在PyCharm编写分类代码时遇TypeError: 'int'对象不可订阅错误求助

Let's break down the problem and fix your code step by step. The core error TypeError: 'int' object is not subscriptable comes from a mismatch between your variable types and how you're trying to access them, plus several other structural issues in your code.

1. Root Cause of the int Subscript Error #

Looking at your code:

net1 = np.matmul(W1, Xi[i]) + b1
y1 = AF(net1)
net2 = np.matmul(W2, y1[i]) + b2

• W1 is defined as a 1D array (np.zeros(2)), and Xi[i] is a 1D list. np.matmul(W1, Xi[i]) returns a scalar value.
• Your AF function returns either 1 or -1 (integers), so y1 is a single integer, not an array. Trying to access y1[i] is invalid—integers don't have subscripts!

2. Fixing Variable Dimensions & Network Structure #

It looks like you're trying to implement a two-layer perceptron with a hidden layer (judging by the for K in range(3) loop). Here's how to correct your variable definitions to match that structure:

• Input layer: 2 features
• Hidden layer: 3 neurons (so W1 should be a (3, 2) array, b1 a (3,) array)
• Output layer: 1 neuron (so W2 should be a (1, 3) array, b2 a scalar)

3. Full Corrected Code with Explanations #

import numpy as np
import matplotlib.pyplot as plt

def AF(x):
    if x >= 0:
        return 1
    else:
        return -1

Q1 = 3
Q2 = 1
# Convert Xi to numpy array for proper indexing later
Xi = np.array([[-1, -1], [0, 1], [1, -1], [0, 0]])
T = np.concatenate((np.ones((1, Q1)), -1 * np.ones((1, Q2))), axis=1)
T = np.squeeze(T)

# Correct dimensions for a 2-layer perceptron (3 hidden neurons)
W1 = np.zeros((3, 2))  # 3 hidden neurons, each with 2 input weights
W2 = np.ones((1, 3))   # Output neuron with 3 hidden layer inputs
b1 = np.zeros(3)       # Bias for each hidden neuron
b2 = 0                 # Bias for output neuron

N = Q1 + Q2
k = 0

for i in range(4):
    # Calculate net input for hidden layer (3 values, one per neuron)
    net1 = np.matmul(W1, Xi[i]) + b1
    # Apply activation function to each hidden neuron output
    y1 = np.array([AF(x) for x in net1])
    # Calculate net input for output layer
    net2 = np.matmul(W2, y1) + b2
    output = AF(net2[0])  # net2 is a 1D array, extract scalar value

    if output != T[i]:
        if T[i] == 1:
            # Find the hidden neuron with maximum net input
            J = np.argmax(net1)
            # Update bias and weights for that hidden neuron
            b1[J] += (1 - net1[J])
            W1[J] += (1 - net1[J]) * Xi[i]
        elif T[i] == -1:
            # Update all hidden neurons' bias and weights
            for K in range(3):
                b1[K] += (-1 - net2[0])
                W1[K] += (-1 - net2[0]) * Xi[i]

# Uncomment and adjust if you want to plot decision boundaries
# xx1 = np.arange(-3, 3)
# Example: Plot one of the hidden layer's decision boundary
# xx2 = (-b1[0] - W1[0, 0] * xx1) / W1[0, 1]

# Plot the data points
plt.plot(Xi[T == -1, 0], Xi[T == -1, 1], 'go', label='Class -1')
plt.plot(Xi[T == 1, 0], Xi[T == 1, 1], 'r^', label='Class 1')
# plt.plot(xx1, xx2, 'g', label='Decision Boundary')
plt.legend()
plt.show()

Key Fixes Made: #

• Converted Xi to a numpy array to enable advanced indexing for plotting.
• Corrected the dimensions of W1, W2, and b1 to match a 2-layer perceptron with 3 hidden neurons.
• Modified y1 to be an array of activation values (one per hidden neuron) instead of a single integer.
• Fixed weight/bias update logic: replaced invalid W[i,J] and b[J] with the correct variables (W1[J], b1[J]), and used scalar multiplication instead of unnecessary np.matmul for scalar-array operations.
• Fixed access to net2 (it's a 1D array, so we use net2[0] to get the scalar value).

Additional Notes: #

• The decision boundary code was commented out because your original code references W and b which no longer exist (we have W1, W2, b1, b2 instead). You can adjust it to plot boundaries for individual hidden neurons or the output layer as needed.
• Make sure your training loop logic aligns with the learning rule you're trying to implement (e.g., perceptron learning rule for the hidden layer).

内容的提问来源于stack exchange，提问作者rezvaneh zahedi