R中的幂回归类似于excel

I have a simple dataset and I am trying to use the power trend to best fit the data. The sample data is very small and is as follows:

我有一个简单的数据集，我正在尝试使用功率趋势来最好地拟合数据。样本数据非常小，如下：

structure(list(Discharge = c(250, 300, 500, 700, 900), Downstream = c(0.3, 
0.3, 0.3, 0.3, 0.3), Age = c(1.32026239202165, 1.08595138888889, 
0.638899189814815, 0.455364583333333, 0.355935185185185)), .Names = c("Discharge", 
"Downstream", "Age"), row.names = c(NA, 5L), class = "data.frame")

Data looks as follows:

数据如下：

> new
  Discharge Downstream       Age
1       250        0.3 1.3202624
2       300        0.3 1.0859514
3       500        0.3 0.6388992
4       700        0.3 0.4553646
5       900        0.3 0.3559352

I tried to plot the above data using ggplot2

我试图使用ggplot2绘制上述数据

ggplot(new)+geom_point(aes(x=Discharge,y=Age))

I could add the linear line using geom_smooth(method="lm") but I am not sure what code do I need to show the power line.

我可以使用geom_smooth（method =“lm”）添加线性线，但我不确定显示电源线需要什么代码。

The output is as follows:

输出如下：

How Can I add a power linear regression line as done in excel ? The excel figure is shown below:

如何在excel中添加幂线性回归线？ excel图如下所示：

3 个解决方案

#1

Use nls (nonlinear least squares) as your smoother

使用nls（非线性最小二乘）作为您的平滑器

eg

例如

ggplot(DD,aes(x = Discharge,y = Age)) +
  geom_point() + 
  stat_smooth(method = 'nls', formula = 'y~a*x^b', start = list(a = 1,b=1),se=FALSE)

Noting Doug Bates comments on R-squared values and non-linear models here, you could use the ideas in Adding Regression Line Equation and R2 on graph

注意Doug Bates在这里评论R平方值和非线性模型，你可以使用添加回归线方程和R2的图表中的想法

to append the regression line equation

附加回归线方程

# note that you have to give it sensible starting values
# and I haven't worked out why the values passed to geom_smooth work!
power_eqn = function(df, start = list(a =300,b=1)){
  m = nls(Discharge ~ a*Age^b, start = start, data = df);
  eq <- substitute(italic(y) == a  ~italic(x)^b, 
               list(a = format(coef(m)[1], digits = 2), 
                    b = format(coef(m)[2], digits = 2)))
  as.character(as.expression(eq));                 
}

ggplot(DD,aes(x = Discharge,y = Age)) +
  geom_point() + 
  stat_smooth(method = 'nls', formula = 'y~a*x^b', start = list(a = 1,b=1),se=FALSE) +  
  geom_text(x = 600, y = 1, label = power_eqn(DD), parse = TRUE)

#2

While mnel's answer is correct for a nonlinear least squares fit, note that Excel isn't actually doing anything nearly that sophisticated. It's really just log-transforming the response and predictor variables, and doing an ordinary (linear) least squares fit. To reproduce this in R, you would do:

虽然mnel的答案对于非线性最小二乘拟合是正确的，但请注意Excel实际上并没有做任何复杂的事情。它实际上只是对响应和预测变量进行对数变换，并进行普通（线性）最小二乘拟合。要在R中重现这一点，您可以：

lm(log(Age) ~ log(Discharge), data=df)

Call:
lm(formula = log(Age) ~ log(Discharge), data = df)

Coefficients:
   (Intercept)  log(Discharge)  
         5.927          -1.024  

As a check, the coefficient for log(Discharge) is identical to that from Excel while exp(5.927) ~ 375.05.

作为检查，log（Discharge）的系数与Excel中的系数相同，而exp（5.927）~375.05。

While I'm not sure how to use this as a trendline in ggplot2, you can do it in base graphics thusly:

虽然我不确定如何在ggplot2中使用它作为趋势线，但你可以在基本图形中这样做：

m <- lm(log(y) ~ log(x), data=df)

newdf <- data.frame(Discharge=seq(min(df$Discharge), max(df$Discharge), len=100))
plot(Age ~ Discharge, data=df)
lines(newdf$Discharge, exp(predict(m, newdf)))

text(600, .8, substitute(b0*x^b1, list(b0=exp(coef(m)[1]), b1=coef(m)[2])))
text(600, .75, substitute(plain("R-square: ") * r2, list(r2=summary(m)$r.squared)))

#3

2018 Update: The call "start" now seems to be depreciated. It is not in the stat_smooth function information either.

2018年更新：呼叫“开始”现在似乎已折旧。它也不在stat_smooth函数信息中。

If you want to choose starting values, you need to use "method.args" option now.

如果要选择起始值，则需要立即使用“method.args”选项。

See changes below:

请参阅以下更改：

ggplot(DD,aes(x = Discharge,y = Age)) +
  geom_point() + 
  stat_smooth(method = 'nls', formula = 'y~a*x^b', method.args = list(start= c(a = 1,b=1)),se=FALSE) + geom_text(x = 600, y = 1, label = power_eqn(DD), parse = TRUE)