Jupyter 无法启动内核

Question

所以我刚刚安装了 anaconda 3 并在 anaconda 提示符下运行了“jupyter notebook”，它打开了 jupyter。 但是，当我打开一个新的 python 3 notebook 时，我无法运行任何代码，因为内核错误“无法启动内核”并显示以下错误消息

Traceback (most recent call last):
  File "C:\Users\Janko\Anaconda3\lib\site-packages\tornado\web.py", line 1699, in _execute
    result = await result
  File "C:\Users\Janko\Anaconda3\lib\site-packages\tornado\gen.py", line 742, in run
    yielded = self.gen.throw(*exc_info)  # type: ignore
  File "C:\Users\Janko\Anaconda3\lib\site-packages\notebook\services\sessions\handlers.py", line 72, in post
    type=mtype))
  File "C:\Users\Janko\Anaconda3\lib\site-packages\tornado\gen.py", line 735, in run
    value = future.result()
  File "C:\Users\Janko\Anaconda3\lib\site-packages\tornado\gen.py", line 742, in run
    yielded = self.gen.throw(*exc_info)  # type: ignore
  File "C:\Users\Janko\Anaconda3\lib\site-packages\notebook\services\sessions\sessionmanager.py", line 88, in create_session
    kernel_id = yield self.start_kernel_for_session(session_id, path, name, type, kernel_name)
  File "C:\Users\Janko\Anaconda3\lib\site-packages\tornado\gen.py", line 735, in run
    value = future.result()
  File "C:\Users\Janko\Anaconda3\lib\site-packages\tornado\gen.py", line 742, in run
    yielded = self.gen.throw(*exc_info)  # type: ignore
  File "C:\Users\Janko\Anaconda3\lib\site-packages\notebook\services\sessions\sessionmanager.py", line 101, in start_kernel_for_session
    self.kernel_manager.start_kernel(path=kernel_path, kernel_name=kernel_name)
  File "C:\Users\Janko\Anaconda3\lib\site-packages\tornado\gen.py", line 735, in run
    value = future.result()
  File "C:\Users\Janko\Anaconda3\lib\site-packages\tornado\gen.py", line 209, in wrapper
    yielded = next(result)
  File "C:\Users\Janko\Anaconda3\lib\site-packages\notebook\services\kernels\kernelmanager.py", line 168, in start_kernel
    super(MappingKernelManager, self).start_kernel(**kwargs)
  File "C:\Users\Janko\Anaconda3\lib\site-packages\jupyter_client\multikernelmanager.py", line 110, in start_kernel
    km.start_kernel(**kwargs)
  File "C:\Users\Janko\Anaconda3\lib\site-packages\jupyter_client\manager.py", line 240, in start_kernel
    self.write_connection_file()
  File "C:\Users\Janko\Anaconda3\lib\site-packages\jupyter_client\connect.py", line 547, in write_connection_file
    kernel_name=self.kernel_name
  File "C:\Users\Janko\Anaconda3\lib\site-packages\jupyter_client\connect.py", line 212, in write_connection_file
    with secure_write(fname) as f:
  File "C:\Users\Janko\Anaconda3\lib\contextlib.py", line 112, in __enter__
    return next(self.gen)
  File "C:\Users\Janko\Anaconda3\lib\site-packages\jupyter_client\connect.py", line 102, in secure_write
    with os.fdopen(os.open(fname, open_flag, 0o600), mode) as f:
PermissionError: [Errno 13] Permission denied: 'C:\\Users\\Janko\\AppData\\Roaming\\jupyter\\runtime\\kernel-624a372e-0a69-4eda-8068-85cfa1e7c4bd.json'

我尝试谷歌搜索，但找不到解决办法。 请记住，我对内核、环境和我在对类似问题的回复中读到的其他行话并不十分熟悉，因此我的故障排除能力相当有限（抱歉）。 这是我的 kernels.json 文件（其中包含 python 可执行文件的正确路径），对类似问题的响应通常与之相关。 我感到困惑的另一件事是文件路径具有双反斜杠，但如果我改为单反，jupyter 会说未找到内核。

{
 "argv": [
  "C:\\Users\\Janko\\Anaconda3\\python.exe",
  "-m",
  "ipykernel_launcher",
  "-f",
  "{connection_file}"
 ],
 "display_name": "Python 3",
 "language": "python"
}

谢谢！

Answer 1

堆栈跟踪的最后一行告诉您 Jupyter 无法写入的目录：

Permission denied: 'C:\\Users\\Janko\\AppData\\Roaming\\jupyter\\runtime\\kernel-624a372e-0a69-4eda-8068-85cfa1e7c4bd.json'

向运行 Jupyter 的用户授予写入该目录的权限，或者将环境变量JUPYTER_RUNTIME_DIR设置为可写目录。

https://jupyter.readthedocs.io/en/latest/projects/jupyter-directories.html