3

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 脚本重现该问题：

# 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, 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)

但这不起作用，因为我获得了一个包含每日时间步长的非常长的常规时间序列。我想获得一个频率 = 23 的 ts 对象，正确指示其数据不可用作 NA 的位置。

我一直在尝试基于此处列出的年度数据示例 将不规则时间序列转换为规则时间序列

但它不适用于频率为 23 的数据（即一年 23 个值）。我想我可以通过避免设置dd$DATUM as.Date()但作为一个动物园对象来解决它，该对象可以作为一个时间序列排序，每年有 23 个值。

有任何想法吗？

谢谢你的帮助

4

1 回答 1

8

23 并不能均匀地划分为一年中的天数，因此您必须综合自己的时间尺度，以便将每年划分为 23 个相等的部分。将dd（具有“日期”课程时间的版本）转换为动物园并根据由年份加上分数组成的新比例创建一个新系列。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)

给予：

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
于 2011-01-27T17:10:59.340 回答