分解時間序列數據

Question

任何人都可以幫助破譯ucm的輸出。 我的主要目標是檢查ts數據是否為季節性數據。 但是我無法每次都進行繪圖和觀察。 我需要使整個過程自動化，並提供季節性指標。

我想了解以下輸出

ucmxmodel$s.season

# Time Series:
#     Start = c(1, 1) 
#     End = c(4, 20) 
#     Frequency = 52 
#       [1] -2.391635076 -2.127871717 -0.864021134  0.149851212 -0.586660213 -0.697838635 -0.933982269  0.954491859 -1.531715424 -1.267769820 -0.504165631
#      [12] -1.990792301  1.273673437  1.786860414  0.050859315 -0.685677002 -0.921831488 -1.283081922 -1.144376739 -0.964042949 -1.510837956  1.391991657
#      [23] -0.261175626  5.419494363  0.543898305  0.002548125  1.126895943  1.474427901  2.154721023  2.501352782  0.515453691 -0.470886132  1.209419689

ucmxmodel$vs.season

# [1] 1.375832 1.373459 1.371358 1.369520 1.367945 1.366632 1.365582 1.364795 1.364270 1.364007 1.364007 1.364270 1.364795 1.365582 1.366632 1.367945
# [17] 1.369520 1.371358 1.373459 1.375816 1.784574 1.784910 1.785223 1.785514 1.785784 1.786032 1.786258 1.786461 1.786643 1.786802 1.786938 1.787052
# [33] 1.787143 1.787212 1.787257 1.787280 1.787280 1.787257 1.787212 1.787143 1.787052 1.786938 1.786802 1.786643 1.786461 1.786258 1.786032 1.785784
# [49] 1.785514 1.785223 1.784910 1.784578 1.375641 1.373276 1.371175 1.369337 1.367762 1.366449 1.365399 1.364612 1.364087 1.363824 1.363824 1.364087
# [65] 1.364612 1.365399 1.366449 1.367762 1.369337 1.371175 1.373276 1.375636 1.784453 1.784788 1.785101 1.785392 1.785662 1.785910 1.786136 1.786339

ucmxmodel$est.var.season

# Season_Variance 
#     0.0001831373 

我如何使用以上信息而無需查看圖表來確定季節性以及在哪個級別（每周，每月，每季度或每年）？

另外，我在est中得到NULL

ucmxmodel$est

# NULL

數據

測試數據為：

structure(c(44, 81, 99, 25, 69, 42, 6, 25, 75, 90, 73, 65, 55, 
9, 53, 43, 19, 28, 48, 71, 36, 1, 66, 46, 55, 56, 100, 89, 29, 
93, 55, 56, 35, 87, 77, 88, 18, 32, 6, 2, 15, 36, 48, 80, 48, 
2, 22, 2, 97, 14, 31, 54, 98, 43, 62, 94, 53, 17, 45, 92, 98, 
7, 19, 84, 74, 28, 11, 65, 26, 97, 67, 4, 25, 62, 9, 5, 76, 96, 
2, 55, 46, 84, 11, 62, 54, 99, 84, 7, 13, 26, 18, 42, 72, 1, 
83, 10, 6, 32, 3, 21, 100, 100, 98, 91, 89, 18, 88, 90, 54, 49, 
5, 95, 22), .Tsp = c(1, 3.15384615384615, 52), class = "ts")

和

structure(c(40, 68, 50, 64, 26, 44, 108, 90, 62, 60, 90, 64, 120, 82, 68, 60,
26, 32, 60, 74, 34, 16, 22, 44, 50, 16, 34, 26, 42, 14, 36, 24, 14, 16, 6, 6,
12, 20, 10, 34, 12, 24, 46, 30, 30, 46, 54, 42, 44, 42, 12, 52, 42, 66, 40,
60, 42, 44, 64, 96, 70, 52, 66, 44, 64, 62, 42, 86, 40, 56, 50, 50, 62, 22,
24, 14, 14, 18, 18, 10, 20, 10, 4, 18, 10, 10, 14, 20, 10, 32, 12, 22, 20, 20,
26, 30, 36, 28, 56, 34, 14, 54, 40, 30, 42, 36, 52, 30, 32, 52, 42, 62, 46,
64, 70, 48, 40, 64, 40, 120, 58, 36, 40, 34, 36, 26, 18, 28, 16, 32, 18, 12,
20), .Tsp = c(1, 4.36, 52), class = "ts")

Answer 1

我認為最直接的方法是遵循Rob Hyndman的方法 （他是R中許多時間序列包的作者）。 對於您的數據，它的工作方式如下：

require(fma)
# Create a model with multiplicative errors (see https://www.otexts.org/fpp/7/7).
fit1 <- stlf(test2)
# Create a model with additive errors.
fit2 <- stlf(data, etsmodel = "ANN")

deviance <- 2 * c(logLik(fit1$model) - logLik(fit2$model))
df <- attributes(logLik(fit1$model))$df - attributes(logLik(fit2$model))$df

# P-value
1 - pchisq(deviance, df)

# [1] 1

根據此分析，我們發現p值為 1，這將使我們得出結論，沒有季節性。

Answer 2

我非常喜歡R中提供的stl()函數。嘗試以下最小示例：

# some random data
x <- rnorm(200)
# as a time series object
xt <- ts(x, frequency = 10)
# do the decomposition
xts <- stl(xt, s.window = "periodic")
# plot the results
plot(xts)

現在，您可以通過比較方差來估算“季節性”。

vars <- apply(xts$time.series, 2, var)
vars['seasonal'] / sum(vars)

現在，您有了季節性方差作為分解后方差總和的一部分。

我強烈建議您閱讀原始論文，以便您了解此處的實際情況。 它非常易於訪問，我喜歡這種方法，因為它非常直觀。