查找直线方程以拟合 R 中的数据

Question

I have just started learning R and trying to show the relationship between the mpg of a car and the horsepower. I was trying to get a line of best fit to match the data so I can use it to predict other values.

 #read data from csv file
 data <- read.csv("auto-mpg.csv")

 #get data
 df <- head(data, 350)

 #set x and y 
 x <- df$horsepower
 y <- df$mpg

 #change x from string to int
 x <- strtoi(x)

 #create scatterplot 
 scatter.smooth(x, y, main = "mpg vs horsepower",
 xlab = "horsepower", ylab = "mpg",
 pch = 19)

This is the graph that I want the exponential decay for

mpg vs horsepower graph

Answer 1

Here are two approaches. First by using linear fit to log(y) as suggested by @Gregor Thomas and then a nonlinear fit to the original data. You did not provide reproducible data so I'm using similar data from mtcars :

data(mtcars)
plot(mpg~hp, mtcars, pch=20)
mpg.lm <- lm(log(mpg)~hp, mtcars)  # Model is log(mpg) = a + b * hp
mpg.lm
# 
# Call:
# lm(formula = log(mpg) ~ hp, data = mtcars)
# 
# Coefficients:
# (Intercept)           hp  
#    3.460467    -0.003429  
# 
X <- seq(50, 350)
Y1 <- exp(predict(mpg.lm, data.frame(hp=X)))
lines(X, Y1, col="red")

The red line shows the fit. Now using nls to fit a nonlinear model directly. We'll use the coefficients from the first model to estimate the starting values for the second:

a1 <- coef(mpg.lm)[1]
b1 <- coef(mpg.lm)[2]
mpg.nls <- nls(mpg ~ a * hp^b, mtcars, start=list(a=exp(a1), b=b1))
mpg.nls    # Model is mpg = a * hp^b
# Nonlinear regression model
#   model: mpg ~ a * hp^b
#    data: mtcars
#        a        b 
# 272.1171  -0.5404 
#  residual sum-of-squares: 288.7
# 
# Number of iterations to convergence: 7 
# Achieved convergence tolerance: 1.127e-07
Y2 <- predict(mpg.nls, data.frame(hp=X))
lines(X, Y2, col="blue")

The blue line shows the non-linear fit.