简单线性回归循环

如何解决简单线性回归循环

我的数据集中有很多预测变量。我想对每个预测变量执行简单的线性回归。所以我做了一个循环。我的代码如下：

m = ncol(finalmev)
predictorlist = colnames(finalmev)[2:m]

for (i in predictorlist){
    model <- summary(lm(paste("ODR ~",i[[1]]),data=finalmev))
}

但是，在我运行循环后，我收到如下错误：

> for (i in predictorlist){
+     model <- summary(lm(paste("ODR ~",data=finalmev))
+ } Error in str2lang(x) : <text>:1:25: unexpected numeric constant 1: ODR ~ Overnight.Deposit 1

这个错误是什么意思？我的代码或数据有问题吗？

解决方法

当前代码在每次迭代时都会覆盖 model。您可能想要创建一个列表来存储它们。

predictorlist = colnames(finalmev)[-1]
model_list <- vector('list',length(predictorlist))

for (i in seq_along(predictorlist)) {
  model_list[[i]] <- summary(lm(paste("ODR ~",predictorlist[i]),data=finalmev))
}

或者使用 lapply -

result <- lapply(predictorlist,function(x) summary(lm(paste("ODR ~",x),data=finalmev))

您似乎有一个名称中带有空格的列。因此，您需要如下所示的引号：

# create a data set
set.seed(1)
finalmev <- data.frame(ODR = 1:4,`Overnight.Deposit 1` = rnorm(4),`Overnight.Deposit 2` = rnorm(4),check.names = FALSE)

# reproduce the error
predictorlist <- colnames(finalmev)[2:NCOL(finalmev)]

for (i in predictorlist){
  model <- summary(lm(paste("ODR ~",i[[1]]),data=finalmev))
}
#R> Error in str2lang(x) : <text>:1:25: unexpected numeric constant
#R> 1: ODR ~ Overnight.Deposit 1
#R>                             ^

# fix the error using quotes
for (i in predictorlist)
  model <- summary(lm(sprintf("ODR ~ `%s`",data=finalmev))

# actually save all the output as pointed out by Ronak Shah
res <- lapply(
  tail(colnames(finalmev),-1),function(x) eval(bquote(summary(lm(.(sprintf("ODR ~ `%s`",x)),data=finalmev)))))

# show the result
res
#R> [[1]]
#R> 
#R> Call:
#R> lm(formula = "ODR ~ `Overnight.Deposit 1`",data = finalmev)
#R> 
#R> Residuals:
#R>       1       2       3       4 
#R> -0.9534 -0.5809  1.2087  0.3256 
#R> 
#R> Coefficients:
#R>                       Estimate Std. Error t value Pr(>|t|)  
#R> (Intercept)             2.4386     0.5950   4.098   0.0547 .
#R> `Overnight.Deposit 1`   0.7746     0.6213   1.247   0.3387  
#R> ---
#R> Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
#R> 
#R> Residual standard error: 1.186 on 2 degrees of freedom
#R> Multiple R-squared:  0.4374,Adjusted R-squared:  0.156 
#R> F-statistic: 1.555 on 1 and 2 DF,p-value: 0.3387
#R> 
#R> 
#R> [[2]]
#R> 
#R> Call:
#R> lm(formula = "ODR ~ `Overnight.Deposit 2`",data = finalmev)
#R> 
#R> Residuals:
#R>       1       2       3       4 
#R> -1.6293  0.3902  0.2308  1.0083 
#R> 
#R> Coefficients:
#R>                       Estimate Std. Error t value Pr(>|t|)  
#R> (Intercept)             2.3372     0.7282   3.209   0.0849 .
#R> `Overnight.Deposit 2`   0.8865     1.1645   0.761   0.5260  
#R> ---
#R> Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
#R> 
#R> Residual standard error: 1.392 on 2 degrees of freedom
#R> Multiple R-squared:  0.2247,Adjusted R-squared:  -0.163 
#R> F-statistic: 0.5795 on 1 and 2 DF,p-value: 0.526
#R>

我使用 eval(bquote(...)) 来获得不错的输出。请注意，您可以将 colnames(finalmev)[2:ncol(finalmev)] 更改为 tail(colnames(finalmev),-1)。如上所述，Ronak Shah 表明您实际上只保存了 for 循环中的最后一个输出。

另外两个选择是：

# move out sprintf
res1 <- lapply(sprintf("ODR ~ `%s`",tail(colnames(finalmev),-1)),function(frm) eval(bquote(summary(lm(.(frm),data = finalmev)))))

# in R 4.1.0 or greater
res2 <- tail(colnames(finalmev),-1) |> 
  sprintf(fmt = "ODR ~ `%s`") |> 
  lapply(\(frm) eval(bquote(summary(lm(.(frm),data = finalmev)))))

# we get the same 
all.equal(res,res2)
#R> [1] TRUE
all.equal(res1,res2)
#R> [1] TRUE