从具有多个模型的 lm lapply 循环中提取最终 p 值统计

我认为没有存储 p 值，您需要从 fstatistics 中计算它，可能是这样的：

set.seed(111)
testdata = data.frame(Kpl = rnorm(100),Y1 = rnorm(100),Y2 = rnorm(100),Y3 = rnorm(100))

IV = colnames(testdata)[-1]
DV = "Kpl"

linear_sumMary <- lapply(IV,function(X){
         sumMary(lm(reformulate(response=DV,termlabels=X),data=testdata))
                         })

names(linear_sumMary) = IV

tab = lapply(IV,function(X){
  p = with(
       linear_sumMary[[x]],pf(fstatistic[1],fstatistic[2],fstatistic[3],lower.tail=falSE)
          )
  data.frame(IV = x,p = p)
})

do.call(rbind,tab)

       IV         p
value  Y1 0.5757187
value1 Y2 0.4922582
value2 Y3 0.4009439

检查例如第一个摘要：

linear_sumMary[[1]]

Call:
lm(formula = reformulate(response = DV,termlabels = X),data = testdata)

Residuals:
     Min       1Q   Median       3Q      Max 
-2.94515 -0.73325  0.05448  0.57901  2.76026 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)
(Intercept) -0.01382    0.10747  -0.129    0.898
Y1          -0.05950    0.10597  -0.562    0.576

Residual standard error: 1.075 on 98 degrees of freedom
Multiple R-squared:  0.003207,Adjusted R-squared:  -0.006964 
F-statistic: 0.3153 on 1 and 98 DF,p-value: 0.5757

好的，我按以下方式编辑了我的代码：

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

df %>%    # Solution 1
  pivot_longer(-X) %>%
  group_split(Name) %>%
  set_names(nm = map(.,~ first(.x$Name))) %>%
  map(~ tidy(lm(X ~ value,data = .))) %>%
  bind_rows(.id = "var") %>%
  filter(term == "value")

# A tibble: 4 x 6
  var   term  estimate std.error statistic p.value
  <chr> <chr>    <dbl>     <dbl>     <dbl>   <dbl>
1 Y1    value  -0.155      0.271   -0.571    0.579
2 Y2    value   0.883      0.877    1.01     0.335
3 Y3    value   0.0341     0.552    0.0618   0.952
4 Y4    value  -0.158      0.469   -0.337    0.743

或者你可以使用这个：

df %>%    # Solution 2
  pivot_longer(Y1:Y4) %>%
  group_by(Name) %>%
  arrange(.by_group = TRUE) %>% 
  nest() %>%
  mutate(models = map(data,~ lm(X ~ value,data = .)),glance = map(models,glancE)) %>%
  unnest(glancE)

# A tibble: 4 x 15
# Groups:   name [4]
  name  data    models r.squared adj.r.squared sigma statistic p.value    df logLik   AIC   BIC
  <chr> <list>  <list>     <dbl>         <dbl> <dbl>     <dbl>   <dbl> <dbl>  <dbl> <dbl> <dbl>
1 Y1    <tibbl~ <lm>    0.0288        -0.0595  0.772   0.327     0.579     1  -14.0  34.0  35.7
2 Y2    <tibbl~ <lm>    0.0845         0.00124 0.750   1.01      0.335     1  -13.6  33.2  34.9
3 Y3    <tibbl~ <lm>    0.000348      -0.0905  0.783   0.00382   0.952     1  -14.2  34.4  36.1
4 Y4    <tibbl~ <lm>    0.0102        -0.0798  0.779   0.113     0.743     1  -14.1  34.2  35.9
# ... with 3 more variables: deviance <dbl>,df.residual <int>,nobs <int>

我知道您已经得到了答案，但在这里我提出了另外 2 个解决方案。认为学习处理问题的替代方法可能没问题，并感谢您的提问，非常好。