首页 \ 问答 \ Python:从多个进程写入单个文件(ZMQ)(Python: write to single file from multiple processes (ZMQ))

Python:从多个进程写入单个文件(ZMQ)(Python: write to single file from multiple processes (ZMQ))

我想从多个进程写入单个文件。 确切地说,我宁愿不使用多处理队列解决方案进行多处理,因为有几个子模块由其他开发人员编写。 但是,对此类子模块的每次写入都与写入zmq队列相关联。 有没有办法将zmq消息重定向到文件? 具体来说,我正在寻找http://www.huyng.com/posts/python-logging-from-multiple-processes/的内容,而不使用logging模块。


I want to write to a single file from multiple processes. To be precise, I would rather not use the Multiple processing Queue solution for multiprocessing as there are several submodules written by other developers. However, each write to the file for such submodules is associated with a write to a zmq queue. Is there a way I can redirect the zmq messages to a file? Specifically I am looking for something along the lines of http://www.huyng.com/posts/python-logging-from-multiple-processes/ without using the logging module.


原文:https://stackoverflow.com/questions/18624516
更新时间:2023-07-06 17:07

最满意答案

您可以将伴随对象与自定义“apply”方法一起使用:

case class MyReturnedModel(reason: String)

object MyReturnedModel {
  def apply(mod: ReturnedModel) = MyReturnedModel(mod.reason.toString)
}

val data: ReturnedModel = ... // Some instance of ReturnedModel
val mr = MyReturnModel(data)

请注意,case类及其伴随对象需要位于同一个文件中才能使用。


You can use a companion object with a custom "apply" method:

case class MyReturnedModel(reason: String)

object MyReturnedModel {
  def apply(mod: ReturnedModel) = MyReturnedModel(mod.reason.toString)
}

val data: ReturnedModel = ... // Some instance of ReturnedModel
val mr = MyReturnModel(data)

Just note that the case class and its companion object need to be in the same file for this to work.

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