多步预测与dplyr和do的结合

Question

dplyr中的do函数可以让您快速轻松地创建许多酷炫的模型，但我正在努力将这些模型用于良好的滚动预测。

# Data illustration

require(dplyr)
require(forecast)

df <- data.frame(
  Date = seq.POSIXt(from = as.POSIXct("2015-01-01 00:00:00"), 
                    to = as.POSIXct("2015-06-30 00:00:00"), by = "hour"))

  df <- df %>% mutate(Hour = as.numeric(format(Date, "%H")) + 1, 
                      Wind = runif(4320, min = 1, max = 5000), 
                      Temp = runif(4320, min = - 20, max = 25), 
                      Price = runif(4320, min = -15, max = 45)
                      )

我的因子变量是Hour，我的外生变量是Wind和temp，我想预测的是Price。基本上，我有24个模型，希望能够进行滚动预测。

现在，我的数据框包含180天。我想要回溯100天，并进行一天的滚动预测，然后能够将其与实际的Price进行比较。

如果采用暴力方法，代码看起来会像这样：

# First I fit the data frame to be exactly the right length
# 100 days to start with (2015-03-21 or so), then 99, then 98.., etc. 
n <- 100 * 24

# Make the price <- NA so I can replace it with a forecast
df$Price[(nrow(df) - n): (nrow(df) - n + 24)] <- NA

# Now I make df just 81 days long, the estimation period + the first forecast
df <- df[1 : (nrow(df) - n + 24), ]

# The actual do & fit, later termed fx(df)

result <- df %>% group_by(Hour) %>% do ({
  historical <- .[!is.na(.$Price), ]
  forecasted <- .[is.na(.$Price), c("Date", "Hour", "Wind", "Temp")]
  fit <- Arima(historical$Price, xreg = historical[, 3:4], order = c(1, 1, 0))
  data.frame(forecasted[], 
             Price = forecast.Arima(fit, xreg = forecasted[3:4])$mean )
})

result

现在我想将 n 修改为99 * 24。但是如果能用循环或应用来完成这个操作并保存每个新的预测，那就太好了。不过我实在想不出怎么做。

我尝试了以下循环，但还没有成功：

# 100 days ago, forecast that day, then the next, etc.
for (n in 1:100) { 
  nx <- n * 24 * 80         # Because I want to start after 80 days
  df[nx:(nx + 23), 5] <- NA # Set prices to NA so I can forecast them
  fx(df) # do the function
  df.results[n] <- # Write the results into a vector / data frame to save them
    # and now rinse and repeat for n + 1
  }

真正令人惊叹的奖励积分，对于类似于的解决方案 :)

- Thorst

Answer 1

我先指出你的for循环中存在错误。你可能想要用(n+80)*24代替n*24*80。如果你想要包括对第81天的预测，那么你的循环计数器也应该从0到99而不是从1到100。

我将尝试为你的问题提供一个优雅的解决方案。首先，我们以与你在帖子中所做的完全相同的方式定义我们的测试数据框：

set.seed(2)
df <- data.frame(
Date = seq.POSIXt(from = as.POSIXct("2015-01-01 00:00:00"), 
                    to = as.POSIXct("2015-06-30 00:00:00"), by = "hour"))
df <- df %>% mutate(Hour = as.numeric(format(Date, "%H")) + 1, 
                    Wind = runif(4320, min = 1, max = 5000), 
                    Temp = runif(4320, min = - 20, max = 25), 
                    Price = runif(4320, min = -15, max = 45)
)

接下来，我们定义一个函数，用于对某一特定天进行预测。输入参数是考虑的数据帧和应在训练集中包含的最小训练天数(在本例中为80)。minTrainingDays+offSet+1表示我们正在预测的实际日期。请注意，我们从偏移量0开始计数。

forecastOneDay <- function(theData,minTrainingDays,offset)
{
  nrTrainingRows <- (minTrainingDays+offset)*24

  theForecast <- theData %>% 
    filter(min_rank(Date) <= nrTrainingRows+24) %>% # Drop future data that we don't need
    group_by(Hour) %>%
    do ({
      trainingData <- head(.,-1) # For each group, drop the last entry from the dataframe
      forecastData <- tail(.,1) %>% select(Date,Hour,Wind,Temp) # For each group, predict the last entry
      fit <- Arima(trainingData$Price, xreg=trainingData[,3:4], order=c(1,1,0))
      data.frame(forecastData, realPrice = tail(.,1)$Price, predictedPrice = forecast.Arima(fit,xreg=forecastData[3:4])$mean)
    })
}

我们希望预测第81-180天。换句话说，我们的训练集至少需要80天，并且想要计算偏移量为0:99的函数结果。这可以通过简单的lapply调用来完成。最后，我们将所有结果合并到一个数据框中：

# Perform one day forecasts for days 81-180
resultList <- lapply(0:99, function(x) forecastOneDay(df,80,x))
# Merge all the results
mergedForecasts <- do.call("rbind",resultList)

编辑 在审查了您的帖子和同时发布的另一个答案后，我注意到我的答案存在两个潜在问题。首先，您想要一个滚动的80天训练数据窗口。然而，在我的先前代码中，所有可用的训练数据都被用来拟合模型，而不是仅返回80天。其次，该代码无法应对夏令时变化等情况。

下面的代码解决了这两个问题。现在函数的输入更加直观：可以使用训练天数和实际预测天作为输入参数。请注意，当对日期进行操作时，POSIXlt数据格式会正确处理夏令时、闰年等问题。因为您数据框中的日期类型是POSIXct，所以我们需要进行小的类型转换，以便正确处理这些问题。

以下是新代码：

forecastOneDay <- function(theData,nrTrainingDays,predictDay) # predictDay should be greater than nrTrainingDays
{
  initialDate <- as.POSIXlt(theData$Date[1]); # First day (midnight hour)
  startDate <- initialDate # Beginning of training interval
  endDate <- initialDate # End of test interval

  startDate$mday <- initialDate$mday + (predictDay-nrTrainingDays-1) # Go back 80 days from predictday
  endDate$mday <- startDate$mday + (nrTrainingDays+1) # +1 to include prediction day

  theForecast <- theData %>% 
    filter(Date >= as.POSIXct(startDate),Date < as.POSIXct(endDate)) %>% 
    group_by(Hour) %>%
    do ({
      trainingData <- head(.,-1) # For each group, drop the last entry from the dataframe
      forecastData <- tail(.,1) %>% select(Date,Hour,Wind,Temp) # For each group, predict the last entry
      fit <- Arima(trainingData$Price, xreg=trainingData[,3:4], order=c(1,1,0))
      data.frame(forecastData, realPrice = tail(.,1)$Price, predictedPrice = forecast.Arima(fit,xreg=forecastData[3:4])$mean)
    })
}

# Perform one day forecasts for days 81-180
resultList <- lapply(81:180, function(x) forecastOneDay(df,80,x))
# Merge all the results
mergedForecasts <- do.call("rbind",resultList)

结果看起来像这样：

> head(mergedForecasts)
Source: local data frame [6 x 6]
Groups: Hour

                 Date Hour     Wind      Temp  realPrice predictedPrice
1 2015-03-22 00:00:00    1 1691.589 -8.722152 -11.207139       5.918541
2 2015-03-22 01:00:00    2 1790.928 18.098358   3.902686      37.885532
3 2015-03-22 02:00:00    3 1457.195 10.166422  22.193270      34.984164
4 2015-03-22 03:00:00    4 1414.502  4.993783   6.370435      12.037642
5 2015-03-22 04:00:00    5 3020.755  9.540715  25.440357      -1.030102
6 2015-03-22 05:00:00    6 4102.651  2.446729  33.528199      39.607848
> tail(mergedForecasts)
Source: local data frame [6 x 6]
Groups: Hour

                 Date Hour      Wind       Temp  realPrice predictedPrice
1 2015-06-29 18:00:00   19 1521.9609 13.6414797  12.884175     -6.7789109
2 2015-06-29 19:00:00   20  555.1534  3.4758159  37.958768     -5.1193514
3 2015-06-29 20:00:00   21 4337.6605  4.7242352  -9.244882     33.6817379
4 2015-06-29 21:00:00   22 3140.1531  0.8127839  15.825230     -0.4625457
5 2015-06-29 22:00:00   23 1389.0330 20.4667234 -14.802268     15.6755880
6 2015-06-29 23:00:00   24  763.0704  9.1646139  23.407525      3.8214642

- Jellen Vermeir

---

非常好的答案，正是我所需要的！ - Thorst

Answer 2

你可以使用dplyr创建可能的“滚动”数据框，方法如下：

library(dplyr)
library(lubridate)

WINDOW_SIZE_DAYS <- 80

df2 <- df %>%
  mutate(Day = yday(Date)) %>%
  replicate( n = WINDOW_SIZE_DAYS, simplify = FALSE ) %>% 
  bind_rows %>%
  group_by(Date) %>%
  mutate(Replica_Num = 1:n() ) %>%
  mutate(Day_Group_id = Day + Replica_Num - 1 ) %>%
  ungroup() %>%
  group_by(Day_Group_id) %>%
  filter( n() >= 24*WINDOW_SIZE_DAYS - 1 ) %>%
  select( -Replica_Num ) %>%
  arrange(Date) %>%
  ungroup()

基本上，这段代码会根据需要复制观察结果，并为每个80天的块分配相应的Day_Group_id。 这样做可以使用group_by(Day_Group_id)，以便在每个80天的块上单独运行模型。

随后，您可以根据需要使用它。例如，只需从上述arima代码中复制/粘贴如下：

df3 <- df2 %>%
  group_by(Day_Group_id, Hour) %>%
  arrange(Date) %>%
  do ({
    trainingData <- head(.,-1) # For each group, drop the last entry from the dataframe
    forecastData <- tail(.,1) %>% select(Date,Hour,Wind,Temp) # For each group, predict the last entry
    fit <- Arima(trainingData$Price, xreg=trainingData[,3:4], order=c(1,1,0))
    data.frame(forecastData, realPrice = tail(.,1)$Price, predictedPrice = forecast.Arima(fit,xreg=forecastData[3:4])$mean)
  })

请注意：

这里使用的是filter(n() >= 24*WINDOW_SIZE_DAYS - 1)而不是filter(n() == 24*WINDOW_SIZE_DAYS)，以选择完整的80天窗口。这是由于在2015-03-08进行夏令时调整。小时2015-03-08 02:00:00在数据集中不存在，因为它直接跳到了2015-03-08 03:00:00。

- akhmed