Question

如何才能找到最准确的变化率？

这是我的情况：

有一个空的目录，直到几千个文件转储到其中
处理这些文件，直到一切都消失
我想知道根据文件从目录中删除的时间来处理这些文件的速度有多快。

我知道变化的速度是:( newamount - oldamount）/ time但是我在“时间”片段上苦苦挣扎。

这是我到目前为止所做的：

prev = ""

loop do

  # Let's find the number of files in the input queue, this is a method that basically does a wc -l on a directory.  
  num = number_of_input_files

  # Sleep if there's nothing in 0
  if num.to_i == 0 or prev == ""
    prev = num
    puts "Current: 0"
    sleep 60
    next
  end

  # Find the difference between the current queue number and the previous queue number
  sum = prev.to_i - num.to_i

  puts "Current: #{num}, Previous: #{prev}, Rate #{sum} per minute"

  prev = num

  sleep 60
end

我想我会在睡眠中拍摄自己的脚60，但我不确定如何捕获更准确的数字。我该怎么做？

Answer 1

存储脚本开始处理的Time，您可以随时查看差异。简而言之：

start = Time.now
# do work
diff = Time.now - start

如果这个脚本同时计算速率并进行处理，那么睡眠就会成为一个问题，但是由于另一个脚本正在处理，你应该没问题。

从脚本的结构来看，只要目录变空，您就会想要重置start。您还可以计算完成的预计完成时间，在我运行这些类型的东西时，我觉得很方便。

prev = nil

loop do

  # Let's find the number of files in the input queue, this is a method that basically does a wc -l on a directory.  
  num = number_of_input_files.to_i

  # Sleep if there's nothing in 0
  if num == 0 || prev.nil?
    prev = num
    puts "Current: 0"
    start = Time.now
    sleep 60
    next
  end

  # Find the difference between the current queue number and the previous queue number
  sum = prev - num
  diff = (Time.now - start) / 60.0 # per minute
  rate = sum / diff
  remaining = sum / rate

  puts "Current: #{num}, Previous: #{prev}, Rate #{rate} per minute"
  puts "#{remaining} minutes remaining"

  prev = num

  sleep 60
end

只要将一批工作转储到目录中，

sum将在一个周期内为负，因此您可能也想要考虑到这一点。

试图计算Ruby

1 个答案: