Question

我正在使用优先级矩阵优化作业订单。例如;在作业6完成之前不能执行作业8。如果是这种情况，我在位置[8,6]上的优先矩阵显示一个。如果没有优先关系，例如对于位置[1,2]，则它显示零。

在我当前的代码中，如果优先级矩阵显示为1，我使用'next'命令进入下一次迭代。但是，现在它的编程方式只是转到下一个j，而我想要它转到下一个'我'（下一个要考虑的工作）。但是，i是if循环之前的一个循环，其中使用了'next'。有没有办法让它转到for（i in n：jobs）循环而不是下一个for（j in n：jobs）？

在我当前的代码下面

# Initialisatie
nJobs = 10
jobDone = rep(FALSE, nJobs) 

currentJob = 1;
jobDone[currentJob] = TRUE;
totalFine = 0; 
finishingTime = durations[currentProject]

jobOrder = c(currentJob);
jobFines = c(totalFine);
jobIndFines = c(totalFine)
jobTimes = c(finishingTime)
jobIndTimes = c(durations[currentJob])

for(iter in 1:(nJobs-1))
{
  bestFine = -1;
  bestOrder = -1;

    for(i in 1:nJobs)
    {
      for(j in 1:nJobs){
      if(precedences[i,j] == 1){ next } #hoe kom ik nu naar de volgende i ipv de volgende j?
        else{

    if( ! jobDone[i]  )
    {
      # If this is the first option or the cheapest option, we update the best option
      if(bestOrder == -1 || ((finishingTime+durations[i])-deadlines[i])*fines[i] > bestFine) 
      {
        bestFine = ((finishingTime+durations[i])-deadlines[i])*fines[i];
        bestOrder = i;
      }
        }
      }


    }
  }
  # Add the best job to the order of jobs
  jobDone[bestOrder] = TRUE;
  jobOrder[iter+1] = bestOrder;
  # Update the current order of projects
  fine = ((finishingTime+durations[bestOrder])-deadlines[bestOrder])*fines[bestOrder]
  jobIndFines[iter+1] = fine
  if (fine>0){
    totalFine = totalFine + fine;
  }
  jobFines[iter+1] = totalFine;
  #jobIndFines[iter+1] = totalFine - jobFines[iter];
  finishingTime = finishingTime + durations[bestOrder];
  jobTimes[iter+1] = finishingTime
  jobIndTimes[iter+1] = durations[bestOrder]
}
totalFine
finishingTime
jobOrder
jobFines
jobTimes
jobIndTimes
jobDone
jobIndFines

非常感谢!!

Answer 1

尝试break。

这些代码段说明了next和break之间的区别。

next在 next 索引值处重新启动当前循环：

for (i in 1:2) {
  for (j in 1:2) {
    print(paste("j =", j))
    next
    print("Python is better than R at everything.") # Never prints, obviously.
  }
  print(paste("i =", i))
}
#> [1] "j = 1"
#> [1] "j = 2"
#> [1] "i = 1"
#> [1] "j = 1"
#> [1] "j = 2"
#> [1] "i = 2"

break完全让你退出当前循环并将控制传递给循环外的下一个语句：

for (i in 1:2) {
  for (j in 1:2) {
    print(paste("j =", j))
    break
    print("Python is better than R at everything.") # Never prints, obviously.
  }
  print(paste("i =", i))
}
#> [1] "j = 1"
#> [1] "i = 1"
#> [1] "j = 1"
#> [1] "i = 2"

R在循环中使用next进入下一次迭代（在当前循环之前是两次for循环）

1 个答案: