在管道中使用 dplyr 中的 group_by() 和 predict.lm 和 do() 进行年份线性外推

Question

我想在管道内应用一年的线性外推。我想做的和这个simple example without grouping非常相似。但在管道内并使用dplyr::group_by()。有一些示例like this one、this one 或this one。但我无法获得理想的输出。

可重现的例子：

test.frame <- data.frame(Country = 
rep(c("Austria", "Brazil", "Canada"), each = 3, times = 3), 
  Entity = rep(c("CO2","CH4","N2O"), times = 9),
  Year = rep(c(1990:1992), each = 9),
  value = runif(27, 1,5))

test.frame2 <- data.frame(Country = 
rep(c("Austria", "Brazil", "Canada"), each = 3), 
    Entity =  rep(c("CO2","CH4","N2O"), times = 3),
    Year = rep(c(1993), each = 3),
    value = 0)

results_frame <- test.frame %>% 
  dplyr::bind_rows(test.frame2)

我有两个分组类别（国家和实体），我想使用 1990 年到 1992 年的值来使用线性外推法填充 1993 年的值。 在this的基础上，我可以估计线性模型：

linear_model <- test.frame %>%  
dplyr::group_by(Country, Entity) %>% 
lm(value ~ Year, data=.)

results <- predict.lm(linear_model, test.frame2)

但是，results 没有显示理想的输出。所以按照here提出的解决方案，我尝试以下方法：

results_frame <- test.frame %>%
  dplyr::group_by(Country, Entity) %>% 
  do(lm( value ~ Year , data = test.frame)) %>%
  predict.lm(linear_model, test.frame2) %>% 
  bind_rows(test.frame)

但它不起作用，而是我得到了

错误：Results 1, 2, 3, 4, 5, ... 必须是数据帧，而不是 lm

任何帮助将不胜感激！

Answer 1

在拟合和预测时必须小心使用正确的数据：

library(dplyr)
set.seed(42)
test.frame <- tibble(Country = rep(c("Austria", "Brazil", "Canada"), each = 3, times = 3), 
                         Entity = rep(c("CO2","CH4","N2O"), times = 9),
                         Year = rep(c(1990:1992), each = 9),
                         value = runif(27, 1,5))

test.frame %>%
  group_by(Country, Entity) %>% 
  do(lm( value ~ Year , data = .) %>% 
       predict(., tibble(Year = 1993)) %>%
       tibble(Year = 1993, value = .)) %>%
  bind_rows(test.frame)
#> # A tibble: 36 x 4
#> # Groups:   Country, Entity [9]
#>    Country Entity  Year value
#>    <fct>   <fct>  <dbl> <dbl>
#>  1 Austria CH4     1993 2.10 
#>  2 Austria CO2     1993 2.03 
#>  3 Austria N2O     1993 6.02 
#>  4 Brazil  CH4     1993 4.90 
#>  5 Brazil  CO2     1993 0.771
#>  6 Brazil  N2O     1993 5.28 
#>  7 Canada  CH4     1993 4.69 
#>  8 Canada  CO2     1993 0.729
#>  9 Canada  N2O     1993 1.49 
#> 10 Austria CO2     1990 4.66 
#> # ... with 26 more rows

Answer 2

您可以使用嵌套的 data.frames 执行以下操作。这种解决方案更通用，因为不需要在预测后重新创建test.frame2，并且可能存在多个自变量：

library(tidyverse)
test.frame %>%
  group_by(Country, Entity) %>%
  nest() %>%
  inner_join(test.frame2 %>% select(-value) %>% group_by(Country, Entity) %>% nest(),
             by = c("Country", "Entity")) %>%
  mutate(model = data.x %>% map(~lm(value ~ Year, data=.)),
         value = map2(model, data.y, predict)) %>%
  select(-data.x, -model) %>%
  unnest() %>%
  bind_rows(test.frame, .)

结果：

   Country Entity Year      value
1  Austria    CO2 1990  3.6245955
2  Austria    CH4 1990  3.3857752
3  Austria    N2O 1990  1.4798741
4   Brazil    CO2 1990  2.5865668
5   Brazil    CH4 1990  1.3271481
6   Brazil    N2O 1990  4.4537926
7   Canada    CO2 1990  4.7295768
8   Canada    CH4 1990  4.5255033
9   Canada    N2O 1990  2.3129381
10 Austria    CO2 1991  4.8810838
11 Austria    CH4 1991  4.9950455
12 Austria    N2O 1991  2.1288504
13  Brazil    CO2 1991  4.7767443
14  Brazil    CH4 1991  2.0315449
15  Brazil    N2O 1991  1.9307966
16  Canada    CO2 1991  4.6831029
17  Canada    CH4 1991  2.2761538
18  Canada    N2O 1991  3.0856428
19 Austria    CO2 1992  3.1223000
20 Austria    CH4 1992  4.7715588
21 Austria    N2O 1992  1.5733608
22  Brazil    CO2 1992  2.9463442
23  Brazil    CH4 1992  1.9569259
24  Brazil    N2O 1992  1.4428006
25  Canada    CO2 1992  3.0750847
26  Canada    CH4 1992  1.4635521
27  Canada    N2O 1992  2.8061861
28 Austria    CO2 1993  3.3736976
29 Austria    CH4 1993  5.7699101
30 Austria    N2O 1993  1.8208485
31  Brazil    CO2 1993  3.7963291
32  Brazil    CH4 1993  2.4016508
33  Brazil    N2O 1993 -0.4018621
34  Canada    CO2 1993  2.5080960
35  Canada    CH4 1993 -0.3068815
36  Canada    N2O 1993  3.2281704