Luigi-运行时未完成的％s

问题描述

我正在尝试以一种非常简单的方式来学习luigi的工作原理.就像一个新手一样，我想出了这段代码

I am trying to learn in a very simple way how luigi works. Just as a newbie I came up with this code

import luigi

class class1(luigi.Task):

  def requires(self):
     return class2()

  def output(self):
    return luigi.LocalTarget('class1.txt')

 def run(self):
    print 'IN class A'


class class2(luigi.Task):

  def requires(self):
     return []

  def output(self):
     return luigi.LocalTarget('class2.txt')


if __name__ == '__main__':
  luigi.run()

在命令提示符下运行此命令会提示错误

Running this in command prompt gives error saying

raise RuntimeError('Unfulfilled %s at run time: %s' % (deps, ',

推荐答案

之所以会发生这种情况，是因为您为class2定义了一个输出，但是从不创建它.

This happens because you define an output for class2 but never create it.

让我们分解一下...

Let's break it down...

运行时

python file.py class2 --local-scheduler

luigi会问:

• class2的输出是否已经在磁盘上?否
• 检查class2的依赖项:无
• 执行run方法(默认情况下为空方法pass)
• run方法未返回错误，因此作业成功完成.

• is the output of class2 already on disk? NO
• check dependencies of class2: NONE
• execute the run method (by default it's and empty method pass)
• run method didn't return errors, so job finishes successfully.

但是，在运行时

python file.py class1 --local-scheduler

路易吉将:

• class1的输出是否已经在磁盘上?否
• 检查任务依赖性:是:class2
• 暂停以检查class2的状态
  • class2的输出是否在磁盘上?否
  • 运行class2-> 运行->完成，没有错误
  • class2的输出是否在磁盘上?否->引发错误
  • is the output of class1 already on disk? NO
  • check task dependencies: YES: class2
  • pause to check status of class2
    • is the output of class2 on disk? NO
    • run class2 -> running -> done without errors
    • is the output of class2 on disk? NO -> raise error

    luigi永远不会运行任务，除非满足其先前的所有依赖关系. (即它们的输出在文件系统上)

    luigi never runs a task unless all of its previous dependencies are met. (i.e. their output is on the file system)

    这篇关于Luigi-运行时未完成的％s的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！