如何使用循环从线性回归中提取系数输出 [英] How to extract coefficients outputs from a linear regression with loop

问题描述

我想知道如何循环执行n次回归，并且每次使用一组不同的变量时，提取一个data.frame，其中每一列都是回归，每一行代表一个变量.

I would like to know how I can loop a regression n times, and in each time with a different set of variables, extract a data.frame where each column is a regression and each row represent a variable.

对于我来说，我的data.frame为:

dt_deals <- data.frame(Premium=c(1,3,4,5),Liquidity=c(0.2,0.3,1.5,0.8),Leverage=c(1,3,0.5,0.7))

但是我还有另一个解释性的哑变量，称为hubris，它是二项分布的乘积，均值为0.25.像这样:

But I have another explanatory dummy variable called hubris, that is the product of a binomial distribution, with 0.25 of mean. Like that:

n <- 10 
hubris_dataset <- data.frame(replicate(n, rbinom(4,1,0.25))

从这个意义上讲，我需要对hubris进行n次模拟，因此我可以对n个具有不同随机二项式分布的集合进行回归，并且每个分布的输出都必须放在data.frame 到目前为止，我可以做到这一点:

In this sense, what I need is to make n simulation of hubris, so I can, make n regression each one with a different set of random binomial distribution and the output of each distribution needs to be put in a data.frame So far I could reach this:

# define n as the number of simulations i want
n=10
# define beta as a data.frame to put every coefficient from the lm regression
beta=NULL

for(i in 1:n) {
  dt_deals2 <- dt_deals
  beta[[i]] <- coef(lm(dt_deals$Premium ~ dt_deals$Liquidity + dt_deals$Leverage + hubris_dataset[,i], data=dt_deals2))
  beta <- cbind(reg$coefficients)
}

但是以这种方式，它仅生成第一组系数，而不会为data.frame再创建十列.

But this way it only generate the first set of coefficient, and doesn't make another ten columns for the data.frame.

推荐答案

@jogo给出了一种更改for循环方法并使用sapply的想法，并将对象beta更改为list().结果就是这样:

@jogo give an idea to change the for-loop method and use sapply, and change the object beta to list(). This was the result:

beta <- sapply(1:n, function(i) coef(lm(Premium ~ Liquidity +Leverage+ hubris_dataset[,i], data=dt_deals2)))

它奏效了

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