为何调用gas_price(100.0,120.0)返回'3/4 Tank'而非'Half Tank'？

核心原因 #

你遇到的问题是浮点数精度误差导致的。虽然数学上(120.0 / 100.0) - 1等于0.2，但计算机在存储和计算浮点数时，可能会将这个值表示为一个非常接近0.2但略小于0.2的数（例如0.19999999999999996）。这个微小的误差会让代码触发percent < .2的条件分支，从而返回'3/4 Tank'而非预期的'Half Tank'。

解决方案 #

针对浮点数精度问题，你可以选择以下几种修复方式：

1. 引入误差范围（Epsilon）

通过定义一个极小的误差阈值，允许浮点数在这个范围内被视为相等：

def gas_price(previous, new):
    percent = (new / previous) - 1
    epsilon = 1e-9  # 极小的误差容忍值
    if abs(new - previous) < epsilon:
        decision = 'Full Tank'
    elif percent < 0.2 - epsilon:
        decision = '3/4 Tank'
    elif percent < 0.8 - epsilon:
        decision = 'Half Tank'
    elif percent < 1.0 - epsilon:
        decision = '1/4 Tank'
    else:
        decision = 'Go Home'
    return decision

2. 使用精确十进制计算

借助Python的decimal模块进行精确的十进制运算，彻底避免浮点数误差：

from decimal import Decimal

def gas_price(previous, new):
    prev_dec = Decimal(str(previous))
    new_dec = Decimal(str(new))
    percent = (new_dec / prev_dec) - Decimal(1)
    
    if new_dec == prev_dec:
        decision = 'Full Tank'
    elif percent < Decimal('0.2'):
        decision = '3/4 Tank'
    elif percent < Decimal('0.8'):
        decision = 'Half Tank'
    elif percent < Decimal('1.0'):
        decision = '1/4 Tank'
    else:
        decision = 'Go Home'
    return decision

3. 调整条件判断顺序与逻辑

通过嵌套判断确保范围覆盖的准确性，减少精度误差的影响：

def gas_price(previous, new):
    percent = (new / previous) - 1
    if new == previous:
        decision = 'Full Tank'
    elif percent < 0.8:
        # 先判断是否小于0.2，否则直接归为0.2~0.8区间
        decision = '3/4 Tank' if percent < 0.2 else 'Half Tank'
    elif percent < 1.0:
        decision = '1/4 Tank'
    else:
        decision = 'Go Home'
    return decision

内容的提问来源于stack exchange，提问作者Miller25