Julia:如何使用 Channel 进行迭代

问题描述

当我运行以下代码时，我收到一条弃用消息，说生产已被频道替换.

When I run the following code, I get a deprecation saying produce has been replace with channels.

function source(dir)
    filelist = readdir(dir)

    for filename in filelist
        name,ext = splitext(filename)
        if ext == ".jld"
            produce(filename)
        end
    end
end

path = "somepathdirectoryhere"

for fname in Task(source(path))
    println(fname)
end

我找不到有关如何使用频道执行此操作的示例.我试过创建一个全局频道并使用 put！而不是没有运气的生产.

I cannot find an example on how to do this with channels. I've tried creating a global channel and using put! instead of produce with no luck.

有什么想法吗?

推荐答案

这是一种方法.修改您的函数以接受通道参数，并在其中 put! 数据:

Here's one way. Modify your function to accept a channel argument, and put! data in it:

function source(dir, chnl)
  filelist = readdir(dir)
  for filename in filelist
    name, ext = splitext(filename)
    if ext == ".jld"
      put!(chnl, filename)  % this blocks until "take!" is used elsewhere
    end
  end
end

然后使用 Channel 构造函数隐式创建您的任务(该构造函数接受一个仅代表通道的单个参数的函数，因此我们需要包装 source 围绕匿名函数的函数):

Then create your task implicitly using the Channel constructor (which takes a function with a single argument only representing the channel, so we need to wrap the source function around an anonymous function):

my_channel = Channel( (channel_arg) -> source( pwd(), channel_arg) )

然后，要么检查通道是否仍然打开(即任务尚未完成)，如果是这样，请输入参数:

Then, either check the channel is still open (i.e. task hasn't finished) and if so take an argument:

julia> while isopen( my_channel)
         take!( my_channel) |> println;
       end
no.jld
yes.jld

或者，将 channel 本身用作迭代器(迭代 Tasks 以及生产/消费功能已被弃用)

or, use the channel itself as an iterator (iterating over Tasks is becoming deprecated, along with the produce / consume functionality)

julia> for i in my_channel
         i |> println
       end
no.jld
yes.jld

或者，您可以根据文档将 @schedule 与 bind 等一起使用，但上面似乎是最直接的方法.

Alternatively you can use @schedule with bind etc as per the documentation, but it seems like the above is the most straightforward way.

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