我在R中有以下问题。

您可以通过以下数据集和R脚本复制问题:

# 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,因为我获得了一个很长的常规时间序列,其中包含日常时间步骤。我想获得一个频率= 23的TS对象,以指示无法作为Na的数据可用的位置。

我一直在根据此处列出的示例来尝试所有内容以获取年度数据将不规则的时间序列转换为常规时间序列

但是它不适用于23个频率(即每年23个值)的数据。我想我可以通过避免设置来解决它 dd$DATUM as.Date() 但是,作为一个可以作为时间序列订购的动物园对象,每年为23个值。

有任何想法吗?

谢谢你的帮助

有帮助吗?

解决方案

23不会平均分为一年中的天数,因此您必须综合自己的时间尺度,以使每年分为23个相等的零件。兑换 dd (具有“日期”课程时间的版本)到动物园,并根据由年度组成的新量表创建一个新系列,加上一部分。最终将其转换为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
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