具有任意系数 r 的 predict.lm [英] predict.lm with arbitrary coefficients r

问题描述

我正在尝试使用 predict.lm 来预测 lm 对象.但是，我想使用手动插入的系数.为此，我尝试过:

I'm trying to predict an lm object using predict.lm. However, I would like to use manually inserted coefficients. To do this I tried:

model$coefficients <- coeff

(其中coeff"是正确系数的向量)这确实会根据我的需要修改系数.尽管如此，当我执行

(where "coeff" is a vector of correct coefficients) which would indeed modify the coefficients as I want. Nevertheless, when I execute

 predict.lm(model, new.data)

我只是得到使用旧"参数计算的预测.有没有办法强制 predict.lm 使用新的?

I just get predictions calculated with the "old" parameters. Is there a way I could force predict.lm to use the new ones?

Post Scriptum:我需要这样做以适应 bin-smooth(也称为回归图).此外，当我手动"预测(即使用矩阵乘法)时，结果很好，因此我很确定问题在于 predict.lm 无法识别我的新系数.

Post Scriptum: I need to do this to fit a bin-smooth (also called regressogram). In addition, when I predict "by hand" (i.e. using matrix multiplication) the results are fine, hence I'm quite sure that the problem lies in the predict.lm not recognizing my new coefficients.

预先感谢您的帮助！

推荐答案

破解 $coefficients 元素似乎确实有效.你能说明什么对你不起作用吗?

Hacking the $coefficients element does indeed seem to work. Can you show what doesn't work for you?

dd <- data.frame(x=1:5,y=1:5)
m1 <- lm(y~x,dd)
m1$coefficients <- c(-2,1)
m1
## Call:
## lm(formula = y ~ x, data = dd)
## 
## Coefficients:
## [1]  -2   1

predict(m1,newdata=data.frame(x=7))  ## 5  = -2+1*7

predict.lm(...) 给出相同的结果.

我会非常谨慎地使用这种方法，每次你对被黑模型做不同的事情时都会检查.

I would be very careful with this approach, checking each time you do something different with the hacked model.

一般来说，如果 predict 和 simulate 方法接受一个 newparams 参数会很好，但它们通常不会......

In general it would be nice if predict and simulate methods took a newparams argument, but they don't in general ...

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