提问人：recurso 提问时间：10/30/2022 更新时间：10/30/2022 访问量：79
如何在 snakemake 中访问嵌套列表的每个元素以命名输出？
How to access each element of a nested list for naming output in snakemake?
问：
这是一个类似的问题： Snakemake：使用 python 嵌套列表推导进行条件分析
我有以下几点：
RUN_ID = ["run1", "run2"]
SAMPLES = [["A", "B", "C"], ["D","E","F"]]
rule all:
input:
summary = expand("foo/{run}/{sample}/outs/{sample}_report.html", run=RUN_ID, sample=SAMPLES)
问题 1：每个运行应仅与相应的样本相关联（基于索引）。所以 run1 只与 A、B、C 配对，run2 只与 D、E、F 配对。RUN_IDSAMPLES
问题 2：每个输出文件的命名应反映这种基于索引的配对。目前，我正在努力让每个嵌套列表的每个元素与每个元素配对SAMPLESRUN_ID
基于上述内容，我想要以下输出：
"foo/run1/A/outs/A_report.html"
"foo/run1/B/outs/B_report.html"
"foo/run1/C/outs/C_report.html"
"foo/run2/D/outs/D_report.html"
"foo/run2/E/outs/E_report.html"
"foo/run2/F/outs/F_report.html"
最初我得到的是这个：
"foo/run1/["A", "B", "C"]/outs/["A", "B", "C"]_report.html"
"foo/run1/["D", "E", "F"]/outs/["D", "E", "F"]_report.html"
"foo/run2/["A", "B", "C"]/outs/["A", "B", "C"]_report.html"
"foo/run2/["D", "E", "F"]/outs/["D", "E", "F"]_report.html"
我在扩展函数中使用zip克服了不需要的配对：
summary= expand(["foo/{run}/{sample}/outs/{sample}_report.html", "foo/{run}/{sample}/outs/{sample}_report.html"], zip, run=RUN_ID, sample=SAMPLES)
给我留下 RUN_ID 和 SAMPLES 之间所需的配对：
"foo/run1/["A", "B", "C"]/outs/["A", "B", "C"]_report.html"
"foo/run2/["D", "E", "F"]/outs/["D", "E", "F"]_report.html"
但是，如上所示，每个嵌套列表都传递到输出路径中，而不是每个嵌套列表的每个元素。我可以通过将 SAMPLES 分成两个不同的列表来实现我想要的，但想要一种更优雅和自动化的方法。
我也不拘泥于使用嵌套列表;感谢对修复或更好方法的任何见解。谢谢！
答：
2赞
SultanOrazbayev
10/30/2022
#1
expand是一个方便的实用程序，对于更复杂的情况，直接使用 Python 生成所需的列表通常更快：
RUN_ID = ["run1", "run2"]
SAMPLES = [["A", "B", "C"], ["D","E","F"]]
desired_files = []
for run, SAMPLE in zip(RUN_ID, SAMPLES):
for sample in SAMPLE:
file = f"foo/{run}/{sample}/outs/{sample}_report.html"
desired_files.append(file)
rule all:
input: desired_files
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