绘制非线性决策边界

Question

我试图绘制一个非线性决策边界，应该看起来像这样：

我已经拟合了形式的正则非线性逻辑回归：

这是我数据的摘录：

  ones     test1   test2 use
1    1  0.051267 0.69956   1
2    1 -0.092742 0.68494   1
3    1 -0.213710 0.69225   1
4    1 -0.375000 0.50219   1
5    1 -0.513250 0.46564   1
6    1 -0.524770 0.20980   1

这些是我使用optim（）函数计算的参数：

[1]  0.377980476 -0.085951551  0.445140731
[4] -1.953080687 -0.506554404 -0.330330236
[7]  0.414649938  0.270281786  0.183804530
[10] -0.155359467 -0.753665545  0.351880543
[13]  0.238052214  0.619714119 -0.582420943
[16]  0.150625144  0.266319363 -0.331130949
[19]  0.177759335 -0.005402135 -0.124253913
[22]  0.085607070  0.580258782  0.973785263
[25]  0.387313615  0.237754576 -0.011198804
[28] -0.514447404

我还是R的新手，我对如何解决这个问题真的不知道，有人能帮帮我吗？

    ones      test1     test2 use
1      1  0.0512670  0.699560   1
2      1 -0.0927420  0.684940   1
3      1 -0.2137100  0.692250   1
4      1 -0.3750000  0.502190   1
5      1 -0.5132500  0.465640   1
6      1 -0.5247700  0.209800   1
7      1 -0.3980400  0.034357   1
8      1 -0.3058800 -0.192250   1
9      1  0.0167050 -0.404240   1
10     1  0.1319100 -0.513890   1
11     1  0.3853700 -0.565060   1
12     1  0.5293800 -0.521200   1
13     1  0.6388200 -0.243420   1
14     1  0.7367500 -0.184940   1
15     1  0.5466600  0.487570   1
16     1  0.3220000  0.582600   1
17     1  0.1664700  0.538740   1
18     1 -0.0466590  0.816520   1
19     1 -0.1733900  0.699560   1
20     1 -0.4786900  0.633770   1
21     1 -0.6054100  0.597220   1
22     1 -0.6284600  0.334060   1
23     1 -0.5938900  0.005117   1
24     1 -0.4210800 -0.272660   1
25     1 -0.1157800 -0.396930   1
26     1  0.2010400 -0.601610   1
27     1  0.4660100 -0.535820   1
28     1  0.6733900 -0.535820   1
29     1 -0.1388200  0.546050   1
30     1 -0.2943500  0.779970   1
31     1 -0.2655500  0.962720   1
32     1 -0.1618700  0.801900   1
33     1 -0.1733900  0.648390   1
34     1 -0.2828300  0.472950   1
35     1 -0.3634800  0.312130   1
36     1 -0.3001200  0.027047   1
37     1 -0.2367500 -0.214180   1
38     1 -0.0639400 -0.184940   1
39     1  0.0627880 -0.163010   1
40     1  0.2298400 -0.411550   1
41     1  0.2932000 -0.228800   1
42     1  0.4832900 -0.184940   1
43     1  0.6445900 -0.141080   1
44     1  0.4602500  0.012427   1
45     1  0.6273000  0.158630   1
46     1  0.5754600  0.268270   1
47     1  0.7252300  0.443710   1
48     1  0.2240800  0.524120   1
49     1  0.4429700  0.670320   1
50     1  0.3220000  0.692250   1
51     1  0.1376700  0.575290   1
52     1 -0.0063364  0.399850   1
53     1 -0.0927420  0.553360   1
54     1 -0.2079500  0.355990   1
55     1 -0.2079500  0.173250   1
56     1 -0.4383600  0.217110   1
57     1 -0.2194700 -0.016813   1
58     1 -0.1388200 -0.272660   1
59     1  0.1837600  0.933480   0
60     1  0.2240800  0.779970   0
61     1  0.2989600  0.619150   0
62     1  0.5063400  0.758040   0
63     1  0.6157800  0.728800   0
64     1  0.6042600  0.597220   0
65     1  0.7655500  0.502190   0
66     1  0.9268400  0.363300   0
67     1  0.8231600  0.275580   0
68     1  0.9614100  0.085526   0
69     1  0.9383600  0.012427   0
70     1  0.8634800 -0.082602   0
71     1  0.8980400 -0.206870   0
72     1  0.8519600 -0.367690   0
73     1  0.8289200 -0.521200   0
74     1  0.7943500 -0.557750   0
75     1  0.5927400 -0.740500   0
76     1  0.5178600 -0.594300   0
77     1  0.4660100 -0.418860   0
78     1  0.3508100 -0.579680   0
79     1  0.2874400 -0.769740   0
80     1  0.0858290 -0.755120   0
81     1  0.1491900 -0.579680   0
82     1 -0.1330600 -0.448100   0
83     1 -0.4095600 -0.411550   0
84     1 -0.3922800 -0.258040   0
85     1 -0.7436600 -0.258040   0
86     1 -0.6975800  0.041667   0
87     1 -0.7551800  0.290200   0
88     1 -0.6975800  0.684940   0
89     1 -0.4038000  0.706870   0
90     1 -0.3807600  0.918860   0
91     1 -0.5074900  0.904240   0
92     1 -0.5478100  0.706870   0
93     1  0.1031100  0.779970   0
94     1  0.0570280  0.918860   0
95     1 -0.1042600  0.991960   0
96     1 -0.0812210  1.108900   0
97     1  0.2874400  1.087000   0
98     1  0.3968900  0.823830   0
99     1  0.6388200  0.889620   0
100    1  0.8231600  0.663010   0
101    1  0.6733900  0.641080   0
102    1  1.0709000  0.100150   0
103    1 -0.0466590 -0.579680   0
104    1 -0.2367500 -0.638160   0
105    1 -0.1503500 -0.367690   0
106    1 -0.4902100 -0.301900   0
107    1 -0.4671700 -0.133770   0
108    1 -0.2885900 -0.060673   0
109    1 -0.6111800 -0.067982   0
110    1 -0.6630200 -0.214180   0
111    1 -0.5996500 -0.418860   0
112    1 -0.7263800 -0.082602   0
113    1 -0.8300700  0.312130   0
114    1 -0.7206200  0.538740   0
115    1 -0.5938900  0.494880   0
116    1 -0.4844500  0.999270   0
117    1 -0.0063364  0.999270   0
118    1  0.6326500 -0.030612   0

Answer 1

虽然不是一个理想的答案，但您可以使用SVM模型来形象化它（它会产生约0.83的样本内错误）：

require(e1071)
data = data[, c("use", "test1", "test2")]

fit = svm(use ~ ., data = data)
plot(fit, data = data)

使用简单的转换，我们可以尝试获得线性可分的数据集：

data2 = data.frame(
    y = factor(data[, "use"]),
    x1 = data[, "test1"]^2,
    x2 = data[, "test2"]^2 )

require(MASS)
fit = glm(y ~ x2 + x1, data = data2, family = binomial(link = "logit"))
plot(x2 ~ x1, data = data2, bg = as.numeric(y) + 1, pch = 21, main = "Logistic regression on Y ~ X1 + X2")
abline(-fit$coefficients[1]/fit$coefficients[2], -fit$coefficients[3]/fit$coefficients[2], col = 'blue', lwd = 2)

这给你这个（约0.73样本内错误）：

所以现在你有了

Y = w0 + w1 * test1^2 + w2 * test2^2

您可以使用它来隔离test2 = f(test1)并绘制非线性边界。