在Perceptron中手动绘制直线边界线 [英] Manually plotting straight boundary line in Perceptron

问题描述

[该材料属于

红点被录取到一所大学，其余的则未被录取.

问题不是如何在情节上获得这条线，而是为什么代码中的以下代码适用于课程材料中适用于R的行:

y = c((-1/coefs [3])*(coefs [2] * x + coefs [1]))

所以实际上是在问这个命令的数学基础.这些系数与逻辑回归系数相对应.

这是数据集，这里是完整的代码:

  dat = read.csv("perceptron.txt"，标头= F)is.data.frame(日期)colnames(dat)= c("test1"，"test2"，"y")头(日期)图(test2〜test1，col = as.factor(y)，pch = 20，data = dat)适合= glm(y〜test1 + test2，家庭=二项式"，数据= dat)系数=系数(适合)(x = c(min(dat[,1])-2, max(dat[,1])+2))(y = c((-1/coefs [3])*(coefs [2] * x + coefs [1])))线(x，y) 

解决方案

答案实际上很简单:

边界位于:

  0 = theta_0 + theta_1 x_1 + theta_2 x_2 

因此，由于图是 x_2 与 x_1 的关系图，因此我们沿着 x_1 选择两个极限点，然后计算期望的 x_2在决策边界:

  x_2 =(-1/theta_2)*(theta_0 + theta_1 x_1) 

[The material belongs to the Coursera Machine Learning course by Andrew Ng]

I got one of the exercises to work in R (I could have opted for Python - not essential to the question), using different methodology, and got the following plot with the boundary decision line on it:

The red points were admitted to a college, while the rest were not.

The question is not a how-to get the line on the plot, but rather why does the following line in the code adapted to R from the course materials works:

y = c((-1/coefs[3]) * (coefs[2] * x + coefs[1]))

So it is in reality asking about the math underpinning this command. The coefficients correspond to the logistic regression coefficients.

Here is the dataset, and here is the entire code:

dat = read.csv("perceptron.txt", header=F)
is.data.frame(dat)
colnames(dat) = c("test1","test2","y")
head(dat)

plot(test2 ~ test1, col = as.factor(y), pch = 20, data=dat)

fit = glm(y ~ test1 + test2, family = "binomial", data = dat)
coefs = coef(fit)
(x = c(min(dat[,1])-2,  max(dat[,1])+2))
(y = c((-1/coefs[3]) * (coefs[2] * x + coefs[1])))
lines(x, y)

解决方案

The answer is actually quite easy:

The boundary is at:

0 = theta_0 + theta_1 x_1 + theta_2 x_2

Hence, since the plot is of x_2 versus x_1, we choose two extreme points along x_1 and calculate the expected x_2 at the decision boundary:

x_2 = (- 1 / theta_2) * (theta_0 + theta_1 x_1)

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