我有一个带有ChIP-seq单端fastq文件名allfiles=['/path/file1.fastq','/path/file2.fastq','/path/file3.fastq']
的列表对象。我正在尝试将该对象allfiles
设置为通配符(我希望输入fastqc规则(以及其他诸如映射的规则,但让我们保持简单)。我尝试了下面代码中的内容([ C0])。但是,这给了我一个错误
lambda wildcards: data.loc[(wildcards.sample),'read1']
有人确切地知道如何定义它吗?看来我很接近,我有大致的了解,但我无法正确执行语法。谢谢!
代码:
"InputFunctionException in line 118 of Snakefile:
AttributeError: 'Wildcards' object has no attribute 'sample'
Wildcards:
"
当前import pandas as pd
import numpy as np
# Read in config file parameters
configfile: 'config.yaml'
sampleFile = config['samples'] # three columns: sample ID , /path/to/chipseq_file_SE.fastq , /path/to/chipseq_input.fastq
outputDir = config['outputdir'] # output directory
outDir = outputDir + "/MyExperiment"
qcDir = outDir + "/QC"
# Read in the samples table
data = pd.read_csv(sampleFile, header=0, names=['sample', 'read1', 'inputs']).set_index('sample', drop=False)
samples = data['sample'].unique().tolist() # sample IDs
read1 = data['read1'].unique().tolist() # ChIP-treatment file single-end file
inplist= data['inputs'].unique().tolist() # the ChIP-input files
inplistUni= data['inputs'].unique().tolist() # the ChIP-input files (unique)
allfiles = read1 + inplistUni
# Target rule
rule all:
input:
expand(f'{qcDir}' + '/raw/{sample}_fastqc.html', sample=samples),
expand(f'{qcDir}' + '/raw/{sample}_fastqc.zip', sample=samples),
# fastqc report generation
rule fastqc:
input: lambda wildcards: data.loc[(wildcards.sample), 'read1']
output:
html=expand(f'{qcDir}' + '/raw/{sample}_fastqc.html',sample=samples) ,
zip=expand(f'{qcDir}' + '/raw/{sample}_fastqc.zip',sample=samples)
log: expand(f'{logDir}' + '/qc/{sample}_fastqc_raw.log',sample=samples)
threads: 4
wrapper: "fastqc {input} 2>> {log}"
的output
文件在解析后没有任何通配符。也就是说,snakefile中当前有一个作业,其中rule fastqc
尝试为所有样本生成一个输出文件。
但是,您似乎希望为每个样本分别运行rule fastqc
。在这种情况下,需要将其概括如下,其中rule fastqc
是通配符:
{sample}