Question

我正在使用每日时间系列，我需要根据我的历史记录建立90天（或更长时间）的预测 - 当前时间系列大约有298个数据点。

我所遇到的问题是最终预测中的着名扁线 - 是的，我可能没有季节性，但我正在努力解决这个问题。另一个问题是如何找到最佳模型并从这里适应这种行为。

我创建了一个测试用例以进一步调查此问题，并且感谢任何帮助。

谢谢，

从

开始

x <- day_data  # My time serie
z <- 90        # Days to forecast

low_bound_date <- as.POSIXlt(min(x$time), format = "%m/%d/%Y") # oldest date in the DF.

> low_bound_date
[1] "2015-12-21 PST"

low_bound_date$yday 
> low_bound_date$yday  # Day in Julian
[1] 354

lbyear <- as.numeric(substr(low_bound_date, 1, 4))
> lbyear
[1] 2015

这是我的时间内容

> ts
Time Series:
Start = c(2065, 4) 
End = c(2107, 7) 
Frequency = 7 
  [2] 20.73 26.19 27.51 26.11 26.28 27.58 26.84 27.00 26.30 28.75 28.43 39.03 41.36 45.42 44.80 45.33 47.79 44.70 45.17
 [20] 34.90 32.54 32.75 33.35 34.76 34.11 33.59 33.60 38.08 30.45 29.66 31.09 31.36 31.96 29.30 30.04 30.85 31.13 25.09
 [39] 17.88 23.73 25.31 31.30 35.18 34.13 34.96 35.12 27.36 38.33 38.59 38.14 38.54 41.72 37.15 35.92 37.37 32.39 30.64
 [58] 30.57 30.66 31.16 31.50 30.68 32.21 32.27 32.55 33.61 34.80 33.53 33.09 20.90  6.91  7.82 15.78  7.25  6.19  6.38
 [77] 38.06 39.82 35.53 38.63 41.91 39.76 37.26 38.79 37.74 35.61 39.70 35.79 35.36 29.63 22.07 35.39 35.99 37.35 38.82
 [96] 25.80 21.31 18.85  9.52 20.75 36.83 44.12 37.79 34.45 36.05 16.39 21.84 31.39 34.26 31.50 30.87 28.88 42.83 41.52
[115] 42.34 47.35 44.47 44.10 44.49 26.89 18.17 40.44 43.93 41.56 39.98 40.31 40.59 40.17 40.22 40.50 32.68 35.89 36.06
[134] 34.30 22.67 12.56 13.29 12.34 28.00 35.27 36.57 33.78 32.15 33.58 34.62 30.96 32.06 33.05 30.66 32.47 30.42 32.83
[153] 31.74 29.39 22.39 12.58 16.46  5.36  4.01 15.32 32.79 31.66 32.02 27.60 31.47 31.61 34.96 27.77 31.91 33.94 33.43
[172] 26.94 28.38 21.42 24.51 23.82 31.71 26.64 27.96 29.29 29.25 28.70 27.02 27.62 30.90 27.46 27.37 26.46 27.77 13.61
[191]  5.87 12.18  5.68  4.15  4.35  4.42 16.42 25.18 26.06 27.39 27.57 28.86 15.18  5.19  5.61  8.28  7.78  5.13  4.90
[210]  5.02  5.27 16.31 25.01 26.19 25.96 24.93 25.53 25.56 26.39 26.80 26.73 26.00 25.61 25.90 25.89 13.80  6.66  6.41
[229]  5.28  5.64  5.71  5.38  5.76  7.20  7.27  5.55  5.31  5.94  5.75  5.93  5.77  6.57  5.52  5.51  5.47  5.69 19.75
[248] 29.22 30.75 29.63 30.49 29.48 31.83 30.42 29.27 30.40 29.91 32.00 30.09 28.93 14.54  7.75  5.63 17.17 22.27 24.93
[267] 35.94 37.42 33.13 25.88 24.27 37.64 37.42 38.33 35.20 21.32  7.32  4.81  5.17 17.49 23.77 23.36 27.60 26.53 24.99
[286] 24.22 23.76 24.10 24.22 27.06 25.53 23.40 37.07 26.52 25.19 28.02 28.53 26.67

第一步，我在ts

中获取数据

day_data_ts <- ts(x$avg_day, start = c(lbyear,low_bound_date$yday), frequency=7)

plot(day_data_ts)

plot_ts

acf(day_data_ts)

acf_ts

第二步，我在msts

中获取数据

day_data_msts <- msts(x$avg_day, seasonal.periods=c(7,365.25), start = c(lbyear,low_bound_date$yday))

plot(day_data_msts)

acf(day_data_msts)

我做了几次拟合迭代，试图找出最佳拟合和预测模型。

首次验证测试仅适用于ts。

fit1 <- HoltWinters(day_data_ts)
> fit1
    Holt-Winters exponential smoothing with trend and additive seasonal component.
    Call: HoltWinters(x = day_data_ts)
    Smoothing parameters: alpha: 1   beta : 0.006757112  gamma: 0

    Coefficients:
             [,1]
    a  28.0922449
    b   0.1652477
    s1  0.6241837
    s2  1.9084694
    s3  0.9913265
    s4  0.8198980
    s5 -1.7015306
    s6 -1.2201020
    s7 -1.4222449


fit2 <- tbats(day_data_ts)
> fit2
    BATS(1, {0,0}, 0.8, -)
    Parameters:   Alpha: 1.309966     Beta: -0.3011143    Damping Parameter: 0.800001
    Seed States:
              [,1]
    [1,] 15.282259
    [2,]  2.177787
    Sigma: 5.501356     AIC: 2723.911


fit3 <- ets(day_data_ts)
> fit3
    ETS(A,N,N) 
      Smoothing parameters: alpha = 0.9999 
      Initial states:       l = 25.2275 
      sigma:  5.8506
         AIC     AICc      BIC 
    2756.597 2756.678 2767.688 


fit4 <- auto.arima(day_data_ts)
> fit4
    ARIMA(1,1,2)                    
    Coefficients:
             ar1      ma1      ma2
          0.7396  -0.6897  -0.2769
    s.e.  0.0545   0.0690   0.0621
    sigma^2 estimated as 30.47:  log likelihood=-927.9
    AIC=1863.81   AICc=1863.94   BIC=1878.58

第二次测试正在使用msts。我还将ets模型更改为MAM。

fit5 <- tbats(day_data_msts)
> fit5
    BATS(1, {0,0}, 0.8, -)
    Parameters:   Alpha: 1.309966     Beta: -0.3011143    Damping Parameter: 0.800001
    Seed States:
              [,1]
    [1,] 15.282259
    [2,]  2.177787
    Sigma: 5.501356     AIC: 2723.911


fit6 <- ets(day_data_msts, model="MAN")
> fit6
    ETS(M,A,N) 
      Smoothing parameters:     alpha = 0.9999      beta  = 9e-04 
      Initial states:           l = 52.8658         b = 3.9184 
      sigma:  0.3459
         AIC     AICc      BIC 
    3042.744 3042.949 3061.229 


fit7 <- auto.arima(day_data_msts)
> fit7
    ARIMA(1,1,2)                    
    Coefficients:
             ar1      ma1      ma2
          0.7396  -0.6897  -0.2769
    s.e.  0.0545   0.0690   0.0621
    sigma^2 estimated as 30.47:  log likelihood=-927.9
    AIC=1863.81   AICc=1863.94   BIC=1878.58

Answer 1

您可以按如下方式预测之前估算的模型（使用内置时间序列LakeHuron）：

library(forecast)
y <- LakeHuron
tsdisplay(y)
# estimate ARMA(1,1)
mod_2 <- Arima(y, order = c(1, 0, 1))
#make forecast for 5 periods (years in this case)
fHuron <- forecast(mod_2, h = 5)
#show results in table
fHuron
#plot results
plot(fHuron)

这会给你：注意ARIMA模型将其预测基于先前的值，因此如果我们在许多时段进行预测，模型将使用已经预测的值来预测下一个。这将降低准确性。

要使用最佳ARIMA模型，请使用此功能：

library(R.utils) #for the function 'withTimeout'
fitARIMA<-function(timeseriesObject, timout)
{
    final.aic <- Inf
    final.order <- c(0,0,0)
    for (p in 0:5) for (q in 0:5) {
        if ( p == 0 && q == 0) {
        next
        }

        arimaFit = tryCatch( 
        withTimeout(arima(timeseriesObject
                            ,order=c(p, 0, q))
                    ,timeout = timeout)
        ,error=function( err ) FALSE
        ,warning=function( err ) FALSE )

        if( !is.logical( arimaFit ) ) {
        current.aic <- AIC(arimaFit)
        if (current.aic < final.aic) {
            final.aic <- current.aic
            final.order <- c(p, 0, q)
            final.arima <- arima(timeseriesObject, order=final.order)
        }
        } else {
        next
        }
    }
    final.order<-c(final.order,final.aic)
    final.order
}

R拟合和预测每日时间序列

1 个答案: