Use Matplotlib to simulate linear regression

First, we need two modules ：

1.numpy
2.matplotlib.pylab

Installation command

pip install numpy

pip install matplotlib

The main function of linear regression is to use a linear function or expression to fit random points in discrete space , It is a predictive modeling technology .

Import module after installation ：

import numpy as np
import matplotlib.pylab as plt

1. First, create data in space ：

In linear space -1 To 1 In the middle of 100 A digital , And pass reshape Function to 100 That's ok 1 Columns of data , And break it up .

x = np.linspace(-1,1,100).reshape((100,1))
m = len(x)
y = 0.8*x + 0.1* np.random.randn(m).reshape((m,1))

linespace The function of is to -1 and 1 In the middle of 100 A digital , And this 100 The difference between the two numbers is the same .

STEP 2

np.hstack Stack the element array of the parameter tuple in the horizontal direction
ones_like Method returns an array of the same type as the given array , This is defined as X

w = np.zeros(2).reshape((2,1))
X = np.hstack((x,np.ones_like(x)))

STEP 3

n = 200
lr = 0.05
J = list()

plt.ion()			#ion()  No parameters are accepted . It's just used to turn on interactive mode 
plt.figure(1,figsize=(8,6)) # Create a shape 
 
for t in range(n):
    pred_y = np.dot(X,w)
    cost = 1/(2*m) * np.sum((np.dot(X,w)-y)**2)
    J.append(cost)
    dw = 1/m * np.dot(X.T,np.dot(X,w)-y)
    w = w - lr * dw;
    if t % 5 == 0:
        plt.subplot(2,1,1)
        plt.cla()
        plt.scatter(x,y)			# Draw on the graph x,y The coordinates are discrete points 
        plt.plot(x,pred_y,'r-',lw=5)	# Red line , Rough as 5 Fit 
        plt.text(0.5,0,'Loss=%.4f' %cost)	# Display the loss value 
        plt.subplot(2,1,2)			# Add a new sub module for displaying graphics 
        plt.cla()
        plt.plot(J)
        plt.pause(0.2)
plt.ioff()
plt.show()

Combine all the code , You can get the running diagram of linear regression .
The figure is dynamically displayed ,

When the number of repetitions is large enough , The loss function becomes less , The resulting straight line （ curve ） It will better fit the discrete points on the interval .

Theoretical knowledge reference ：https://blog.csdn.net/qq_45771939/article/details/119800382?ops_request_misc=&request_id=&biz_id=102&utm_term=%E7%BA%BF%E6%80%A7%E5%9B%9E%E5%BD%92&utm_medium=distribute.pc_search_result.none-task-blog-2_allsobaiduweb~default-3-119800382.142^v32pc_rank_34,185^v2control&spm=1018.2226.3001.4187