我想我在 plt.plot 的尺寸上有问题

Question

在此处输入图像描述

我的 plt.plot 看起来像这样。

我想像这样在plt中的plot衬线行。 在此处输入图像描述

我的代码就像

import numpy as np
import pandas as pd
import matplotlib.pyplot as plt

# Preprocessing Input data
data = pd.read_csv('advertising.csv')
plt.rcParams['figure.figsize'] = (12.0, 9.0)
X = data.iloc[:,0]
Y = data.iloc[:,-1]

plt.scatter(X,Y)
# Building the model
m = 2
c = 1

L = 0.01  # The learning Rate
epochs = 10  # The number of iterations to perform gradient descent

n = float(len(X)) # Number of elements in X

# Performing Gradient Descent 
for i in range(epochs): 
    Y_pred = m*X + c  # The current predicted value of Y
    D_m = (-2/n) * sum(X *(Y - Y_pred))  # Derivative wrt m
    D_c = (-2/n) * sum(Y - Y_pred)  # Derivative wrt c
    m = m - L * D_m  # Update m
    c = c - L * D_c  # Update c
    
print (m, c)

# Making predictions
Y_pred = m*X + c

plt.scatter(X, Y) 
plt.plot([min(X), max(X)], [min(Y_pred), max(Y_pred)], color='red')  # regression line
plt.show()

Answer 1

我猜你应该在第二个plt.scatter中使用 plot Y_pred 而不是 Y

import matplotlib.pyplot as plt

# Preprocessing Input data
data = pd.read_csv('advertising.csv')
plt.rcParams['figure.figsize'] = (12.0, 9.0)
X = data.iloc[:,0]
Y = data.iloc[:,-1]

# plt.scatter(X,Y) # Not required
# Building the model
m = 2
c = 1

L = 0.01  # The learning Rate
epochs = 10  # The number of iterations to perform gradient descent

n = float(len(X)) # Number of elements in X

# Performing Gradient Descent 
for i in range(epochs): 
    Y_pred = m*X + c  # The current predicted value of Y
    D_m = (-2/n) * sum(X *(Y - Y_pred))  # Derivative wrt m
    D_c = (-2/n) * sum(Y - Y_pred)  # Derivative wrt c
    m = m - L * D_m  # Update m
    c = c - L * D_c  # Update c
    
print (m, c)

# Making predictions
Y_pred = m*X + c 

plt.scatter(X, Y_pred) #########  Y ---> Y_pred
plt.plot([min(X), max(X)], [min(Y_pred), max(Y_pred)], color='red')  # regression line
plt.show()