Jupyter:即使前一个单元出现故障也可以运行下一个单元的技巧

问题描述

如果长时间运行的单元发生故障，我想发送警报，但是我不想尝试/除外，因为那样的话，当我查看错误时，我将发送不必要的消息.有办法吗?

I want to send alerts if a long running cell fails, but I don't want to try/except because then I will send needless messages when I am looking at the error. Is there a way to do this?

所需的工作流程:

1)运行status=train()单元格

2)在最初的15秒内没有看到错误

2) see no error in first 15 seconds

3)执行下一个单元格send_alert('done or error')，无论单元格1的结果如何，该单元格都将执行.

3) execute next cell send_alert('done or error') that will execute regardless of the outcome of cell 1.

4)去做别的事情

这里是一个单细胞解决方案，每次都会令人讨厌:

Here is a one cell solution that is annoying to code every time:

try:
    start = time.time()
    train(...)
except Exception as e:
    pass
end = time.time()
if end - start > 60: send_alert('done')

推荐答案

这是一个解决方案，它具有很小但可扩展的自定义iPython魔术.

Here's one solution, with a pretty small but extensible custom iPython magic.

您可以将其保存在名为magics.py的文件中，或具有pip可安装的软件包.我使用了可点子安装的东西:

You can keep it in a file named magics.py somewhere, or have a pip-installable package. I used something pip-installable:

.
├── magics
│   ├── __init__.py
│   └── executor.py
└── setup.py

# magics/executor.py

import time
from IPython.core.magic import Magics, magics_class, cell_magic

@magics_class
class Exceptor(Magics):

    @cell_magic
    def exceptor(self, line, cell):
        timeout = 2
        try:
            start = time.time()
            self.shell.ex(cell)
        except:
            if time.time() - start > timeout:
                print("Slow fail!")
        else:
            if time.time() - start > timeout:
                print("done")

# magics/__init__.py

from .exceptor import Exceptor

def load_ipython_extension(ipython):
    ipython.register_magics(Exceptor)

这里是使用此示例.请注意，%load_ext magics使用了程序包的名称，然后为您提供了名为%exceptor的单元格魔术.

Here is an example of using this. Notice that %load_ext magics takes the name of the package, and then gives you the cell magic named %exceptor.

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