Question

我的任务是对虚拟变量进行一些研究。这是一个R代码：

parameters_estimation2 <- function(n) 
{
  age <- sample(20:40, n, replace=TRUE)
  male <- sample(0:1, n, replace=TRUE)
  education <- sample(0:6, n, replace=TRUE)

  experience <- floor(rexp(n, 0.2))
  for(i in 1:n) 
  {if(experience[i]>15) {
    experience[i] <- floor(rexp(1, 0.2))
    if(experience[i]>15) { i <- i-1  }
  }}

  sqexperience <- experience*experience
  e <- rnorm(n, 0, 4)

  B0 <- -200; B1 <- 15; B2 <- 100; B3 <-10; B4 <- 5; B5 <-20;
  wage <- B0 + B1*age + B2*male + B3*education+ B4*experience+ B5*sqexperience+e

  #Dummy making
  expe1 <- c(rep(0,n)); expe2 <- c(rep(0,n)); expe3 <- c(rep(0,n)); expe4 <- c(rep(0,n));
  expe5 <- c(rep(0,n)); expe6 <- c(rep(0,n)); expe7 <- c(rep(0,n)); expe8 <- c(rep(0,n)); 
  expe9 <- c(rep(0,n)); expe10 <- c(rep(0,n)); expe11 <- c(rep(0,n)); expe12 <- c(rep(0,n));
  expe13 <- c(rep(0,n)); expe14 <- c(rep(0,n)); expe15 <- c(rep(0,n));  
  for(i in 1:n) 
    {
      if(experience[i]==1) { expe1[i] <-1
      } else if(experience[i]==2) { expe2[i] <-1
      } else if(experience[i]==3) { expe3[i] <-1
      } else if(experience[i]==4) { expe4[i] <-1
      } else if(experience[i]==5) { expe5[i] <-1
      } else if(experience[i]==6) { expe6[i] <-1
      } else if(experience[i]==7) { expe7[i] <-1
      } else if(experience[i]==8) { expe8[i] <-1
      } else if(experience[i]==9) { expe9[i] <-1
      } else if(experience[i]==10) { expe10[i] <-1
      } else if(experience[i]==11) { expe11[i] <-1
      } else if(experience[i]==12) { expe12[i] <-1
      } else if(experience[i]==13) { expe13[i] <-1
      } else if(experience[i]==14) { expe14[i] <-1
      } else if(experience[i]==15) { expe15[i] <-1
      }}

  regression<-lm(wage~age+male+education+expe1+expe2+expe3+expe4+expe5+expe6+expe7+expe8+expe9+expe10+expe11+expe12+expe13+expe14+expe15)
  return(summary(regression)$coefficients[,"Estimate"])  
}

times <- 1000
size <- rep(200, times)
koeficientai1 <-mapply(parameters_estimation2, size)
blah <- as.data.table(koeficientai1)
beta0sample200d <- mean(koeficientai1[,"(Intercept)"])

问题是在最后一行我得到了：

Error in koeficientai1[, "(Intercept)"] : incorrect number of dimensions

我认为问题是koeficientai1是大型列表。但是后来我只用5个变量尝试了另一个lm回归，代码正常工作，我得到了简单的数据框。

Answer 1

尝试用

替换最后一行

beta0sample200d <- mean(sapply(koeficientai1, function(x) x["(Intercept)"]))

koeficientai1是一个列表，但您尝试将其作为data.frame访问，因此出现错误消息。

sapply从(Intercept)中的每个列表元素中提取名为koeficientai1的元素（在您的情况下，每个列表元素都是一个命名向量）并返回包含结果的向量。

lm回归和大列表中有很多变量

1 个答案: