具有不同系数的模型的存储系数估计值

Question

我试图存储不同模型的系数估计值。为了说明我的问题，下面是一个例子。

library(fpp) 

creditlog <- data.frame(score=credit$score, 
    log.savings=log(credit$savings+1), 
    log.income=log(credit$income+1), 
    log.address=log(credit$time.address+1), 
    log.employed=log(credit$time.employed+1)) 

fit_1 <-lm(score ~ log.income + log.address + log.employed , data=creditlog) 
fit_2 <-lm(score ~ log.savings + log.employed , data=creditlog) 
fit_3 <-lm(score ~ log.address + log.employed , data=creditlog) 
fit_4 <- lm(score ~ log.income + log.address , data=creditlog) 


coef_1 <-summary(fit_1)$coef[,1] 
coef_2 <-summary(fit_2)$coef[,1] 
coef_3 <-summary(fit_3)$coef[,1] 
coef_4 <-summary(fit_4)$coef[,1] 
> coef_1 
(Intercept) log.income log.address log.employed 
    -14.957037 10.082396  3.353521  1.049130 
> coef_2 
(Intercept) log.savings log.employed 
    24.34323  11.28698  1.92655 
> coef_3 
(Intercept) log.address log.employed 
    26.115064  3.438382  1.213017 
> coef_4 
(Intercept) log.income log.address 
    -13.38037 10.23459  3.58023 

如果我尝试rbind，我得到

 (Intercept) log.income log.address log.employed 
coef_1 -14.95704 10.082396 3.353521  1.04913 
coef_2 24.34323 11.286978 1.926550  24.34323 
coef_3 26.11506 3.438382 1.213017  26.11506 
coef_4 -13.38037 10.234590 3.580230 -13.38037 
Warning message: 
In rbind(coef_1, coef_2, coef_3, coef_4) : 
    number of columns of result is not a multiple of vector length (arg 2) 
> 

这是不正确的答案。我需要的是类似的，

 (Intercept) log.savings log.income log.address log.employed 
fit_1 -14.957037 NA   10.082396 3.353521  1.04913 
fit_2 24.34323 11.28698  NA   NA    1.92655 
fit_3 26.115064 NA   NA   3.438382  1.213017 
fit_4 -13.38037 NA   10.23459 3.58023  NA 

在此先感谢。

Answer 1

你可以改变矢量data.frames和使用dplyr的rbind_all：

library(dplyr) 
# transforming in data.frames 
coef_1 <- as.data.frame(t(summary(fit_1)$coef[,1])) 
coef_2 <- as.data.frame(t(summary(fit_2)$coef[,1])) 
coef_3 <- as.data.frame(t(summary(fit_3)$coef[,1])) 
coef_4 <- as.data.frame(t(summary(fit_4)$coef[,1])) 

# binding them all 
coefs <- rbind_all(list(coef_1, coef_2, coef_3, coef_4)) 
row.names(coefs) <- c("fit_1", "fit_2", "fit_3", "fit_4") 
coefs 


     (Intercept) log.income log.address log.employed log.savings 
fit_1 -14.95704 10.08240 3.353521  1.049130   NA 
fit_2 24.34323   NA   NA  1.926550 11.28698 
fit_3 26.11506   NA 3.438382  1.213017   NA 
fit_4 -13.38037 10.23459 3.580230   NA   NA 

Answer 2

下面就可以利用基础R以列表和部分列组合在一起使用：

c1 = data.frame(a=1,b=2,d=3) 
c2 = data.frame(b=2,c=3) 
c3 = data.frame(a=4,d=5) 

cc = data.frame(a=numeric(), b=numeric(), c=numeric(), d=numeric()) 
ff = function(vect, cc){ 
    n = nrow(cc)+1 
    for(i in 1:length(vect)){ 
     cc[n,names(vect)[i]] = vect[i] 
    } 
    cc 
} 

cc=ff(c1, cc) 
cc=ff(c2, cc) 
cc=ff(c3, cc) 
cc 
    a b c d 
1 1 2 NA 3 
2 NA 2 3 NA 
3 4 NA NA 5 

Answer 3

这里有一个办法用相对较短的一段代码来完成。它使用coef直接从模型对象中提取系数，并使用lapply来避免为每个模型对象重复相同的代码。 rbind.fill负责将每个系数值在正确的列：

library(plyr) # For the rbind.fill function 

fits = rbind.fill(lapply(list(fit_1, fit_2, fit_3, fit_4), 
        function(x) as.data.frame(t(coef(x))))) 

fits 
    (Intercept) log.income log.address log.employed log.savings 
1 -14.95704 10.08240 3.353521  1.049130   NA 
2 24.34323   NA   NA  1.926550 11.28698 
3 26.11506   NA 3.438382  1.213017   NA 
4 -13.38037 10.23459 3.580230   NA   NA 

如果你有很多超过四个模型对象，不想键入他们的名字，你可以通过编程引用对象的名称。例如，如果您有模型对象fit_1至fit_20，则将list(fit_1, fit_2, fit_3, fit_4)替换为mget(paste0("fit_", 1:20))。 mget接受文本字符串的向量并返回具有这些名称的对象。