在我正在上的机器学习课程中，我有100条数据条目，并将其用于感知机算法。我想要展示这样的一个图表。

如上图所示，我们用红点和蓝点表示数据，以及不同计算出的线条，这些线条最小化了误差。这就是我想要的输出。以下是我的数据和代码。

data.csv

0.78051,-0.063669,10.28774,0.29139,10.40714,0.17878,10.2923,0.4217,10.50922,0.35256,10.27785,0.10802,10.27527,0.33223,10.43999,0.31245,10.33557,0.42984,10.23448,0.24986,10.0084492,0.13658,10.12419,0.33595,10.25644,0.42624,10.4591,0.40426,10.44547,0.45117,10.42218,0.20118,10.49563,0.21445,10.30848,0.24306,10.39707,0.44438,10.32945,0.39217,10.40739,0.40271,10.3106,0.50702,10.49638,0.45384,10.10073,0.32053,10.69907,0.37307,10.29767,0.69648,10.15099,0.57341,10.16427,0.27759,10.33259,0.055964,10.53741,0.28637,10.19503,0.36879,10.40278,0.035148,10.21296,0.55169,10.48447,0.56991,10.25476,0.34596,10.21726,0.28641,10.67078,0.46538,10.3815,0.4622,10.53838,0.32774,10.4849,0.26071,10.37095,0.38809,10.54527,0.63911,10.32149,0.12007,10.42216,0.61666,10.10194,0.060408,10.15254,0.2168,10.45558,0.43769,10.28488,0.52142,10.27633,0.21264,10.39748,0.31902,10.5533,1,00.44274,0.59205,00.85176,0.6612,00.60436,0.86605,00.68243,0.48301,01,0.76815,00.72989,0.8107,00.67377,0.77975,00.78761,0.58177,00.71442,0.7668,00.49379,0.54226,00.78974,0.74233,00.67905,0.60921,00.6642,0.72519,00.79396,0.56789,00.70758,0.76022,00.59421,0.61857,00.49364,0.56224,00.77707,0.35025,00.79785,0.76921,00.70876,0.96764,00.69176,0.60865,00.66408,0.92075,00.65973,0.66666,00.64574,0.56845,00.89639,0.7085,00.85476,0.63167,00.62091,0.80424,00.79057,0.56108,00.58935,0.71582,00.56846,0.7406,00.65912,0.71548,00.70938,0.74041,00.59154,0.62927,00.45829,0.4641,00.79982,0.74847,00.60974,0.54757,00.68127,0.86985,00.76694,0.64736,00.69048,0.83058,00.68122,0.96541,00.73229,0.64245,00.76145,0.60138,00.58985,0.86955,00.73145,0.74516,00.77029,0.7014,00.73156,0.71782,00.44556,0.57991,00.85275,0.85987,00.51912,0.62359,0

现在这是我的代码。第一部分

import numpy as npimport pandas as pd# 设置随机种子，可以随意更改以查看不同的解决方案。np.random.seed(42)import matplotlib.pyplot as pltdef stepFunction(t):    return 1 if t >= 0 else 0def prediction(X, W, b):    return stepFunction((np.matmul(X, W) + b)[0])# TODO: 填写下面的代码以实现感知机技巧。# 输入# 数据X，标签y，# 权重W（作为数组）和偏置b，# 根据感知机算法调整权重和偏置W, b，# 并返回W和b。def perceptronStep(X, y, W, b, learn_rate=0.01):    for i in range(len(X)):        y_hat = prediction(X[i], W, b)        if y[i] - y_hat == 1:            W[0] += X[i][0] * learn_rate            W[1] += X[i][1] * learn_rate            b += learn_rate        elif y[i] - y_hat == -1:            W[0] -= X[i][0] * learn_rate            W[1] -= X[i][1] * learn_rate            b -= learn_rate    return W, b# 此函数在数据集上重复运行感知机算法，# 并返回迭代中获得的一些边界线，# 用于绘图目的。# 可以随意调整学习率和迭代次数，# 并在下方查看结果图表。def trainPerceptronAlgorithm(X, y, learn_rate=0.01, num_epochs=25):    x_min, x_max = min(X.T[0]), max(X.T[0])    y_min, y_max = min(X.T[1]), max(X.T[1])    W = np.array(np.random.rand(2, 1))    b = np.random.rand(1)[0] + x_max    # 这些是下方绘制的解决方案线。    boundary_lines = []    for i in range(num_epochs):        # 在每个迭代中，我们应用感知机步骤。        W, b = perceptronStep(X, y, W, b, learn_rate)        # 这里我有一个疑问。为什么如果y = W0*x1 + W1*x2 + b        # 那么我们可以得到 x2 = y/W1 - (W0*x1)/W1 - b/W1 + y/W1)        # 如果我们去掉y/W1，我们只得到截距和斜率        # 但为什么我们不使用最后一个项y/W1        boundary_lines.append((-W[0] / W[1], -b / W[1]))    return boundary_lines# 获取数据并绘制点data = pd.read_csv('data.csv', header = None)X = data.iloc[:, :2].valuesy = data.iloc[:, -1].valuesx1 = X[:, 0]x2 = X[:, 1]color = ['red' if value == 1 else 'blue' for value in y]plt.scatter(x1, x2, marker='o', color=color)plt.xlabel('X1输入特征')plt.ylabel('X2输入特征')plt.title('X1, X2的感知机回归')plt.show()

当你运行这段代码时，你会正确地得到

所以现在我想在同一个图表中绘制代表每次迭代最佳函数的线条。为此，我注释了上面的最后一行plt.show()，并做了以下操作

# 现在让我们绘制代表每次迭代最佳函数的线条boundary_lines = trainPerceptronAlgorithm(X, y)x_lin = np.linspace(0, 1, 100)for line in boundary_lines:    Θo, Θ1  = line    Θ1 = Θ1[0]    Θo = Θo[0]    # TODO: 误差函数的方程是    # y = W0*x1 + W1*x2 + b    # 所以我们可以得到 x2 = y/W1 - (W0*x1)/W1 - b/W1 + y/W1)    # 如果我们去掉y/W1，我们只得到截距和斜率    # boundary_lines.append((-W[0] / W[1], -b / W[1])    # plt.axes([-0.5, -0.5, 1.5, 1.5])    plt.plot(x_lin, (Θ1 * x_lin / Θo))    plt.draw()    plt.pause(5)    input("按回车键继续")    plt.close()

但这并没有得到我期望的结果。为什么没有得到期望的结果？

回答：

错误在于plt.plot(x_lin, (Θ1 * x_lin / Θo))，这里应该使用Θo * x_lin + Θ1而不是Θ1 * x_lin / Θo。

fig, ax = plt.subplots(1, 1, figsize=(8,5))ax.set_xlim(0, 1)ax.set_ylim(0, 1)ax.scatter(x1, x2, marker='o', color=color)for i, line in enumerate(boundary_lines):    Θo, Θ1  = line    if i == len(boundary_lines) - 1:        c, ls, lw = 'k', '-', 2    else:        c, ls, lw = 'g', '--', 1.5    ax.plot(x_lin, Θo * x_lin + Θ1, c=c, ls=ls, lw=lw)plt.show()

结果：