我有几个问题，我有兴趣解决。我想通过一个值（例如：
）对数组中的conc列进行采样和存储newdata <- data[ which(data$conc > 8), ]
但是，我想用它保存相关的日期时间戳。最后在另一个数组中，当conc值超过8.00然后降到8.00以下时，我想存储这一集的持续时间。因此，例如，21：30将记录为15分钟，另一个时间将记录在00:15和03:00之间，导致存储值为165分钟。
datetime conc
20/08/2012 21:00 7.29
20/08/2012 21:15 7.35
20/08/2012 21:30 35.23
20/08/2012 21:45 7.44
20/08/2012 22:00 13.30
20/08/2012 22:15 7.60
20/08/2012 22:30 7.65
20/08/2012 22:45 7.70
20/08/2012 23:00 7.83
20/08/2012 23:15 8.07
20/08/2012 23:30 8.30
20/08/2012 23:45 22.44
21/08/2012 00:00 7.81
21/08/2012 00:15 10.67
21/08/2012 00:30 11.07
21/08/2012 00:45 8.29
21/08/2012 01:00 8.17
21/08/2012 01:15 8.29
21/08/2012 01:30 8.26
21/08/2012 01:45 8.93
21/08/2012 02:00 9.74
21/08/2012 02:15 9.69
21/08/2012 02:30 9.15
21/08/2012 02:45 9.52
21/08/2012 03:00 9.10
21/08/2012 03:15 7.10
答案 0 :(得分：0)
也许一种形式是在数据中添加两个列，一个表示浓度高于8，另一个表示在返回8之前的累计时间。
#generating data
data <- read.table(text="datetime conc
'20/08/2012 21:00' 7.29
'20/08/2012 21:15' 7.35
'20/08/2012 21:30' 35.23
'20/08/2012 21:45' 7.44
'20/08/2012 22:00' 13.30
'20/08/2012 22:15' 7.60
'20/08/2012 22:30' 7.65
'20/08/2012 22:45' 7.70
'20/08/2012 23:00' 7.83
'20/08/2012 23:15' 8.07
'20/08/2012 23:30' 8.30
'20/08/2012 23:45' 22.44
'21/08/2012 00:00' 7.81
'21/08/2012 00:15' 10.67
'21/08/2012 00:30' 11.07
'21/08/2012 00:45' 8.29
'21/08/2012 01:00' 8.17
'21/08/2012 01:15' 8.29
'21/08/2012 01:30' 8.26
'21/08/2012 01:45' 8.93
'21/08/2012 02:00' 9.74
'21/08/2012 02:15' 9.69
'21/08/2012 02:30' 9.15
'21/08/2012 02:45' 9.52
'21/08/2012 03:00' 9.10
'21/08/2012 03:15' 7.10", sep=" ", header=TRUE, stringsAsFactors=FALSE)
#converting to date
data$datetime<-as.POSIXct(data$datetime, format="%d/%m/%Y %H:%M")
#creating stamps
data$stamp <- NA
data$stamp[which(data$conc<8)] <- "less.than.8"
data$stamp[which(data$conc>8)] <- "greater.than.8"
#calculating cumulative durationg in the episodes of sequencies of conc>8
for (i in 1:nrow(data)){
if(data$stamp[i] =="less.than.8"){
data$cum.duration[i] <- 0}
if(data$stamp[i] =="greater.than.8"){
data$cum.duration[i] <- (data$datetime[i]-data$datetime[i-1])+data$cum.duration[i-1]}
}
这将产生下表，然后你可以用它做任何你想做的事情：
datetime conc stamp cum.duration
1 2012-08-20 21:00:00 7.29 less.than.8 0
2 2012-08-20 21:15:00 7.35 less.than.8 0
3 2012-08-20 21:30:00 35.23 greater.than.8 15
4 2012-08-20 21:45:00 7.44 less.than.8 0
5 2012-08-20 22:00:00 13.30 greater.than.8 15
6 2012-08-20 22:15:00 7.60 less.than.8 0
7 2012-08-20 22:30:00 7.65 less.than.8 0
8 2012-08-20 22:45:00 7.70 less.than.8 0
9 2012-08-20 23:00:00 7.83 less.than.8 0
10 2012-08-20 23:15:00 8.07 greater.than.8 15
11 2012-08-20 23:30:00 8.30 greater.than.8 30
12 2012-08-20 23:45:00 22.44 greater.than.8 45
13 2012-08-21 00:00:00 7.81 less.than.8 0
14 2012-08-21 00:15:00 10.67 greater.than.8 15
15 2012-08-21 00:30:00 11.07 greater.than.8 30
16 2012-08-21 00:45:00 8.29 greater.than.8 45
17 2012-08-21 01:00:00 8.17 greater.than.8 60
18 2012-08-21 01:15:00 8.29 greater.than.8 75
19 2012-08-21 01:30:00 8.26 greater.than.8 90
20 2012-08-21 01:45:00 8.93 greater.than.8 105
21 2012-08-21 02:00:00 9.74 greater.than.8 120
22 2012-08-21 02:15:00 9.69 greater.than.8 135
23 2012-08-21 02:30:00 9.15 greater.than.8 150
24 2012-08-21 02:45:00 9.52 greater.than.8 165
25 2012-08-21 03:00:00 9.10 greater.than.8 180
26 2012-08-21 03:15:00 7.10 less.than.8 0
要仅选择结束剧集，您可以写：
lines <- which(data$conc>8)
lines <- lines[(lines[2:length(lines)] - lines[1:(length(lines)-1)])>1]
data[lines,]
哪个会给你：
datetime conc stamp cum.duration
3 2012-08-20 21:30:00 35.23 greater.than.8 15
5 2012-08-20 22:00:00 13.30 greater.than.8 15
12 2012-08-20 23:45:00 22.44 greater.than.8 45
25 2012-08-21 03:00:00 9.10 greater.than.8 180