Datetime 从频率为23的不规则(as.Date)时间序列创建规则时间序列
我在R中有以下问题。我想从不规则时间序列(即日期和数据值列表)创建一个ts()对象(即规则时间序列) 您可以使用以下数据集和R脚本重现问题:Datetime 从频率为23的不规则(as.Date)时间序列创建规则时间序列,datetime,r,time-series,zoo,Datetime,R,Time Series,Zoo,我在R中有以下问题。我想从不规则时间序列(即日期和数据值列表)创建一个ts()对象(即规则时间序列) 您可以使用以下数据集和R脚本重现问题: # dput(dd) result dd <- structure(list(NDVI = structure(c(14L, 4L, 11L, 12L, 20L, 17L, 5L, 7L, 21L, 23L, 25L, 19L, 15L, 9L, 3L, 24L, 2L, 6L, 22L, 16L, 13L, 18L, 1
# dput(dd) result
dd <- structure(list(NDVI = structure(c(14L, 4L, 11L, 12L, 20L, 17L,
5L, 7L, 21L, 23L, 25L, 19L, 15L, 9L, 3L, 24L, 2L, 6L, 22L, 16L,
13L, 18L, 10L, 8L, 1L), .Names = c("1", "2", "3", "4", "5", "6",
"7", "8", "9", "10", "11", "12", "13", "14", "15", "16", "17",
"18", "19", "20", "21", "22", "23", "24", "25"), .Label = c("0.4186",
"0.5452", "0.5915", "0.5956", "0.6010", "0.6860", "0.6966", "0.7159",
"0.7161", "0.7264", "0.7281", "0.7523", "0.7542", "0.7701", "0.7751",
"0.7810", "0.7933", "0.8075", "0.8113", "0.8148", "0.8207", "0.8302",
"0.8305", "0.8369", "0.9877"), class = "factor"), DATUM = structure(c(11005,
11021, 11037, 11085, 11101, 11117, 11133, 11149, 11165, 11181,
11197, 11213, 11229, 11245, 11261, 11277, 11293, 11309, 11323,
11339, 11355, 11371, 11387, 11403, 11419), class = "Date")), .Names = c("NDVI",
"DATUM"), row.names = c("1", "2", "3", "4", "5", "6", "7", "8",
"9", "10", "11", "12", "13", "14", "15", "16", "17", "18", "19",
"20", "21", "22", "23", "24", "25"), class = "data.frame")
require(zoo)
dd$DATUM <- as.Date(dd$DATUM,"A%Y%j") # Ayear,julianday
z <- zoo(dd$NDVI,dd$DATUM,frequency=23)
z # this is a regular time series with a frequency=23 and start=c(2000,1)
# there are 5 measurements in 2000 (2 jan, 1 feb, 2 apr) for which no data is available
# this should be marked as an NA is the final regular time series
ts.z <- as.ts(z,start=c(2000,1),frequency=23)
#dput(dd)结果
dd23不能平均地划分一年中的天数,因此你必须综合你自己的时间尺度,以便将每年划分为23个相等的部分。将dd
(有“日期”上课时间的版本)转换为zoo,并根据由年份加分数组成的新比例创建新系列。最后将其转换为ts系列:
library(zoo)
z <- zoo(as.numeric(as.character(dd[[1]])), dd[[2]])
lt <- unclass(as.POSIXlt(time(z)))
yr <- lt$year + 1900
jul <- lt$yday
delta <- min(unlist(tapply(jul, yr, diff))) # 16
zz <- aggregate(z, yr + jul / delta / 23)
as.ts(zz)
对这看起来很好,但由于变量zt没有定义,因此它在这里还不起作用。为什么你要除以16?谢谢你的帮助。@Jan我相信他会修好的,但我猜是的zt@Jan,正如Bill指出的,zt是时间(z),但从那时起,我简化并清理了代码,消除了它。16是因为diff(time(z))
显示点之间相隔16天或16天的倍数。效果很好!漂亮的解决方案。谢谢你。数据实际上是MODIS卫星数据,该数据汇总(合成)为16天的时间步长。这是一个创建常规时间序列(ts类)的智能解决方案。Cheeswa添加了一个计算,它给出了16
,因此该值更合理。
Time Series:
Start = c(2000, 4)
End = c(2001, 7)
Frequency = 23
[1] 0.7701 0.5956 0.7281 NA NA 0.7523 0.8148 0.7933 0.6010 0.6966
[11] 0.8207 0.8305 0.9877 0.8113 0.7751 0.7161 0.5915 0.8369 0.5452 0.6860
[21] 0.8302 0.7810 0.7542 0.8075 0.7264 0.7159 0.4186