Question

我有兴趣制作一个带有最小二乘回归线的图和将数据点连接到回归线的线段，如图所示，称为垂直偏移： http://mathworld.wolfram.com/LeastSquaresFitting.html alt text http://mathworld.wolfram.com/images/eps-gif/LeastSquaresOffsets_1000.gif

我在这里完成了情节和回归线：

## Dataset from http://www.apsnet.org/education/advancedplantpath/topics/RModules/doc1/04_Linear_regression.html

## Disease severity as a function of temperature

# Response variable, disease severity
diseasesev<-c(1.9,3.1,3.3,4.8,5.3,6.1,6.4,7.6,9.8,12.4)

# Predictor variable, (Centigrade)
temperature<-c(2,1,5,5,20,20,23,10,30,25)

## For convenience, the data may be formatted into a dataframe
severity <- as.data.frame(cbind(diseasesev,temperature))

## Fit a linear model for the data and summarize the output from function lm()
severity.lm <- lm(diseasesev~temperature,data=severity)

# Take a look at the data
plot(
 diseasesev~temperature,
        data=severity,
        xlab="Temperature",
        ylab="% Disease Severity",
        pch=16,
        pty="s",
        xlim=c(0,30),
        ylim=c(0,30)
)
abline(severity.lm,lty=1)
title(main="Graph of % Disease Severity vs Temperature")

我应该使用某种for循环和段http://www.iiap.res.in/astrostat/School07/R/html/graphics/html/segments.html来做垂直偏移吗？有更有效的方法吗？如果可能，请提供一个例子。

Answer 1

首先需要找出垂直线段底部的坐标，然后调用segments函数，该函数可以将坐标向量作为输入（不需要循环）。

perp.segment.coord <- function(x0, y0, lm.mod){
 #finds endpoint for a perpendicular segment from the point (x0,y0) to the line
 # defined by lm.mod as y=a+b*x
  a <- coef(lm.mod)[1]  #intercept
  b <- coef(lm.mod)[2]  #slope
  x1 <- (x0+b*y0-a*b)/(1+b^2)
  y1 <- a + b*x1
  list(x0=x0, y0=y0, x1=x1, y1=y1)
}

现在只需拨打电话：

ss <- perp.segment.coord(temperature, diseasesev, severity.lm)
do.call(segments, ss)
#which is the same as:
segments(x0=ss$x0, x1=ss$x1, y0=ss$y0, y1=ss$y1)

请注意，除非确保绘图的x单位和y单位具有相同的表观长度（等距刻度），否则结果将不会垂直。您可以使用pty="s"获取方形图并将xlim和ylim设置为相同的范围。

在R中的最小二乘回归图中绘制垂直偏移

1 个答案: