【发布时间】：2020-05-03 19:44:34
【问题描述】：
我正在尝试通过受益于 R 的矢量化来改进我的代码，例如使用更多的应用族函数而不是 for 循环，因为我使用的数据集达到 30 万条记录，我希望能够缩短时间在正在运行的脚本上。
我已经准备了一个 repex 以及实际的 for 循环，我只是不知道是否可以将其转换为非循环结构。
这里是：
df <- structure(list(time = structure(c(1500697800, 1500698100, 1500698400,
1500698700, 1500699000, 1500699300, 1500699600, 1500699900, 1500700200,
1500700500, 1500700800, 1500701100, 1500701400, 1500701700, 1500702000,
1500702300, 1500702600, 1500702900, 1500703200, 1500703500, 1500703800,
1500704100, 1500704400, 1500704700, 1500705000, 1500705300, 1500705600,
1500705900, 1500706200, 1500706500, 1500706800, 1500707100, 1500707400,
1500707700, 1500708000, 1500708300, 1500708600, 1500708900, 1500709200,
1500709500, 1500709800, 1500710100, 1500710400, 1500710700, 1500711000,
1500711300, 1500711600, 1500711900, 1500712200, 1500712500, 1500712800,
1500713100, 1500713400, 1500713700, 1500714000, 1500714300, 1500714600,
1500714900, 1500715200, 1500715500, 1500715800, 1500716100, 1500716400,
1500716700, 1500717000, 1500717300, 1500717600, 1500717900, 1500718200,
1500718500, 1500718800, 1500719100, 1500719400, 1500719700, 1500720000,
1500720300, 1500720600, 1500720900, 1500721200, 1500721500, 1500721800,
1500722100, 1500722400, 1500722700, 1500723000, 1500723300, 1500723600,
1500723900, 1500724200, 1500724500, 1500724800, 1500725100, 1500725400,
1500725700, 1500726000, 1500726300, 1500726600, 1500726900, 1500727200,
1500727500, 1500727800, 1500728100, 1500728400, 1500728700, 1500729000,
1500729300, 1500729600, 1500729900, 1500730200, 1500730500, 1500730800,
1500731100, 1500731400, 1500731700, 1500732000, 1500732300, 1500732600,
1500732900, 1500733200, 1500733500, 1500733800, 1500734100, 1500734400,
1500734700, 1500735000, 1500735300, 1500735600, 1500735900, 1500736200,
1500736500, 1500736800, 1500737100, 1500737400, 1500737700, 1500738000,
1500738300, 1500738600, 1500738900, 1500739200, 1500739500, 1500739800,
1500740100, 1500740400, 1500740700, 1500741000), class = c("POSIXct",
"POSIXt"), tzone = "UTC"), rate = c(8021.22624828867, 8022.17252092756,
4026.57093082574, 0, 0, 0, 0, 0, 0, 0, 0, 1092.48742657481, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2352.47712160156, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), is.rate = c("OFF", "OFF", "OFF",
"OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF",
"OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF",
"OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF",
"OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF",
"OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF",
"OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF",
"OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF",
"OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF",
"OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF",
"OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF",
"OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF",
"OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF",
"OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF",
"OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF",
"OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF",
"OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF")), class = c("tbl_df",
"tbl", "data.frame"), row.names = c(NA, -145L))
为了快速解释数据：它有一个时间变量，一些速率，以及一个当速率不为 0 时的标志 --> ON。
for 循环的想法是，它将拾取高于 0 的速率值，并且从时间的角度来看，它将在接下来的一个小时内“拖尾”is.rate 标志。我知道这听起来很复杂，但是一旦你在 repex 上运行 for 循环，它应该是有意义的。
说到for循环，这里是：
for (i in which(temp_df$rate != 0)) {
temp_df$is.rate[i:(i + 12)] <- "ON" # 12 in this case is a factor of lag-time. Since data is in 5 min intervals, this means the next hour
}
我很想尝试优化这段代码，最好完全去掉for循环并使用类似的东西来应用家庭功能，但我看不到代码结构。
【问题讨论】：
-
如果您为此示例数据的至少一部分提供预期输出，将会有所帮助。
-
你的意思是当
rate > 0,is.rate <- "ON"。但是，当rate为0 时，那么is.rate将在接下来的11 行中保持"ON"？ -
或许
zoo::rollapply(df$rate > 0, 12, any, partial = TRUE)? -
kiyanuDevs，如果其中一个答案解决了您的问题，请accept it；这样做不仅为回答者提供了一些积分，而且还为有类似问题的读者提供了一些关闭。尽管您只能接受一个答案，但您可以选择对您认为有帮助的人进行投票。 （如果仍有问题，您可能需要编辑您的问题并提供更多详细信息。）
标签： r performance for-loop optimization vectorization