在Golang中同时读取文件

Question

读取部分不是并发的,而是处理.我用这种方式表达了标题,因为我最有可能使用该短语再次搜索此问题.:)

在尝试超越示例之后,我陷入了僵局,所以这对我来说是一次学习经历.我的目标是这些:

1. 逐行读取文件(最终使用缓冲区来完成行组).
2. 将文本传递给func()执行某些正则表达式工作的文本.
3. 将结果发送到某处但避免使用互斥锁或共享变量.我正在向一个频道发送整数(总是数字1).这有点愚蠢,但如果它没有引起问题,我想把它留下来,除非你们有一个更整洁的选择.
4. 使用工作池来执行此操作.我不确定我怎么告诉工人们自己重新排队？

这是游乐场的链接.我试着写有用的评论,希望这是有道理的.我的设计可能完全错误,所以不要犹豫重构.

package main

import (
  "bufio"
  "fmt"
  "regexp"
  "strings"
  "sync"
)

func telephoneNumbersInFile(path string) int {
  file := strings.NewReader(path)

  var telephone = regexp.MustCompile(`\(\d+\)\s\d+-\d+`)

  // do I need buffered channels here?
  jobs := make(chan string)
  results := make(chan int)

  // I think we need a wait group, not sure.
  wg := new(sync.WaitGroup)

  // start up some workers that will block and wait?
  for w := 1; w <= 3; w++ {
    wg.Add(1)
    go matchTelephoneNumbers(jobs, results, wg, telephone)
  }

  // go over a file line by line and queue up a ton of work
  scanner := bufio.NewScanner(file)
  for scanner.Scan() {
    // Later I want to create a buffer of lines, not just line-by-line here ...
    jobs <- scanner.Text()
  }

  close(jobs)
  wg.Wait()

  // Add up the results from the results channel.
  // The rest of this isn't even working so ignore for now.
  counts := 0
  // for v := range results {
  //   counts += v
  // }

  return counts
}

func matchTelephoneNumbers(jobs <-chan string, results chan<- int, wg *sync.WaitGroup, telephone *regexp.Regexp) {
  // Decreasing internal counter for wait-group as soon as goroutine finishes
  defer wg.Done()

  // eventually I want to have a []string channel to work on a chunk of lines not just one line of text
  for j := range jobs {
    if telephone.MatchString(j) {
      results <- 1
    }
  }
}

func main() {
  // An artificial input source.  Normally this is a file passed on the command line.
  const input = "Foo\n(555) 123-3456\nBar\nBaz"
  numberOfTelephoneNumbers := telephoneNumbersInFile(input)
  fmt.Println(numberOfTelephoneNumbers)
}

Answer 1

你几乎就在那里,只需要一些关于goroutines同步的工作.您的问题是您正在尝试提供解析器并在同一例程中收集结果,但这是无法完成的.

我建议如下:

1. 在单独的例程中运行扫描程序,一旦读取所有内容,关闭输入通道.
2. 运行单独的例程,等待解析器完成其工作,而不是关闭输出通道.
3. 收集主程序中的所有结果.

相关更改可能如下所示:

// Go over a file line by line and queue up a ton of work
go func() {
    scanner := bufio.NewScanner(file)
    for scanner.Scan() {
        jobs <- scanner.Text()
    }
    close(jobs)
}()

// Collect all the results...
// First, make sure we close the result channel when everything was processed
go func() {
    wg.Wait()
    close(results)
}()

// Now, add up the results from the results channel until closed
counts := 0
for v := range results {
    counts += v
}

在操场上充分运用的例子:http://play.golang.org/p/coja1_w-fY

值得补充的是,您不一定需要WaitGroup实现相同的目标,您需要知道的是何时停止接收结果.这可以通过扫描仪广告(在频道上)读取多少行然后收集器只读取指定数量的结果(尽管你也需要发送零)来实现.