将字典参数传递给Snakemake时报MissingInputException缺失输入文件错误

错误原因 #

Snakemake规则的input块中声明的所有值都会被识别为输入文件路径，Snakemake构建DAG时会自动校验这些文件是否存在，不存在就会抛出MissingInputException。
你的代码中，info_dict_from_sample函数返回的字典同时包含了输入文件路径filepath和HLA分型字符串hla，两个值都被unpack到了input块，Snakemake将HLA分型字符串HLA-A02:01误判为输入文件路径，找不到对应文件就触发了报错。

解决方法 #

将非文件类型的参数从input块移到params块中，params块的内容会被当作普通参数处理，不会校验文件是否存在，修改后的参考代码如下：

import pandas as pd

samples = pd.read_csv("acc_filepath_df.csv").set_index("sample", drop=False)
sids = samples['sample']

# 单独获取输入文件路径的函数
def get_input_file(wildcards):
    return samples.loc[wildcards.sample, "filepath"]

# 单独获取HLA参数的函数
def get_hla_param(wildcards):
    return samples.loc[wildcards.sample, "HLA"]

expected_filenames = "mhctyped/{sample}.txt.gz"

rule all:
    input:
        expand(expected_filenames,
            zip,
            sample=sids)

rule test_run:
    input:
        get_input_file
    output:
        "mhctyped/{sample}.txt.gz"
    params:
        hla = get_hla_param
    shell:
        "cat {input} {params.hla} > {output}"

如果需要保留返回字典的写法，也可以调整为：

import pandas as pd

samples = pd.read_csv("acc_filepath_df.csv").set_index("sample", drop=False)
sids = samples['sample']

def info_dict_from_sample(wildcards):
    return {
        "filepath": samples.loc[wildcards.sample, "filepath"], 
        "hla": samples.loc[wildcards.sample, "HLA"]
    }

expected_filenames = "mhctyped/{sample}.txt.gz"

rule all:
    input:
        expand(expected_filenames,
            zip,
            sample=sids)

rule test_run:
    input:
        filepath = lambda wildcards: info_dict_from_sample(wildcards)["filepath"]
    output:
        "mhctyped/{sample}.txt.gz"
    params:
        hla = lambda wildcards: info_dict_from_sample(wildcards)["hla"]
    shell:
        "cat {input.filepath} {params.hla} > {output}"

内容的提问来源于stack exchange，提问作者Ravin Poudel