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[英]Error in plotting xts object: 'x' must be a time-series object
[英]Subset xts time-series object in R
我有像这样某些月份的时间序列xts
对象
library(xts)
seq<- seq(as.POSIXct("2015-09-01"),as.POSIXct("2015-09-04"), by = "30 mins")
ob<- xts(data.frame(power=1:(length(seq))),seq)
现在,对应于每个观察(比如A
),我想计算过去两个小时观察的平均值。 因此,对应于每个观察( A
),我需要计算在A
两小时之前发生的观察的索引,假设它是B
。 然后我可以计算A
和B
之间观测值的平均值。 因此,
i=10 # dummy
ind_cur<- index(ob[i,]) # index of current observation
ind_back <- ind_cur - 3600 * 2 # index of 2 hours back observation
使用这些指数,我将ob
子集为
ob['ind_cur/ind_back']
它导致以下错误:
Error in if (length(c(year, month, day, hour, min, sec)) == 6 && c(year, :
missing value where TRUE/FALSE needed
In addition: Warning messages:
1: In as_numeric(YYYY) : NAs introduced by coercion
2: In as_numeric(MM) : NAs introduced by coercion
3: In as_numeric(DD) : NAs introduced by coercion
4: In as_numeric(YYYY) : NAs introduced by coercion
5: In as_numeric(MM) : NAs introduced by coercion
6: In as_numeric(DD) : NAs introduced by coercion
谁能帮我对ob
进行子集化! 在链接上找到了一个相关的问题,但不足以解决这个问题。
更新预期输出显示为
2015-09-01 00:00:00 1 NA # as I don't have previous data
2015-09-01 00:30:00 2 NA
2015-09-01 01:00:00 3 NA
2015-09-01 01:30:00 4 NA
2015-09-01 02:00:00 5 10/4 # mean of prevous 4 observations (last two hours)
2015-09-01 02:30:00 6 14/4
2015-09-01 03:00:00 7 18/4
这是一个一般很难解决的问题,因此您需要推出自己的解决方案。 最简单的方法是通过重叠 2 小时间隔使用window
来进行子集化。
# initialize a result object
ob2 <- ob * NA_real_
# loop over all rows and calculate 2-hour mean
for(i in 2:nrow(ob)) {
ix <- index(ob)[i]
ob2[i] <- mean(window(ob, start=ix-3600*2, end=ix))
}
# set incomplete 2-hour intervals to NA
is.na(ob2) <- which(index(ob2) < start(ob2)+3600*2)
我们可以将rollapply()
包与lag()
结合使用,以将产生的滚动mean
偏移一行。
rollapply(lag(ob), 4, mean)
# power
#2015-09-01 00:00:00 NA
#2015-09-01 00:30:00 NA
#2015-09-01 01:00:00 NA
#2015-09-01 01:30:00 NA
#2015-09-01 02:00:00 2.5
#2015-09-01 02:30:00 3.5
#2015-09-01 03:00:00 4.5
# Or if you want it as new variable in your xts object
ob$mean <- rollapply(lag(ob),4,mean)
基于对“预期输出”问题的更新和 RS 的评论:
library(TTR)
head(SMA(ob$power, 4)) # 2 hour moving average
结果
SMA
2015-09-01 00:00:00 NA
2015-09-01 00:30:00 NA
2015-09-01 01:00:00 NA
2015-09-01 01:30:00 2.5
2015-09-01 02:00:00 3.5
2015-09-01 02:30:00 4.5
这假设有问题所述的 30 分钟间隔。
要看起来更像预期输出:
lag(head(SMA(ob$power, 4),7))
SMA
2015-09-01 00:00:00 NA
2015-09-01 00:30:00 NA
2015-09-01 01:00:00 NA
2015-09-01 01:30:00 NA
2015-09-01 02:00:00 2.5
2015-09-01 02:30:00 3.5
2015-09-01 03:00:00 4.5
包data.table
提供了滚动功能,对单个和多个时间序列都很有用:
head(
as.data.table(ob)[, roll_power := frollmean(power, 4, align = 'right')]
)
# at the end of a 4 1/2 hour lag
index power roll_power
1: 2015-09-01 00:00:00 1 NA
2: 2015-09-01 00:30:00 2 NA
3: 2015-09-01 01:00:00 3 NA
4: 2015-09-01 01:30:00 4 2.5 # the rolling mean covers this, and preceding rows
5: 2015-09-01 02:00:00 5 3.5
6: 2015-09-01 02:30:00 6 4.5
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