我参加的一门ML课程中，我有100项数据，并且正在Perceptron算法中使用它。
我要显示的是这样的情节。
如上所示，我们用红色和蓝色的点表示数据，并使用不同的计算线将误差最小化。这是我想要的输出。这是我的数据和代码。
data.csv
0.78051,-0.063669,1
0.28774,0.29139,1
0.40714,0.17878,1
0.2923,0.4217,1
0.50922,0.35256,1
0.27785,0.10802,1
0.27527,0.33223,1
0.43999,0.31245,1
0.33557,0.42984,1
0.23448,0.24986,1
0.0084492,0.13658,1
0.12419,0.33595,1
0.25644,0.42624,1
0.4591,0.40426,1
0.44547,0.45117,1
0.42218,0.20118,1
0.49563,0.21445,1
0.30848,0.24306,1
0.39707,0.44438,1
0.32945,0.39217,1
0.40739,0.40271,1
0.3106,0.50702,1
0.49638,0.45384,1
0.10073,0.32053,1
0.69907,0.37307,1
0.29767,0.69648,1
0.15099,0.57341,1
0.16427,0.27759,1
0.33259,0.055964,1
0.53741,0.28637,1
0.19503,0.36879,1
0.40278,0.035148,1
0.21296,0.55169,1
0.48447,0.56991,1
0.25476,0.34596,1
0.21726,0.28641,1
0.67078,0.46538,1
0.3815,0.4622,1
0.53838,0.32774,1
0.4849,0.26071,1
0.37095,0.38809,1
0.54527,0.63911,1
0.32149,0.12007,1
0.42216,0.61666,1
0.10194,0.060408,1
0.15254,0.2168,1
0.45558,0.43769,1
0.28488,0.52142,1
0.27633,0.21264,1
0.39748,0.31902,1
0.5533,1,0
0.44274,0.59205,0
0.85176,0.6612,0
0.60436,0.86605,0
0.68243,0.48301,0
1,0.76815,0
0.72989,0.8107,0
0.67377,0.77975,0
0.78761,0.58177,0
0.71442,0.7668,0
0.49379,0.54226,0
0.78974,0.74233,0
0.67905,0.60921,0
0.6642,0.72519,0
0.79396,0.56789,0
0.70758,0.76022,0
0.59421,0.61857,0
0.49364,0.56224,0
0.77707,0.35025,0
0.79785,0.76921,0
0.70876,0.96764,0
0.69176,0.60865,0
0.66408,0.92075,0
0.65973,0.66666,0
0.64574,0.56845,0
0.89639,0.7085,0
0.85476,0.63167,0
0.62091,0.80424,0
0.79057,0.56108,0
0.58935,0.71582,0
0.56846,0.7406,0
0.65912,0.71548,0
0.70938,0.74041,0
0.59154,0.62927,0
0.45829,0.4641,0
0.79982,0.74847,0
0.60974,0.54757,0
0.68127,0.86985,0
0.76694,0.64736,0
0.69048,0.83058,0
0.68122,0.96541,0
0.73229,0.64245,0
0.76145,0.60138,0
0.58985,0.86955,0
0.73145,0.74516,0
0.77029,0.7014,0
0.73156,0.71782,0
0.44556,0.57991,0
0.85275,0.85987,0
0.51912,0.62359,0
现在，这是我的代码。第一部分
import numpy as np
import pandas as pd
# Setting the random seed, feel free to change it and see different solutions.
np.random.seed(42)
import matplotlib.pyplot as plt
def stepFunction(t):
return 1 if t >= 0 else 0
def prediction(X, W, b):
return stepFunction((np.matmul(X, W) + b)[0])
# TODO: Fill in the code below to implement the perceptron trick.
# INPUTS
# data X, the labels y,
# the weights W (as an array), and the bias b,
# The function weights and bias W, b, according to the perceptron algorithm,
# and return W and 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
# This function runs the perceptron algorithm repeatedly on the dataset,
# and returns a few of the boundary lines obtained in the iterations,
# for plotting purposes.
# Feel free to play with the learning rate and the num_epochs,
# and see your results plotted below.
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
# These are the solution lines that get plotted below.
boundary_lines = []
for i in range(num_epochs):
# In each epoch, we apply the perceptron step.
W, b = perceptronStep(X, y, W, b, learn_rate)
# Here I have a doubt . Why if y = W0*x1 + W1*x2 + b
# So we can get x2 =y/W1 -(W0*x1)/W1 -b/W1 + y/W1)
# If we remove y/W1 we just get intercept and slope
# But why we are not using the last term y/W1
boundary_lines.append((-W[0] / W[1], -b / W[1]))
return boundary_lines
# Get data and plot the points
data = pd.read_csv('data.csv', header = None)
X = data.iloc[:, :2].values
y = data.iloc[:, -1].values
x1 = 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 input feature')
plt.ylabel('X2 input feature')
plt.title('Perceptron regression for X1, X2')
plt.show()
当您运行此代码时，您将正确获得
所以现在我想在同一图中绘制代表每次迭代的最佳功能的线，为此，我注释了plt.show（）上方的最后一行，并做了
# So now lets plot the lines that represent the best function for each iteration
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: The equation of the error function is
# y = W0*x1 + W1*x2 + b
# So we can get x2 =y/W1 -(W0*x1)/W1 -b/W1 + y/W1)
# If we remove y/W1 we just get intercept and slope
# 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("Press enter to continue")
plt.close()
但这并不能给我带来预期的结果。
为什么没有得到预期的结果？
最佳答案
错误是在plt.plot(x_lin, (Θ1 * x_lin / Θo))中，您应该使用Θ1 * x_lin / Θo而不是Θo * x_lin + Θ1。
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：
关于python - 用matplotlib进行Perceptron算法绘图，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/59581706/