使用指定方法创建Python平方根求解函数时遇无输出问题的技术求助

Hey there! Let's figure out why your code isn't producing output and fix it up. I can see you're trying to implement a square root calculation (looks like a take on Newton-Raphson iteration), so let's break down the issues step by step:

Key Problems in Your Code #

• Infinite Loop: Your while True loop has no exit condition unless the exact check (int((temp+n)/2))**2 == num is met. For non-perfect squares (like 2, 3, 5), this condition will never trigger, so the loop runs forever without returning anything.
• Overly Strict Termination Check: That integer cast and exact equality check only works for perfect squares. Most square roots are irrational, so we need a way to stop when our guess is "close enough" (using a precision threshold) instead of waiting for an exact match.
• Redundant Logic: The if not num/n == n check is unnecessary—if num/n == n, then n is already the square root, and we can just return it immediately instead of proceeding with the temp variable dance.

Fixed Implementation #

Here's a revised version of your function that addresses these issues, with comments explaining each change:

def findSquareRt(num):
    # Handle edge cases first to avoid errors
    if num < 0:
        return "Error: No real square root for negative numbers"
    if num == 0:
        return 0
    
    n = 8  # Initial guess
    precision = 1e-6  # Define how close we need to be to the actual value
    
    while True:
        next_n = (n + num / n) / 2  # Core Newton-Raphson iteration step
        # Stop the loop when our guess is accurate enough
        if abs(next_n - n) < precision:
            return next_n
        # Update our guess for the next iteration
        n = next_n

How This Works #

1. Edge Case Handling: We first check for negative numbers (no real roots) and 0 (trivial case) to avoid runtime errors.
2. Precision-Based Termination: Instead of waiting for an exact match, we stop when the difference between our current and next guess is smaller than 1e-6—you can adjust this value for more or less precision depending on your needs.
3. Simplified Iteration: The Newton-Raphson formula is streamlined to its core form, which converges quickly to the correct square root for any positive number.

Test It Out #

Try these examples to see it in action:

• print(findSquareRt(16)) → Returns 4.0
• print(findSquareRt(2)) → Returns approximately 1.41421356237
• print(findSquareRt(25)) → Returns 5.0

内容的提问来源于stack exchange，提问作者LauJk_78