从S3开始的Luigi管道

Question

我的初始文件在AWS S3中。有人能告诉我如何在Luigi Task中设置这个吗？

我查看了文档并找到了luigi.S3，但不清楚该怎么做，然后我在web上搜索，只从mortar-luigi和luigi顶部的实现获得链接。

更新

在遵循为@matagus提供的示例之后（我也按照建议创建了~/.boto文件）：

# coding: utf-8

import luigi

from luigi.s3 import S3Target, S3Client

class MyS3File(luigi.ExternalTask):
    def output(self):
        return S3Target('s3://my-bucket/19170205.txt')

class ProcessS3File(luigi.Task):

    def requieres(self):
        return MyS3File()

    def output(self):
        return luigi.LocalTarget('/tmp/resultado.txt')

    def run(self):
        result = None

        for input in self.input():
           print("Doing something ...")
           with input.open('r') as f:
               for line in f:
                   result = 'This is a line'

        if result:
            out_file = self.output().open('w')
            out_file.write(result)

当我执行的时候什么也没发生

DEBUG: Checking if ProcessS3File() is complete
INFO: Informed scheduler that task   ProcessS3File()   has status   PENDING
INFO: Done scheduling tasks
INFO: Running Worker with 1 processes
DEBUG: Asking scheduler for work...
DEBUG: Pending tasks: 1
INFO: [pid 21171] Worker Worker(salt=226574718, workers=1, host=heliodromus, username=nanounanue, pid=21171) running   ProcessS3File()
INFO: [pid 21171] Worker Worker(salt=226574718, workers=1, host=heliodromus, username=nanounanue, pid=21171) done      ProcessS3File()
DEBUG: 1 running tasks, waiting for next task to finish
INFO: Informed scheduler that task   ProcessS3File()   has status   DONE
DEBUG: Asking scheduler for work...
INFO: Done
INFO: There are no more tasks to run at this time
INFO: Worker Worker(salt=226574718, workers=1, host=heliodromus, username=nanounanue, pid=21171) was stopped. Shutting down Keep-Alive thread

如您所见，消息Doing something...从不打印。怎么了？

Answer 1

这里的关键是定义一个没有输入的外部任务，哪些输出是您在S3中已经拥有的文件。Luigi文档在Requiring another Task中提到了这一点：

Note that requires() can not return a Target object. If you have a simple Target object that is created externally you can wrap it in a Task class

所以，基本上你会得到这样的结果：

import luigi

from luigi.s3 import S3Target

from somewhere import do_something_with


class MyS3File(luigi.ExternalTask):

    def output(self):
        return luigi.S3Target('s3://my-bucket/path/to/file')

class ProcessS3File(luigi.Task):

    def requires(self):
        return MyS3File()

    def output(self):
        return luigi.S3Target('s3://my-bucket/path/to/output-file')

    def run(self):
        result = None
        # this will return a file stream that reads the file from your aws s3 bucket
        with self.input().open('r') as f:
            result = do_something_with(f)

        # and the you 
        out_file = self.output().open('w')
        # it'd better to serialize this result before writing it to a file, but this is a pretty simple example
        out_file.write(result)

更新：

Luigi使用boto从AWS S3读取文件和/或将文件写入AWS S3，因此为了使此代码工作，您需要在boto配置文件~/boto中提供凭据（查找其他possible config file locations here）：

[Credentials]
aws_access_key_id = <your_access_key_here>
aws_secret_access_key = <your_secret_key_here>