在 R 中查找具有最低 AIC 的模型（从 for 循环返回）

Question

我正在尝试寻找 AIC 最低的模型。模型是 return 从两个 for 循环中编辑出来的，这使得列的组合成为可能。我无法制作具有最低 AIC 的函数 return 模型。下面的代码演示了我被卡住的地方：

rm(list = ls())

data <- iris

data <- data[data$Species %in% c("setosa", "virginica"),]

data$Species = ifelse(data$Species == 'virginica', 0, 1)

mod_headers <- names(data[1:ncol(data)-1])

f <- function(mod_headers){
    for(i in 1:length(mod_headers)){
    tab <- combn(mod_headers,i)
    for(j in 1:ncol(tab)){
      tab_new <- c(tab[,j])
      mod_tab_new <- c(tab_new, "Species")
      model <- glm(Species ~., data=data[c(mod_tab_new)], family = binomial(link = "logit"))
    }
    }
  best_model <- model[which(AIC(model)[order(AIC(model))][1])]
  print(best_model)
}

f(mod_headers)

有什么建议吗？谢谢！

Answer 1

我用向量化替代方案替换了你的 for 循环

library(tidyverse)
library(iterators)
# Column names you want to use in glm model, saved as list
whichcols <- Reduce("c", map(1:length(mod_headers), ~lapply(iter(combn(mod_headers,.x), by="col"),function(y) c(y))))

# glm model results using selected column names, saved as list
models <- map(1:length(whichcols), ~glm(Species ~., data=data[c(whichcols[[.x]], "Species")], family = binomial(link = "logit")))

# selects model with lowest AIC
best <- models[[which.min(sapply(1:length(models),function(x)AIC(models[[x]])))]]

输出

Call:  glm(formula = Species ~ ., family = binomial(link = "logit"), 
data = data[c(whichcols[[.x]], "Species")])

Coefficients:
 (Intercept)  Petal.Length  
       55.40        -17.17  

Degrees of Freedom: 99 Total (i.e. Null);  98 Residual
Null Deviance:      138.6 
Residual Deviance: 1.208e-09    AIC: 4

Answer 2

使用你的循环，将所有模型放在一个列表中。然后计算所有这些模型的 AIC。最后 return 具有最小 AIC 的模型。

f <- function(mod_headers) {

  models <- list()
  k <- 1
  for (i in 1:length(mod_headers)) {
    tab <- combn(mod_headers, i)
    for(j in 1:ncol(tab)) {
      mod_tab_new <- c(tab[, j], "Species")
      models[[k]] <- glm(Species ~ ., data = data[mod_tab_new], 
                         family = binomial(link = "logit"))
      k <- k + 1
    }
  }

  models[[which.min(sapply(models, AIC))]]
}

Answer 3

glm() 使用迭代重新加权最小二乘算法。该算法在收敛之前达到最大迭代次数 - 更改此参数有助于您的情况：

 glm(Species ~., data=data[mod_tab_new], family = binomial(link = "logit"), control = list(maxit = 50))

使用 which 时还有另一个问题，我在每个模型拟合后将其替换为 if，以便与目前为止的最低 AIC 进行比较。但是，我认为有比这种 for-loop 方法更好的解决方案。

f <- function(mod_headers){
  lowest_aic <- Inf     # added
  best_model <- NULL    # added

  for(i in 1:length(mod_headers)){
    tab <- combn(mod_headers,i)
    for(j in 1:ncol(tab)){
      tab_new <- tab[, j]
      mod_tab_new <- c(tab_new, "Species")
      model <- glm(Species ~., data=data[mod_tab_new], family = binomial(link = "logit"), control = list(maxit = 50))
      if(AIC(model) < lowest_aic){ # added
        lowest_aic <- AIC(model)   # added
        best_model <- model        # added
      }
    }
  }
  return(best_model)
}

在 R 中查找具有最低 AIC 的模型（从 for 循环返回）

Finding model (returned from for loops) with lowest AIC in R

for-loop

r

glm

model-comparison