Question

我正在测试可分组数据的不同线性回归模型（grp1 和 grp2）。

library(rcompanion)
library(dplyr)
library(tidyr)
  
#some sample data
sampleDF <- data.frame(grp1=append(rep("foo",6),rep("bar",6)), 
                       grp2=append(rep("A",3),rep("B",3)), 
                       x=seq(3)-rnorm(3), 
                       y=seq(3)*3+1)

#some models
model1 <- y ~ x
model2 <- y ~ x - 1

#do the fitting
sampleDF %>%
  group_by(grp1, grp2) %>% 
  do(model1 = lm(model1, data = .),
     model2 = lm(model2, data = .)) %>% 
  ungroup() -> fittedModels

这产生：

 fittedModels
# A tibble: 4 x 4
  grp1  grp2  model1 model2
* <fct> <fct> <list> <list>
1 bar   A     <lm>   <lm>  
2 bar   B     <lm>   <lm>  
3 foo   A     <lm>   <lm>  
4 foo   B     <lm>   <lm>

使用 broom 可以很好地将模型收集到数据框中，例如：

fittedModels %>% rowwise() %>% broom::tidy(model2)

# A tibble: 4 x 7
# Groups:   grp1, grp2 [4]
  grp1  grp2  term  estimate std.error statistic p.value
  <fct> <fct> <chr>    <dbl>     <dbl>     <dbl>   <dbl>
1 bar   A     x         3.43     0.124      27.7 0.00130
2 bar   B     x         3.43     0.124      27.7 0.00130
3 foo   A     x         3.43     0.124      27.7 0.00130
4 foo   B     x         3.43     0.124      27.7 0.00130

我想比较使用不同方法的模型。库 rcompanion 提供了方法 compareLM() 和 pairwiseModelAnova()。

尝试：

fittedModels %>% summarise(compareLM(model1, model2))
> Error in formula.default(fits[[i]]) : invalid formula

rowwise(fittedModels) %>% summarise(compareLM(model1, model2))
> Error: Column `compareLM(model1, model2)` must be length 1 (a summary value), not 2

仔细检查发现：

fittedModels$model1[[1]]

Call:
lm(formula = y ~ x, data = .)

Coefficients:
(Intercept)            x  
          0            3  

m.1 <- fittedModels$model1[1]
m.2 <- fittedModels$model2[1]
compareLM(m.1,m.2)
> Error in formula.default(fits[[i]]) : invalid formula

这可能表明管道存在问题？！

有什么办法可以解决这个问题吗？

Answer 1

我发现您的数据有错误，但在将随机数添加到 sampleDF$x 时它会起作用。

library(rcompanion)
library(dplyr)
library(tidyr)
library(broom)
library(purrr)

#some sample data
set.seed(1)
sampleDF <- data.frame(grp1=append(rep("foo",6),rep("bar",6)), 
                       grp2=append(rep("A",3),rep("B",3)), 
                       x=seq(3) - rnorm(3), 
                       y=seq(3)*3+1)

#some models
model1 <- y ~ x
model2 <- y ~ x - 1

# Use purrr
sampleDF %>%
  group_by(grp1, grp2) %>%
  nest() %>%
  mutate(model1 = purrr::map(data, ~lm(model1, data = .))) %>%
  mutate(model2 = purrr::map(data, ~lm(model2, data = .))) %>%
  mutate(tidy_model1 = purrr::map(model1, broom::tidy)) %>%
  mutate(tidy_model2 = purrr::map(model2, broom::tidy)) %>%
  mutate(compare_m1m2 = purrr::map2(model1, model2, ~compareLM(.x, .y)))

> # A tibble: 4 x 8
> # Groups:   grp1, grp2 [4]
> grp1  grp2  data             model1 model2 tidy_model1      tidy_model2      > compare_m1m2    
> <chr> <chr> <list>           <list> <list> <list>           <list>           > <list>          
> 1 foo   A     <tibble [3 × 2]> <lm>   <lm>   <tibble [2 × 5]> <tibble [1 × 5]> <named list [2]>
> 2 foo   B     <tibble [3 × 2]> <lm>   <lm>   <tibble [2 × 5]> <tibble [1 × 5]> <named list [2]>
> 3 bar   A     <tibble [3 × 2]> <lm>   <lm>   <tibble [2 × 5]> <tibble [1 × 5]> <named list [2]>
> 4 bar   B     <tibble [3 × 2]> <lm>   <lm>   <tibble [2 × 5]> <tibble [1 × 5]> <named list [2]>

线性回归和模型比较与分组数据

1 个答案: