Question

我正在测试嵌套循环。我们的想法是首先遍历i和k，将它们相加，然后循环遍历j（j必须不等于k）。循环运行了很长时间，我必须手动停止循环。由于它是一个小数据，我知道它不应该花那么长时间，但不能弄清楚它出错的地方。以下是样本数据。

test = data.frame(ID = c(1,2,3), s= c(0.4,0.3,0.3), j1 = c(0.3,0.22,0.15), j2 = c(0.11,0.58, 0.02))

这是代码：

j = 1
k = 1

firstsum = 0
tm1 <- system.time(
  while (j <= nrow(test)){
    while (k <= nrow(test)){
      if (k == j) {
        next
      } else {
        for (i in 3:4){

          normindator = normindator + (test[j,i] * test[k, i])
          denominator = denominator + test[j, i] * test[j, i]
          firstsum = firstsum + norminator/denominator * test[k, 2]
          k = k + 1
          normindator = 0
          denominator = 0
        }
      }
    }
    secondsum = 0
    secondsum = secondsum + firstsum * test[j,2]
    j = j + 1
    k = 1
  }
)

Answer 1

使用next跳过k的增量。试试这个版本的循环

test = data.frame(ID = c(1,2,3), s= c(0.4,0.3,0.3), j1 = c(0.3,0.22,0.15), j2 = c(0.11,0.58, 0.02))


j = 1
k = 1

firstsum = 0
tm1 <- system.time(
  while (j <= nrow(test)){
    while (k <= nrow(test)){
      if (k != j) {
        for (i in 3:4){
          normindator = 0
          denominator = 0

          normindator = normindator + (test[j,i] * test[k, i])
          denominator = denominator + test[j, i] * test[j, i]
          firstsum = firstsum + normindator/denominator * test[k, 2]

        }
      }
      k = k + 1
    }
    secondsum = 0
    secondsum = secondsum + firstsum * test[j,2]
    j = j + 1
    k = 1
  }
)

嵌套循环无法停止

1 个答案: