Python递归函数实现：寻找满足向量差和阈值的迭代向量

Let's break down why your recursive function isn't behaving as expected and fix it properly.

The Problem with Your Current Code #

Your recursive function has two key issues that prevent it from working correctly:

• In the else branch, you call diff_sum(M, np.dot(M, v1)) but don't return the result of this recursive call. This means the recursive work gets discarded, and the function just returns the first iteration's values every time.
• You're unnecessarily recalculating np.dot(M, v1) for the recursive call—v1 is already the result of M*v, so you can just pass v1 directly as the new input vector.

Corrected Recursive Function #

Here's the fixed version with clearer logic and proper recursion handling:

import numpy as np

def diff_sum(M, v):
    v_next = np.dot(M, v)
    # Calculate sum of absolute differences between current and next vector
    total_diff = np.sum(np.absolute(v - v_next))
    
    # Check if we've met the threshold
    if total_diff < 0.45:
        return (total_diff, v, v_next)
    # If not, recurse with the next vector (and return the recursive result!)
    else:
        return diff_sum(M, v_next)

Key Improvements #

• Clearer variable names: v_next makes it obvious this is the next iteration of the vector, instead of vague v1.
• Proper recursion flow: Adding return to the else branch ensures the result from deeper recursive calls is passed back up correctly.
• Efficient summation: Using np.sum() instead of Python's built-in sum() is more reliable for numpy arrays and performs better with larger data.
• Avoid redundant calculations: We pass v_next directly to the recursive call instead of recalculating the matrix dot product.

Testing the Function #

Let's use your sample data to verify it works:

# Your original matrix and initial vector
M = np.array([[0, 0, 0, 0.5, 0, 0], [1, 0, 0, 0, 0.5, 0], [0, 1, 0, 0, 0.5, 1], [0, 0, 1, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0.5, 0, 0]])
M = M.astype(float)
v0 = np.array([1/6, 1/6, 1/6, 1/6, 1/6, 1/6])

# Run the recursive function
diff, vi, vi_plus_1 = diff_sum(M, v0)

print(f"Sum of absolute differences: {diff}")
print(f"Vector v_i: {vi}")
print(f"Vector v_{i+1}: {vi_plus_1}")

This will iterate until it finds the pair of vectors where the sum of absolute differences is below 0.45, just like your manual calculation of v4 and v5.

内容的提问来源于stack exchange，提问作者user152103