Question

我想使用snakemake实用程序（5.6.0）使用EGA上存储的文件。首先，我想尝试使用官方文档中编写的代码，因此我尝试了以下方法：

import snakemake.remote.EGA as EGA

ega = EGA.RemoteProvider()

rule get_remote_file_ega:
   input:
       ega.remote("ega/dataset_id/foo.bam")
   output:
       "data/foo.bam"
   shell:
       "cp {input} {output}"

在执行脚本之前，我根据需要创建了环境变量（EGA_USERNAME和EGA_PASSWORD）。

然后我得到以下错误：

me:~/scripts$ snakemake -s test_ega.smk
Building DAG of jobs...
Traceback (most recent call last):
  File "/home/puissant/miniconda3/lib/python3.7/site-packages/snakemake/__init__.py", line 551, in snakemake
    export_cwl=export_cwl)
  File "/home/puissant/miniconda3/lib/python3.7/site-packages/snakemake/workflow.py", line 433, in execute
    dag.init()
  File "/home/puissant/miniconda3/lib/python3.7/site-packages/snakemake/dag.py", line 122, in init
    job = self.update([job], progress=progress)
  File "/home/puissant/miniconda3/lib/python3.7/site-packages/snakemake/dag.py", line 603, in update
    progress=progress)
  File "/home/puissant/miniconda3/lib/python3.7/site-packages/snakemake/dag.py", line 655, in update_
    missing_input = job.missing_input
  File "/home/puissant/miniconda3/lib/python3.7/site-packages/snakemake/jobs.py", line 396, in missing_input
    for f in self.input
  File "/home/puissant/miniconda3/lib/python3.7/site-packages/snakemake/jobs.py", line 397, in <genexpr>
    if not f.exists and not f in self.subworkflow_input)
  File "/home/puissant/miniconda3/lib/python3.7/site-packages/snakemake/io.py", line 208, in exists
    return self.exists_remote
  File "/home/puissant/miniconda3/lib/python3.7/site-packages/snakemake/io.py", line 119, in wrapper
    v = func(self, *args, **kwargs)
  File "/home/puissant/miniconda3/lib/python3.7/site-packages/snakemake/io.py", line 258, in exists_remote
    return self.remote_object.exists()
  File "/home/puissant/miniconda3/lib/python3.7/site-packages/snakemake/remote/EGA.py", line 173, in exists
    return self.parts.path in self.provider.get_files(self.parts.dataset)
  File "/home/puissant/miniconda3/lib/python3.7/site-packages/snakemake/remote/EGA.py", line 126, in get_files
    "data/metadata/datasets/{dataset}/files".format(dataset=dataset))
  File "/home/puissant/miniconda3/lib/python3.7/site-packages/snakemake/remote/EGA.py", line 96, in api_request
    headers["Authorization"] = "Bearer {}".format(self.token)
  File "/home/puissant/miniconda3/lib/python3.7/site-packages/snakemake/remote/EGA.py", line 77, in token
    self._login()
  File "/home/puissant/miniconda3/lib/python3.7/site-packages/snakemake/remote/EGA.py", line 45, in _login
    "client_id"    : self._client_id(),
  File "/home/puissant/miniconda3/lib/python3.7/site-packages/snakemake/remote/EGA.py", line 151, in _client_id
    return self._credentials("EGA_CLIENT_ID")
NameError: name 'self' is not defined

其中涉及的代码部分在那里（EGA.py第151行）：

1   @classmethod
2   def _client_id(cls):
3         return self._credentials("EGA_CLIENT_ID")

错误可能来自第3行上的“ self”而不是“ cls”吗？因为在将其更改为“ cls”后，错误将移至以相同方式构建的下一个块。我对python对象的了解有限，我希望我不要说很荒谬。

我忘记了任何步骤还是误解了其中任何一个？

Answer 1

您是正确的，您应该使用cls（这里大概代表“ class”），而不是self。 self通常是用于类（即对象）实例的名称。如果您在函数的其他位置使用self，则需要将它们切换为cls。

snakemake.remote.EGA：NameError：名称“ self”未定义

1 个答案: