Python中不使用int()转换且不定义def函数，如何比较两个数值字符串的大小关系？

Let's work through this problem directly—no fancy functions or type conversion allowed, just plain Python string logic. Let's use your example values to walk through the solution:

num1 = "67"
num2 = "1954"

First, leverage string length for quick checks #

For positive integer strings, length tells us a lot right away: a longer string always represents a larger number (since we're dealing with base-10 digits). So we start here:

if len(num1) > len(num2):
    print(f"{num1} is greater than {num2}")
elif len(num1) < len(num2):
    print(f"{num1} is less than {num2}")

If lengths match, compare digit-by-digit #

When the strings are the same length, we can compare individual characters directly. Numeric characters have sequential ASCII values (e.g., '0' < '1' < ... < '9'), so we don't need to convert them to numbers to know their order:

else:
    # Iterate through each pair of digits
    for d1, d2 in zip(num1, num2):
        if d1 > d2:
            print(f"{num1} is greater than {num2}")
            break
        elif d1 < d2:
            print(f"{num1} is less than {num2}")
            break
    else:
        # The loop finished without breaking—all digits are identical
        print(f"{num1} is equal to {num2}")

Full working code for your example #

Putting it all together, here's the complete snippet that will correctly evaluate your test case:

num1 = "67"
num2 = "1954"

if len(num1) > len(num2):
    print(f"{num1} > {num2}")
elif len(num1) < len(num2):
    print(f"{num1} < {num2}")
else:
    for d1, d2 in zip(num1, num2):
        if d1 > d2:
            print(f"{num1} > {num2}")
            break
        elif d1 < d2:
            print(f"{num1} < {num2}")
            break
    else:
        print(f"{num1} == {num2}")

Running this will output 67 < 1954, which is exactly what we expect.

Quick notes #

• This assumes your input strings are positive integers (no negative signs, decimals, or leading zeros). If leading zeros are possible, you'd want to add a quick cleanup step like num1 = num1.lstrip('0') or '0' to handle cases like "0067" vs "67".
• The else attached to the for loop is a handy Python trick—it only runs if the loop completes without hitting a break, meaning every digit matched perfectly.

内容的提问来源于stack exchange，提问作者desperate