R 中已知固定截距的线性回归

发布于 2024-12-03 11:52:00 字数 513 浏览 1 评论 0原文

我想使用 R 中的 lm() 函数计算线性回归。此外，我想获得回归的斜率，其中我明确地将截距赋予 lm()。

我在互联网上找到了一个例子，并尝试阅读 R-help“?lm”（不幸的是我无法理解它），但我没有成功。谁能告诉我我的错误在哪里？

lin <- data.frame(x = c(0:6), y = c(0.3, 0.1, 0.9, 3.1, 5, 4.9, 6.2))
plot (lin$x, lin$y)

regImp = lm(formula = lin$x ~ lin$y)
abline(regImp, col="blue")

# Does not work:
# Use 1 as intercept
explicitIntercept = rep(1, length(lin$x))
regExp = lm(formula = lin$x ~ lin$y + explicitIntercept)
abline(regExp, col="green")

感谢您的帮助。

原文

I want to calculate a linear regression using the lm() function in R. Additionally I want to get the slope of a regression, where I explicitly give the intercept to lm().

I found an example on the internet and I tried to read the R-help "?lm" (unfortunately I'm not able to understand it), but I did not succeed. Can anyone tell me where my mistake is?

lin <- data.frame(x = c(0:6), y = c(0.3, 0.1, 0.9, 3.1, 5, 4.9, 6.2))
plot (lin$x, lin$y)

regImp = lm(formula = lin$x ~ lin$y)
abline(regImp, col="blue")

# Does not work:
# Use 1 as intercept
explicitIntercept = rep(1, length(lin$x))
regExp = lm(formula = lin$x ~ lin$y + explicitIntercept)
abline(regExp, col="green")

Thanls for your help.

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瞳孔里扚悲伤 2024-12-10 11:52:00

您可以从回归中减去显式截距，然后拟合无截距模型：

> intercept <- 1.0
> fit <- lm(I(x - intercept) ~ 0 + y, lin)
> summary(fit)

0 + 通过 lm 抑制截距拟合。

编辑要绘制拟合图，请使用

> abline(intercept, coef(fit))

P.S.模型中的变量看起来是错误的：通常是 y ~ x，而不是 x ~ y （即回归变量应位于左侧，回归变量应位于左侧）在右侧）。

You could subtract the explicit intercept from the regressand and then fit the intercept-free model:

> intercept <- 1.0
> fit <- lm(I(x - intercept) ~ 0 + y, lin)
> summary(fit)

The 0 + suppresses the fitting of the intercept by lm.

edit To plot the fit, use

> abline(intercept, coef(fit))

P.S. The variables in your model look the wrong way round: it's usually y ~ x, not x ~ y (i.e. the regressand should go on the left and the regressor(s) on the right).

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月隐月明月朦胧 2024-12-10 11:52:00

我看到您已接受使用 I() 的解决方案。我原以为基于 offset() 的解决方案会更明显，但口味各不相同，在完成偏移解决方案后，我可以欣赏 I() 解决方案的经济性：

with(lin, plot(y,x) )
lm_shift_up <- lm(x ~ y +0 + 
                       offset(rep(1, nrow(lin))), 
             data=lin)
abline(1,coef(lm_shift_up))

I see that you have accepted a solution using I(). I had thought that an offset() based solution would have been more obvious, but tastes vary and after working through the offset solution I can appreciate the economy of the I() solution:

with(lin, plot(y,x) )
lm_shift_up <- lm(x ~ y +0 + 
                       offset(rep(1, nrow(lin))), 
             data=lin)
abline(1,coef(lm_shift_up))

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