作为 Snakefile 输出的函数或字典理解

Question

我有一个带有高度冗余输出的 Snakemake 规则（此处已简化）：

rule redundant:
    output:
        a_x = "prefix_a_x.ext"
        b_x = "prefix_b_x.ext"
        a_y = "prefix_a_y.ext"
        b_y = "prefix_b_y.ext"
    run:
        process_1(output.a_x)
        process_2(output.a_y)
        process_3(output.b_x)
        process_4(output.b_y)

我可以以可读性为代价删除冗余：

rule less_readable:
    output:
        [f"prefix_{a}_{b}.ext"
         for a in ["a", "b"]
         for b in ["x", "y"]]
    run:
        process_1(output[0])
        process_2(output[1])
        process_3(output[2])
        process_4(output[3])

（我不想冒险使用

expand("prefix_{a}_{b}.ext", a=["a", "b"], b=["x", "y"])

因为输出的顺序很重要，我担心它可能会改变 在未来的 Snakemake 版本中。）

我想要两全其美，比如：

rule more_readable:
    output:
        {f"{a}_{b}": f"prefix_{a}_{b}.ext"
         for a in ["a", "b"]
         for b in ["x", "y"]}
    run:
        process_1(output.a_x)
        process_2(output.a_y)
        process_3(output.b_x)
        process_4(output.b_y)

可悲的是，这不起作用：

$ snakemake -j 1 prefix_a_x.ext
Building DAG of jobs...
MissingRuleException:
No rule to produce prefix_a_x.ext (if you use input functions make sure that they don't raise unexpected exceptions).

如何为自动生成的目标分配标签？

我用

$ snakemake --version
7.14.2

Answer 1

你最后的片段几乎是正确的，确保使用字典扩展（

**

）：

rule more_readable:
    output:
        **{f"{a}_{b}": f"prefix_{a}_{b}.ext"
         for a in ["a", "b"]
         for b in ["x", "y"]}