预测R中的lm函数(多元线性回归) [英] predict lm function in R (multiple linear regression)

问题描述

我使用函数lm在R中进行了多元线性回归，我想用它来预测多个值.因此，我正在尝试使用功能predict(). 这是我的代码:

I did a multiple linear regression in R using the function lm and I want to use it to predict several values. So I'm trying to use the function predict(). Here is my code:

new=data.frame(t=c(10, 20, 30))
v=1/t
LinReg<-lm(p ~ log(t) + v)
Pred=predict(LinReg, new, interval="confidence")

所以我想预测t=c(10,20,30...)时p的值.但是，这不起作用，我不明白为什么.我收到的错误消息是:

So I would like to predict the values of p when t=c(10,20,30...). However, this is not working and I don't see why. The error message I get is:

"model.frame.default中的错误(术语，newdata，na.action = na.action，xlev = object $ xlevels):可变长度不同(适用于"vart") 另外:警告消息: 'newdata'有3行，但找到的变量有132行"

"Error in model.frame.default(Terms, newdata, na.action = na.action, xlev = object$xlevels) : variable lengths differ (found for 'vart') In addition: Warning message: 'newdata' had 3 rows but variables found have 132 rows "

132是我进行回归的变量向量的长度.我检查了我的矢量1/t，它定义明确，并且系数数量正确.奇怪的是，如果我对一个变量进行简单的线性回归，则相同的代码可以很好地工作...

132 is the length of my vector of variables upon which I run the regression. I checked my vector 1/t and it is well-defined and has the right number of coefficients. What is curious is that if I do a simple linear regression (of one variable), the same code works well...

new=data.frame(t=c(10, 20, 30))
LinReg<-lm(p ~ log(t))
Pred=predict(LinReg, new, interval="confidence")

任何人都可以帮助我！预先感谢.

Can anyone help me please! Thanks in advance.

推荐答案

问题是，当您拟合模型时，您将v定义为与t不同的新变量. R不记得如何创建变量，因此在拟合模型时它不知道v是t的函数.因此，当您预测值时，它会使用现有的v值，该值与您指定的新的t值具有不同的长度.

The problem is you defined v as a new, distinct variable from t when you fit your model. R doesn't remember how a variable was created so it doesn't know that v is a function of t when you fit the model. So when you go to predict values, it uses the existing values of v which would have a different length than the new values of t you are specifying.

相反，您想适应

new <- data.frame(t=c(10, 20, 30))
LinReg <- lm(p ~ log(t) + I(1/t))
Pred <- predict(LinReg, new, interval="confidence")

如果您确实希望v是一个完全独立的变量，那么您还需要在new data.frame中提供<​​c2>的值，以便预测p.

If you did want v to be a completely independent variable, then you would need to supply values for v as well in your new data.frame in order to predict p.

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