R：拟合曲线到点：使用什么线性/非线性模型？

debugcn 发表于 Dev

迈卡

我有一个应该遵循幂律分布的数据。

x = distance 
y = %

我想创建一个模型并将拟合线添加到我的图中。

我的目标是重新创建这样的东西：

由于作者使用 R 方；我假设他们应用了线性模型，因为 R^2不适用于非线性模型。http://blog.minitab.com/blog/adventures-in-statistics-2/why-is-there-no-r-squared-for-nonlinear-regression

但是，我不知道如何将我的线“弯曲”到点；如何将公式添加y ~ a*x^(-b)到我的模型中。

而不是卷线，我从简单的线性回归中取回了这条线。

我的问题是：

Do I correctly assume the model y ~ a*x^(-b) used by author is linear?
what type of model to use to recreate my example: lm, glm, nls, etc. ?

I generated the dummy data, including the applied power law formula from the plot above:

set.seed(42)
scatt<-runif(10)

x<-seq(1, 1000, 100)
b = 1.8411
a = 133093
y = a*x^(-b) + scatt  # add some variability in my dependent variable

plot(y ~ x)

and tried to create a glm model.

# formula for non-linear model
m<-m.glm<-glm(y ~ x^2, data = dat) # 

# add predicted line to plot
lines(x,predict(m),col="red",lty=2,lwd=3)

This is my first time to model, so I am really confused and I don't know where to start... thank you for any suggestion or directions, I really appreciate it...

李哲源

I personally think this question a dupe of this: `nls` fails to estimate parameters of my model but I would be cold-blooded if I close it (as OP put a bounty). Anyway, bounty question can not be closed.

So the best I could think of, is to post a community wiki answer (I don't want to get this bounty).

当您想要拟合这种形式的模型时y ~ a*x^(-b)，通常会受益log于在两侧进行变换并拟合线性模型log(y) ~ log(x)。