配符都应在output:
部分中指定,对于该规则,该部分为空。实际上,除非明确地将rule test:
指定为目标,否则rule test:
绝对没有效果:如果没有指定任何输出,Snakemake只会忽略这个没有任何结果的无用规则我猜文件
["rawReads/1_2_Brain_S77_L002", "rawReads/17_6_Brain_S83_L002"]
已经存在,因此Snakemake发现目标存在于磁盘上,并且什么也不做,产生“无输出”我不明白你所说的“最终将它们重命名为一个较短的名称”是什么意思,但这里有一个如何复制文件的方法。将此作为“如何使用通配符访问我的示例名称”的模式:
工作原理:
rule copy:
声明的输出(如果用值"SampleName1"
替换{sample}
)与文件名"path_to_target/foo_SampleName1_bar"
匹配"path_to_source/blablabla_SampleName1_bazz"
存在,则Snakemake满足要求,并且知道如何生成文件"path_to_target/foo_SampleName1_bar"
{sample}
值"SampleName2"
重复步骤2和3李>rule copy:
应该运行两次:每个文件一次李>相关问题 更多 >
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