打破 for 循环

Question

我正在尝试从两种不同的仪器中排列一些时间。我想找到两个时间序列之间的最小时间差，以开始将 x1 与 x2 对齐。

在下面的示例中，我可以得到 x1[1] 和 x2 之间的最小时间差。我想要的是找到 x1[i] 和 x2 之间的最小差异，然后是 x1[i+1] 和 x2，但是如果 x1[i+1] 和 x2 的差异不小于 x1[i] 和 x2然后我想打破 for 循环。这里 x1[4] 和 x1[5] 都与 x2[5] 和 x2[6] 相差 3 秒，所以应该在 i = 5 处中断，但它不会中断并循环遍历整个 x1.

x1 <- c("11:30:27", "11:30:37", "11:30:47", "11:30:57", "11:31:07", "11:31:17", "11:31:27", "11:31:37", "11:31:47",
"11:31:57", "11:32:07", "11:32:17", "11:32:27", "11:32:37", "11:32:47", "11:32:57", "11:33:07", "11:33:17",
"11:33:27", "11:33:37", "11:33:47", "11:33:57", "11:34:07", "11:34:17", "11:34:27", "11:34:37", "11:34:47",
"11:34:57", "11:35:07", "11:35:17", "11:35:27", "11:35:37", "11:35:47", "11:35:57", "11:36:07", "11:36:17",
"11:36:27", "11:36:37", "11:36:47", "11:36:57", "11:37:07", "11:37:17", "11:37:27", "11:37:37", "11:37:47",
"11:37:57", "11:38:07", "11:38:17", "11:38:27", "11:38:37", "11:38:47", "11:38:57", "11:39:07", "11:39:17",
"11:39:27", "11:39:37", "11:39:47", "11:39:57")

x2 <- c("10:59:23", "10:59:33", "10:59:43", "10:59:53", "11:30:54",
"11:31:04", "11:31:14", "11:31:24", "11:31:34", "11:31:44", "11:31:54", "11:32:04", "11:32:14", "11:32:24",
"11:32:34", "11:32:44", "11:32:54", "11:33:04", "11:33:14", "11:33:24", "11:33:34", "11:33:44", "11:33:54",
"11:34:04", "11:34:14", "11:34:24", "11:34:34", "11:34:44", "11:34:54", "11:35:04", "11:35:14", "11:35:24",
"11:35:34", "11:35:44", "11:35:54", "11:36:04", "11:36:14", "11:36:24", "11:36:34", "11:36:44", "11:36:54",
"11:37:04", "11:37:14", "11:37:24", "11:37:34", "11:37:44", "11:37:54", "11:38:04", "11:38:14", "11:38:24",
"11:38:34", "11:38:44", "11:38:54", "11:39:04", "11:39:14", "11:39:24", "11:39:34", "11:39:44", "11:39:54",
"11:40:04", "11:40:14", "11:40:24", "11:40:34", "11:40:44", "11:40:54", "11:41:04", "11:41:14", "11:41:24",
"11:41:34", "11:41:44", "11:41:54", "11:42:04", "11:42:14", "11:42:24", "11:42:34", "11:42:44", "11:42:54",
"11:43:04", "11:43:14", "11:43:24", "11:43:34", "11:43:44", "11:43:54", "11:44:04", "11:44:14", "11:44:24",
"11:44:34", "11:44:44", "11:44:54", "11:45:04", "11:45:14", "11:45:24", "11:45:34", "11:45:44", "11:45:54",
"11:46:04", "11:46:14", "11:46:24", "11:46:34", "11:46:44", "11:46:54", "11:47:04", "11:47:14", "11:47:24")

x2[which(abs(as.numeric(difftime(strptime(x1[1], format = "%H:%M:%S"), strptime(x2, format = "%H:%M:%S")))) == 
                     min(abs(as.numeric(difftime(strptime(x1[1], format = "%H:%M:%S"), strptime(x2, format = "%H:%M:%S"))))))]


for (i in 1:length(x1)){
  Mindf <- 200000000
  MinRow <- min(abs(as.numeric(difftime(strptime(x1[i], format = "%H:%M:%S"), strptime(x2, format = "%H:%M:%S")))))
  if (!MinRow < Mindf){
    j <- i
    break
  } else {
    Mindf <- MinRow
  }
}

Answer 1

问题是您在循环的每次迭代中将 Mindf 重置为其初始（大）值，

for (i in 1:length(x1)){
  Mindf <- 200000000
  ...
}

所以 if (!MinRow < Mindf) 条件从不适用。 else 分支中发生的 Mindf 到 MinRow 的设置只是在循环顶部重置为初始值。

你可能想要

Mindf <- 200000000

for (i in 1:length(x1)) {
  ...
}

这是寻找最小距离点的另一种方法

以矢量化方式转换为时间格式

x1 <- strptime(x1,format = "%H:%M:%S")
x2 <- strptime(x2,format = "%H:%M:%S")

最小值出现在

arrayInd(which.min(abs(outer(x1,x2,"-"))),c(length(x1),length(x2)))

#     [,1] [,2]
#[1,]    4    5

x1 的第 4 个条目和 x2 的第 5 个条目。

打破 for 循环

Breaking a for loop

for-loop

r

break