我想从此数据中选择距离值 <=40 且事件值 <=15 的点组。它们必须是连续的，并且至少有 3 个。我开始使用的代码是 rle() 命令。我不知道如何将该信息链接回数据 - 以提取两个条件 >=3 的“运行”。我使用了 which 命令和 cbind 来查看并检查 rle 是否正常工作。那么如何获取这些组的关联信息呢？然后让事情变得更加复杂......我需要一次完成两天的信息。这意味着，正如您从下面的示例数据中看到的那样，我有一个日期/时间，我需要在 48 小时窗口内找到所有 >=3 个位置组，其中 dist<40 AND activ<15。这些窗口是排他性的，从午夜开始，到 48 小时后的午夜结束。我在下面提供的示例数据对于最后一部分来说是不够的，但如果有必要，我愿意将其过去。
run <- rle(Dist <= 40 & Activ <= 15)
tempdist <- Dist <= 40
tempactive <- Activ <= 15
which(tempdist == TRUE & tempactive == TRUE)
combine <- cbind(tempdist, tempactive)
structure(list(Fix = 1:15, Date = structure(c(1L, 1L, 1L, 2L,
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("2010.06.23",
"2010.06.24", "2010.06.25", "2010.06.26", "2010.06.27", "2010.06.28",
"2010.06.29", "2010.06.30"), class = "factor"), Time = structure(c(86L,
89L, 95L, 6L, 11L, 69L, 106L, 107L, 113L, 121L, 123L, 128L, 136L,
14L, 22L), .Label = c("0:00:36", "0:00:39", "0:00:42", "0:00:48",
"0:01:04", "0:01:28", "0:01:35", "1:00:15", "1:00:19", "1:00:49",
"1:01:05", "1:01:29", "10:00:13", "10:00:35", "10:00:36", "10:00:42",
"10:01:05", "10:01:11", "11:00:15", "11:00:20", "11:00:35", "11:00:42",
"11:00:57", "11:01:05", "12:00:13", "12:00:19", "12:00:36", "12:00:42",
"12:01:39", "12:03:03", "13:00:13", "13:00:32", "13:00:36", "13:00:47",
"13:01:05", "13:01:52", "14:00:18", "14:00:36", "14:00:48", "14:01:04",
"14:03:04", "15:00:35", "15:00:36", "15:01:04", "15:01:28", "15:01:36",
"16:00:18", "16:00:19", "16:00:20", "16:00:36", "16:00:45", "16:03:05",
"17:00:35", "17:00:42", "17:01:05", "17:01:11", "18:00:13", "18:00:35",
"18:00:36", "18:00:37", "18:00:48", "18:01:04", "19:00:13", "19:00:27",
"19:00:30", "19:00:35", "19:00:42", "19:00:43", "2:00:18", "2:00:36",
"2:00:47", "2:00:54", "2:00:59", "2:01:30", "2:02:37", "20:00:13",
"20:00:35", "20:00:36", "20:01:05", "21:00:14", "21:00:19", "21:00:20",
"21:00:21", "21:00:35", "21:00:48", "21:01:08", "21:02:23", "22:00:35",
"22:00:36", "22:00:38", "22:01:36", "22:03:04", "22:03:05", "23:00:15",
"23:00:16", "23:00:20", "23:00:36", "23:01:05", "23:02:05", "23:03:03",
"3:00:35", "3:00:36", "3:00:47", "3:00:48", "3:00:59", "3:01:58",
"4:00:13", "4:00:17", "4:00:19", "4:00:35", "4:00:39", "5:00:16",
"5:00:35", "5:00:36", "5:01:23", "6:00:13", "6:00:16", "6:00:18",
"6:00:19", "6:00:35", "6:00:36", "7:00:18", "7:00:19", "7:00:36",
"7:01:05", "7:01:22", "8:00:18", "8:00:35", "8:00:36", "8:00:42",
"8:01:05", "9:00:13", "9:00:19", "9:00:35", "9:00:36", "9:00:53",
"9:01:36"), class = "factor"), X = c(NA, 6351162.121, 6351137.038,
6351180.514, 6351175.284, 6351189.867, 6351168.96, 6351230.592,
6351187.825, 6351182.809, 6351179.366, 6351181.313, 6351157.538,
6351180.688, 6351178.51), Y = c(NA, 436904.9145, 436928.697,
436925.2068, 436946.703, 436961.3917, 436967.0372, 436934.0528,
436978.3563, 436949.7116, 436963.6412, 436967.4078, 436949.5642,
436964.5655, 436961.398), Temp = c(14L, 19L, 16L, 14L, 18L, 20L,
18L, 14L, 13L, 15L, 17L, 13L, 16L, 18L, 20L), Activ = c(0.5,
0.5, 57.5, 21.5, 0.5, 1, 48, 62.5, 42.5, 2, 9.5, 0.5, 0.5, 0.5,
0.5), DateTime = structure(list(sec = c(8, 36, 16, 28, 5, 18,
58, 13, 35, 36, 19, 35, 53, 35, 42), min = c(1L, 0L, 0L, 1L,
1L, 0L, 1L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L), hour = c(21L, 22L,
23L, 0L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L), mday = c(23L,
23L, 23L, 24L, 24L, 24L, 24L, 24L, 24L, 24L, 24L, 24L, 24L, 24L,
24L), mon = c(5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L,
5L, 5L, 5L), year = c(110L, 110L, 110L, 110L, 110L, 110L, 110L,
110L, 110L, 110L, 110L, 110L, 110L, 110L, 110L), wday = c(3L,
3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L), yday = c(173L,
173L, 173L, 174L, 174L, 174L, 174L, 174L, 174L, 174L, 174L, 174L,
174L, 174L, 174L), isdst = c(0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L,
0L, 0L, 0L, 0L, 0L, 0L, 0L)), .Names = c("sec", "min", "hour",
"mday", "mon", "year", "wday", "yday", "isdst"), class = c("POSIXlt",
"POSIXt"), tzone = "UTC"), DistX = c(NA, -25.0830000005662, 43.4760000007227,
-5.23000000044703, 14.5829999996349, -20.9069999996573, 61.6320000002161,
-42.7669999999925, -5.01599999982864, -3.4429999999702, 1.94699999969453,
-23.7750000003725, 23.1500000003725, -2.17800000030547, -7.12799999956042
), DistY = c(NA, 23.7824999999721, -3.49020000000019, 21.4961999999941,
14.6886999999988, 5.64550000004238, -32.9844000000157, 44.30349999998,
-28.6447000000044, 13.9296000000322, 3.76659999997355, -17.8435999999638,
15.0013000000035, -3.16750000003958, 0.131600000022445), Dist = c(NA,
34.5653612056503, 43.6158694984158, 22.1232799205819, 20.6983525112342,
21.6558149058434, 69.9033194303938, 61.5777265027509, 29.0805621350161,
14.3487980388844, 4.24005714096063, 29.72616164217, 27.5855306584331,
3.84405258179195, 7.12921472209522), LnDist = c(NA, 3.54285205938314,
3.77542106345205, 3.09663044416928, 3.0300541082747, 3.07527400526048,
4.24711313638682, 4.12030022240512, 3.37006998311865, 2.66366617834105,
1.44457674579472, 3.39202752144892, 3.31729138374919, 1.34652716994060,
1.96420109120712), TimeDif = c(59.4666666666667, 59.6666666666667,
61.2, 59.6166666666667, 59.2166666666667, 61.6666666666667, 58.25,
60.3666666666667, 60.0166666666667, 59.7166666666667, 60.2666666666667,
60.3, 59.7, 60.1166666666667, 60), Velocity = c(NA, 0.579307729703637,
0.712677606183265, 0.371092198835593, 0.349535927575022, 0.351175376851514,
1.20005698592951, 1.02006173113337, 0.484541440738952, 0.240281295655334,
0.0703549304362937, 0.492971171511941, 0.462069190258511, 0.0639432090123419,
0.118820245368254), Heading = c(NA, 5.47117955593762, 1.65090330804596,
6.04452351734706, 0.781787182793282, 4.97612800901505, 2.06219261575797,
5.51543193207954, 3.31494594980981, 6.04087053685035, 0.477085335105382,
4.06855550443118, 0.995825324541324, 3.74395368878987, 4.7308492849138
)), .Names = c("Fix", "Date", "Time", "X", "Y", "Temp", "Activ",
"DateTime", "DistX", "DistY", "Dist", "LnDist", "TimeDif", "Velocity",
"Heading"), row.names = c(NA, 15L), class = "data.frame")
最佳答案
我确信有更有效的方法可以做到这一点，但现在是星期五，这种方法很有效。 rle() 返回一个包含长度和值的列表对象。正如 jimmyb 提到的，关键是获取从 rle() 返回的内容并将其转换为对 data.frame 有用的内容。 rep() 是做到这一点的武器。
#Your rle code
run <- rle(dat$Dist <= 40 & dat$Activ <= 15)
#Add columns for the length of the run and the value of that run
dat$run <- rep(run$lengths, run$lengths)
dat$cond <- rep(run$values, run$lengths)
#Subset the data based on the length and value
subset(dat, run >= 3 & cond)
根据提供的示例数据，将返回第 10 - 15 行。这是您所期望的吗？
关于r - 使用 rle 在 R 中选择一组连续数据，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/5682939/