Question

两小时的时间序列xts1和xts2，xts1有一些缺失时间。

xts1
time                   speed power
2010-01-01 00:00:00     0.1  1.1
2010-01-01 01:00:00     0.2  1.2
2010-01-01 05:00:00     0.2  1.2
.....

xts2
time                   speed power
2010-01-01 00:00:00     0.1  1.1
2010-01-01 01:00:00     0.2  1.2
2010-01-01 02:00:00     0.2  1.2
.....

将它们组合成一个文件（获得平均速度，并根据相同的时间戳求和功率），得到不一致的数组错误。使用的命令是：

hourly.data.table = data.table (time = time(xts1), meanspeed=    (coredata(xts1$speed)+coredata(xts2$speed))/2, power= coredata(xts1$power)+coredata(xts2$power))

如何按时间进行此组合？提前致谢。

Answer 1

尝试

library(xts)
xts1 <- xts(df1[-1], order.by = as.POSIXct(df1$time))
xts2 <- xts(df2[-1], order.by = as.POSIXct(df2$time))
res <- xts1+xts2
res[,1] <- res[,1]/2
res
#                     speed power
#2010-01-01 00:00:00   0.1   2.2
#2010-01-01 01:00:00   0.2   2.4

数据

df1 <- structure(list(time = c("2010-01-01 00:00:00", 
"2010-01-01 01:00:00", 
"2010-01-01 05:00:00"), speed = c(0.1, 0.2, 0.2), power = c(1.1, 
1.2, 1.2)), .Names = c("time", "speed", "power"), 
class = "data.frame", row.names = c(NA, -3L))

df2 <- structure(list(time = c("2010-01-01 00:00:00", 
"2010-01-01 01:00:00", 
"2010-01-01 02:00:00"), speed = c(0.1, 0.2, 0.2), power = c(1.1, 
1.2, 1.2)), .Names = c("time", "speed", "power"), 
class = "data.frame", row.names = c(NA, -3L))

r两个xts操作

1 个答案:

数据