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我正在尝试使用多个因变量和多个自变量进行回归。基本上我House Prices在整个美国的县级都有,这是我的 IV。然后我有几个县级的其他变量(GDP, construction employment),这些构成了我的因变量。我想知道是否有一种有效的方法可以同时进行所有这些回归。我试图得到:

lm(IV1 ~ DV11 + DV21)
lm(IV2 ~ DV12 + DV22)

我想为每个自变量和每个因变量执行此操作。

编辑: OP添加了此信息以响应我的回答,现已删除,这误解了问题。

我认为我没有很好地解释这个问题,我很抱歉。每个因变量都有 2 个与之关联的自变量,它们是唯一的。所以如果我有 500 个因变量,我有 500 个唯一的自变量 1,和 500 个唯一的自变量 2。

好的,我会再试一次,如果我不能再次解释自己,我可能会放弃(哈哈)。我不知道mtcarsR中的意思是什么[this is in reference to Metrics's answer],所以让我试试这种方式。我将有 3 个数据向量,每个向量大约 500 行。我正在尝试从每行数据中构建回归。假设向量 1 是我的因变量(我试图预测的变量),向量 2 和 3 构成了我的自变量。因此,第一个回归将包含每个向量的第 1 行值,第二个回归将包含每个向量的第 2 行值,依此类推。再次感谢大家。

4

1 回答 1

3

我假设您将数据框作为 mydata。

mydata<-mtcars #mtcars is the data in R

dep<-c("mpg~","cyl~","disp~") # list of unique dependent variables with ~ 
indep1<-c("hp","drat","wt")  # list of first unique independent variables 
indep2<-c("qsec","vs","am") # list of second unique independent variables 
> myvar<-cbind(dep,indep1,indep2) # matrix of variables
> myvar
     dep     indep1 indep2
[1,] "mpg~"  "hp"   "qsec"
[2,] "cyl~"  "drat" "vs"  
[3,] "disp~" "wt"   "am" 



for (i in 1:dim(myvar)[1]){
print(paste("This is", i, "regression", "with dependent var",gsub("~","",myvar[i,1])))
k[[i]]<-lm(as.formula(paste(myvar[i,1],paste(myvar[i,2:3],collapse="+"))),mydata)
print(k[[i]]
}



 [1] "This is 1 regression with dependent var mpg"

Call:
lm(formula = as.formula(paste(myvar[i, 1], paste(myvar[i, 2:3], 
    collapse = "+"))), data = mydata)

Coefficients:
(Intercept)           hp         qsec  
   48.32371     -0.08459     -0.88658  

[1] "This is 2 regression with dependent var cyl"

Call:
lm(formula = as.formula(paste(myvar[i, 1], paste(myvar[i, 2:3], 
    collapse = "+"))), data = mydata)

Coefficients:
(Intercept)         drat           vs  
     12.265       -1.421       -2.209  

[1] "This is 3 regression with dependent var disp"

Call:
lm(formula = as.formula(paste(myvar[i, 1], paste(myvar[i, 2:3], 
    collapse = "+"))), data = mydata)

Coefficients:
(Intercept)           wt           am  
    -148.59       116.47        11.31  

注意:您可以对大量变量使用相同的过程。

替代方法:

受到哈德利的回答的启发我使用函数Map来解决上述问题:

dep<-list("mpg~","cyl~","disp~") # list of unique dependent variables with ~ 
indep1<-list("hp","drat","wt")  # list of first unique independent variables 
indep2<-list("qsec","vs","am") # list of second unique independent variables
Map(function(x,y,z) lm(as.formula(paste(x,paste(list(y,z),collapse="+"))),data=mtcars),dep,indep1,indep2)
[[1]]

Call:
lm(formula = as.formula(paste(x, paste(list(y, z), collapse = "+"))), 
    data = mtcars)

Coefficients:
(Intercept)           hp         qsec  
   48.32371     -0.08459     -0.88658  


[[2]]

Call:
lm(formula = as.formula(paste(x, paste(list(y, z), collapse = "+"))), 
    data = mtcars)

Coefficients:
(Intercept)         drat           vs  
     12.265       -1.421       -2.209  


[[3]]

Call:
lm(formula = as.formula(paste(x, paste(list(y, z), collapse = "+"))), 
    data = mtcars)

Coefficients:
(Intercept)           wt           am  
    -148.59       116.47        11.31  
于 2013-08-05T23:01:35.443 回答