在 R 中绘制回归线

Question

I want to plot a simple regression line in R. I've entered the data, but the regression line doesn't seem to be right. Can someone help?

x <- c(10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120)
y <- c(10, 18, 25, 29, 30, 28, 25, 22, 18, 15, 11, 8)
df <- data.frame(x,y)
plot(y,x)
abline(lm(y ~ x))

Answer 1

Oh, @GBR24 has nice formatted data. Then I'm going to elaborate a little bit based on my comment.

fit <- lm(y ~ poly(x, 3))   ## polynomial of degree 3
plot(x, y)  ## scatter plot (colour: black)

x0 <- seq(min(x), max(x), length = 20)  ## prediction grid
y0 <- predict.lm(fit, newdata = list(x = x0))  ## predicted values
lines(x0, y0, col = 2)  ## add regression curve (colour: red)

Answer 2

x <- c(10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120)
y <- c(10, 18, 25, 29, 30, 28, 25, 22, 18, 15, 11, 8)

df <- data.frame(x,y)

plot(y ~ x, df)
model <- lm(y ~ x, df)

You're trying to fit a linear function to parabolic data. As such, you won't end up with a pretty line of best fit.

Something like this might work:

model <- lm(y ~ I(x^2), df)

plot(y ~ x, df)
lines(df$x, predict(model), col = 'blue')

Although that doesn't really fit well, we could try 3rd- or 4th-order polynomial models:

model <- lm(y ~ I(x^3), df)
lines(df$x, predict(model), col = 'red')
model <- lm(y ~ I(x^4), df)
lines(df$x, predict(model), col = 'green')

Although those don't fit very well, either. Look to Zheyuan's answer for a better-fitting function.

Answer 3

1. As I said above the graph in the original question switched the x-axis and y-axis
2. The linear model answer is the best for the question since that is what was asked.
3. The other answers address further modeling choices such as best cubic model above or best quadratic below. This just combines the reasoning above.

    x <- c(10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120)
    y <- c(10, 18, 25, 29, 30, 28, 25, 22, 18, 15, 11, 8)
    summary(lm(y~x))
    plot(x,y)
    abline(lm(y ~ x)) # black answer 1
    ########################
    fit <- lm(y ~ poly(x, 2))   ## polynomial of degree 2
    y0 <- predict.lm(fit)  ## predicted values
    lines(x, y0, col = 2)  ##  predicted poly red color
    #y1 <- predict(fit, interval = "prediction")
    [![#lines(x, y1\[,1\], col = 3)  same as y1 green color   # answer 2
    #########################
    w <- 1 + (x-1)^2  # with weights
    wfit <- lm(y ~ poly(x,2), weights = w)
    y2 <- predict(wfit, interval = "prediction")
    lines(x, y2\[,1\], col = 4) # blue    # answer 3