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仅从forecast()[重复]中提取预测值

broom arima forecasting dplyr r

Davide Lorino · 技术社区 · 8 年前

这个问题已经有了答案:

in R, extract part of object from list 1个答案

我有一个这样的数据框架:

> head(forecasts)
$`1_1`
         Point Forecast    Lo 80    Hi 80    Lo 95    Hi 95
Dec 2016       7.370299 7.335176 7.405422 7.316583 7.424015

$`1_10`
         Point Forecast    Lo 80    Hi 80    Lo 95    Hi 95
Dec 2016       7.396656 7.359845 7.433467 7.340359 7.452953

$`1_2`
         Point Forecast    Lo 80    Hi 80    Lo 95    Hi 95
Dec 2016       7.780033 7.752462 7.807605 7.737866 7.822201

$`1_3`
         Point Forecast    Lo 80    Hi 80    Lo 95    Hi 95
Dec 2016       7.216894 7.178896 7.254892 7.158781 7.275007

$`1_4`
         Point Forecast    Lo 80    Hi 80    Lo 95    Hi 95
Dec 2016       7.501195 7.465049 7.537341 7.445915 7.556475

$`1_5`
         Point Forecast    Lo 80    Hi 80    Lo 95    Hi 95
Dec 2016       7.455131 7.424918 7.485345 7.408924 7.501339

我只想提取 Point Forecast

呼叫 str(forecasts) 返回大量输出,这只是“预测”变量中89个列表之一的输出:

$ 9_9  :List of 10
  ..$ method   : chr "ARIMA(0,0,0)(0,1,0)[12] with drift"
  ..$ model    :List of 19
  .. ..$ coef     : Named num 0.00965
  .. .. ..- attr(*, "names")= chr "drift"
  .. ..$ sigma2   : num 0.0047
  .. ..$ var.coef : num [1, 1] 1.24e-06
  .. .. ..- attr(*, "dimnames")=List of 2
  .. .. .. ..$ : chr "drift"
  .. .. .. ..$ : chr "drift"
  .. ..$ mask     : logi TRUE
  .. ..$ loglik   : num 33.4
  .. ..$ aic      : num -62.7
  .. ..$ arma     : int [1:7] 0 0 0 0 12 0 1
  .. ..$ residuals: Time-Series [1:38] from 2014 to 2017: 0.00546 0.00583 0.006 0.00564 0.00563 ...
  .. ..$ call     : language .f(y = .x[[i]], x = list(x = c(5.4677292870219, 5.85045765518954, 6.02852764863892, 5.67941181324485,  5.67526620| __truncated__ ...
  .. ..$ series   : chr ".x[[i]]"
  .. ..$ code     : int 0
  .. ..$ n.cond   : int 0
  .. ..$ nobs     : int 26
  .. ..$ model    :List of 10
  .. .. ..$ phi  : num(0) 
  .. .. ..$ theta: num(0) 
  .. .. ..$ Delta: num [1:12] 0 0 0 0 0 0 0 0 0 0 ...
  .. .. ..$ Z    : num [1:13] 1 0 0 0 0 0 0 0 0 0 ...
  .. .. ..$ a    : num [1:13] 0.0677 5.6916 5.7073 5.692 5.7108 ...
  .. .. ..$ P    : num [1:13, 1:13] 0 0 0 0 0 0 0 0 0 0 ...
  .. .. ..$ T    : num [1:13, 1:13] 0 1 0 0 0 0 0 0 0 0 ...
  .. .. ..$ V    : num [1:13, 1:13] 1 0 0 0 0 0 0 0 0 0 ...
  .. .. ..$ h    : num 0
  .. .. ..$ Pn   : num [1:13, 1:13] 1 0 0 0 0 0 0 0 0 0 ...
  .. ..$ xreg     : int [1:38, 1] 1 2 3 4 5 6 7 8 9 10 ...
  .. .. ..- attr(*, "dimnames")=List of 2
  .. .. .. ..$ : NULL
  .. .. .. ..$ : chr "drift"
  .. ..$ bic      : num -60.2
  .. ..$ aicc     : num -62.2
  .. ..$ x        : Time-Series [1:38] from 2014 to 2017: 5.47 5.85 6.03 5.68 5.68 ...
  .. ..$ fitted   : Time-Series [1:38] from 2014 to 2017: 5.46 5.84 6.02 5.67 5.67 ...
  .. ..- attr(*, "class")= chr [1:2] "ARIMA" "Arima"
  ..$ level    : num [1:2] 80 95
  ..$ mean     : Time-Series [1:1] from 2017 to 2017: 6.32
  ..$ lower    : Time-Series [1, 1:2] from 2017 to 2017: 6.23 6.18
  .. ..- attr(*, "dimnames")=List of 2
  .. .. ..$ : NULL
  .. .. ..$ : chr [1:2] "80%" "95%"
  ..$ upper    : Time-Series [1, 1:2] from 2017 to 2017: 6.4 6.45
  .. ..- attr(*, "dimnames")=List of 2
  .. .. ..$ : NULL
  .. .. ..$ : chr [1:2] "80%" "95%"
  ..$ x        : Time-Series [1:38] from 2014 to 2017: 5.47 5.85 6.03 5.68 5.68 ...
  ..$ series   : chr ".x[[i]]"
  ..$ fitted   : Time-Series [1:38] from 2014 to 2017: 5.46 5.84 6.02 5.67 5.67 ...
  ..$ residuals: Time-Series [1:38] from 2014 to 2017: 0.00546 0.00583 0.006 0.00564 0.00563 ...
  ..- attr(*, "class")= chr "forecast"

1 回复 | 直到 8 年前

akrun 8 年前

我们可以从 list 使用 lapply

lapply(forecasts, function(x) as.numeric(x$mean, na.rm = TRUE)))

如果所有列表元素中的预测数相同,则可以将其转换为 matrix 或 data.frame

sapply(forecasts, `[[`, "mean")

或者使用 tidyverse

library(tidyverse)
forecasts %>% 
      map_df(~ .x$mean %>% 
              as.numeric)