如何计算R中具有相同值的连续变量的持续时间

Question

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我想计算交通灯在每个交通周期中相对于绿色、琥珀色、红色的持续时间（我的示例数据中的列sg.0），例如计算从第一个绿色状态到的所有时间长度每个周期的最后一个绿色状态，我该怎么做？ Data.frame 如下所示：

 time sg. 0
1   2014-09-01 00:00:12.0 green
2   2014-09-01 00:00:13.5 green
3   2014-09-01 00:00:30.0 amber
4   2014-09-01 00:00:30.0 amber
5   2014-09-01 00:00:31.5 amber
6   2014-09-01 00:00:32.0 amber
7   2014-09-01 00:00:32.2 amber
8   2014-09-01 00:00:33.5 amber
9   2014-09-01 00:00:33.0   red
10  2014-09-01 00:00:35.0   red
11  2014-09-01 00:00:35.2   red
12  2014-09-01 00:00:37.0   red
13  2014-09-01 00:00:41.0   red
14  2014-09-01 00:00:42.0   red
15  2014-09-01 00:00:42.2   red
16  2014-09-01 00:00:43.0   red
17  2014-09-01 00:00:44.7   red
18  2014-09-01 00:00:44.2   red
19  2014-09-01 00:00:45.5   red
20  2014-09-01 00:00:47.0   red
21  2014-09-01 00:00:48.7   red
22  2014-09-01 00:00:49.7   red
23  2014-09-01 00:00:49.7   red
24  2014-09-01 00:00:49.9   red
25  2014-09-01 00:00:50.9 green
26  2014-09-01 00:00:50.0 green
27  2014-09-01 00:00:52.0 green
28  2014-09-01 00:00:53.0 green
29  2014-09-01 00:00:54.0 green
30  2014-09-01 00:00:55.0 green
31  2014-09-01 00:00:55.0 green
32  2014-09-01 00:01:02.0 green
33  2014-09-01 00:01:03.7 green
34  2014-09-01 00:01:05.7 green
35  2014-09-01 00:01:07.0 green

原始数据：

structure(list(time = structure(c(1409518812, 1409518813.6, 1409518830, 
1409518830.1, 1409518831.6, 1409518832, 1409518832.2, 1409518833.6, 
1409518833, 1409518835, 1409518835.3, 1409518837, 1409518841, 
1409518842, 1409518842.3, 1409518843, 1409518844.8, 1409518844.2, 
1409518845.6, 1409518847, 1409518848.7, 1409518849.7, 1409518849.8, 
1409518849.9, 1409518850.9, 1409518850, 1409518852, 1409518853, 
1409518854, 1409518855, 1409518855.1, 1409518862, 1409518863.8, 
1409518865.8, 1409518867, 1409518868, 1409518870.7, 1409518870.3, 
1409518884, 1409518884.2, 1409518884.3, 1409518884.5, 1409518890, 
1409518942, 1409518942.1, 1409518943.7, 1409518943.3, 1409518944.9, 
1409518944, 1409518945, 1409518947, 1409518949.5, 1409518949.6, 
1409518953, 1409518954, 1409518957.8, 1409518957.2, 1409518961, 
1409518961.1, 1409518961.2, 1409518962.2, 1409518962.3, 1409518964, 
1409518965, 1409518966, 1409518967, 1409518967.1, 1409518974, 
1409518975.8, 1409518977.8, 1409518979, 1409518980, 1409519068, 
1409519068.1, 1409519068.7, 1409519070, 1409519071, 1409519073, 
1409519073.8, 1409519081, 1409519082, 1409519083.3, 1409519083.8, 
1409519084.7, 1409519086, 1409519087.6, 1409519089.2, 1409519089.3, 
1409519091, 1409519091.1, 1409519091.6, 1409519092, 1409519092.1, 
1409519093, 1409519094, 1409519094.5, 1409519095, 1409519095.1, 
1409519103, 1409519104), class = c("POSIXct", "POSIXt")), `sg. 0` = structure(c(2L, 
2L, 1L, 1L, 1L, 1L, 1L, 1L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 
3L, 3L, 3L, 3L, 3L, 3L, 3L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 
3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 3L, 3L, 3L, 3L, 3L, 
3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 2L, 
2L, 2L, 2L), .Label = c("amber", "green", "red"), class = "factor")), .Names = c("time", 
"sg. 0"), row.names = c(NA, 100L), class = "data.frame")

Answer 1

您可能希望首先唯一地识别每个颜色循环，然后您可以收集每个组的统计数据。你可以找到循环

cycle<-cumsum(c(FALSE, dd[-1,2] != dd[-nrow(dd),2]))

（假设您的 data.frame 名为 dd）。然后你可以找到从头到尾的持续时间

tapply(dd[,1], interaction(dd[,2], cycle, drop=T), function(x) diff(range(x)))

给了

green.0 amber.1   red.2 green.3 amber.4   red.5 green.6 amber.7   red.8 green.9 
    1.6     3.6    16.9    40.0     2.9    16.2    17.8     2.0    23.5     9.0 

或者，如果您指的是格力/琥珀色/红色循环中的循环，您可以这样做

cycle<-cumsum(c(dd[1,2]!="green", dd[-1,2] == "green" & dd[-nrow(dd),2] !="green"))
tapply(dd[,1], cycle, function(x) as.double(diff(range(x)), units="mins"))

给了

        0         1         2         3 
0.6316667 1.8533333 2.2050000 0.1500000

Answer 2

与 MrFlick 的方法类似，您可以使用 rle 首先为每个颜色周期生成一个指标，然后使用它来计算持续时间。

# If you want to calculate the time within each colour
r <- rle(as.numeric(dat$sg.0))
r$values <- seq_along(r$values)
dat$id <- inverse.rle(r)

(a <- aggregate(time ~ sg.0 + id, dat, function(i) diff(as.numeric(range(i)))))
#    sg.0 id time
#1  green  1  1.6
#2  amber  2  3.6
#3    red  3 16.9
# ...

# Use a similar approach, if the cycle is for each green/amber/red
r <- rle(as.numeric(dat$sg.0))
r$values <- rep(seq_along(r$values), each=3, length=length(r$values))
dat$cycle <- inverse.rle(r)

 (b <- aggregate(time ~ cycle, dat, function(i) diff(as.numeric(range(i)))))
#  cycle  time
#1     1  37.9
#2     2 111.2
#3     3 132.3
#4     4   9.0

EDIT 添加了as.numeric 以聚合函数调用以在几秒钟内持续报告