Question

在我的第一轮中，我有两类样本将由不同的参数处理，然后在第二轮中将它们合并在一起。类似于以下示例：

SAMPLES = ['1', '2', '3']
CLASS1 = ['1', '2']
CLASS2 = ['3']

rule all:
    input:
        expand('{class1}.first.txt', class1 = CLASS1),
        expand('{class2}.first.txt', class2 = CLASS2),
        expand('{sample}.second.tx', sample = SAMPLES)

rule first_CLASS1:
    input:
        '{class1}.txt'
    output:
        '{class1}.first.txt'
    shell:
        'touch {wildcards.class1}.first.txt'

rule first_CLASS2:
    input:
        '{class2}.txt'
    output:
        '{class2}.first.txt'
    shell:
        'touch {wildcards.class2}.first.txt'

rule second:
    input:
        '{sample}.first.txt'
    output:
        '{sample}.second.txt'
    shell:
        'touch {wildcards.sample}.second.txt'

但是我得到了这样的AmbiguousRuleException：

AmbiguousRuleException:
Rules first_CLASS2 and first_CLASS1 are ambiguous for the file 1.first.txt.
Consider starting rule output with a unique prefix, constrain your wildcards, or use the ruleorder directive.
Wildcards:
    first_CLASS2: class2=1
    first_CLASS1: class1=1
Expected input files:
    first_CLASS2: 1.txt
    first_CLASS1: 1.txtExpected output files:
    first_CLASS2: 1.first.txt
    first_CLASS1: 1.first.txt

我宣布

expand('{class2}.first.txt', class2 = CLASS2)

全部运行，然后CLASS2 = ['3']。但是它报告了class2 = 1，这的确使我感到困惑。

Answer 1

我认为是因为{class1}.first.txt和{class2}.first.txt共享相同的常数部分，即.first.txt，并且错误消息表明它们不是唯一的。您可以通过将所有规则放在规则前面来限制通配符来解决它：

wildcard_constraints:
    class1= '|'.join([re.escape(x) for x in CLASS1]),
    class2= '|'.join([re.escape(x) for x in CLASS2]),

另请参阅我的问题https://groups.google.com/forum/#!msg/snakemake/wVlJW9X-9EU/gSZh4U0_CQAJ。就个人而言，我希望约束条件是默认行为...

Snakemake中的歧义规则异常，两个分支的parms崩溃

1 个答案: