forループを破る

Question

私は2つの異なる楽器からいくつかのタイミングを整理しようとしています。私は2つの時系列の中で最も小さい時間差を見つけて、x1をx2で上にしたいと思っています。

以下の例では、x1 [1]とx2の最小の時間差を得ることができます。私が望むのは、x1 [i]とx2、そしてx1 [i + 1]とx2の差を見つけることですが、x1 [i + 1]とx2の差がx1 [i]とx2私はforループを壊したい。ここで、x1 [4]とx1 [5]はx2 [5]とx2 [3]と3秒異なりますので、i = 5のところで壊れますが、x1の全体をループしません。

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

問題はあなたがループの各反復で、その最初の（大きな）値にごMindfをリセットしていることで、

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

条件if (!MinRow < Mindf)は適用されませんように。 elseブランチで発生するMindf〜MinRowの設定は、ループの先頭で初期値に戻されます。おそらく、ここで

Mindf <- 200000000 

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

---

を望んでいた

は最短距離点

を見つけるための別のアプローチは、

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

ベクトル化の方法で時刻の形式に最小限の変換でありますは、

にあります。

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

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

第4のエントリはx1であり、第5のエントリはx2である。