Question

# user input
num = int(input("Please enter engine size: "))

# calculations
if num <= 1000:
    print("The motor tax for your vehicle is €150")
elif num >= 1001 <= 1200:
    print("The motor tax for your vehicle is €175")
elif num >= 1201 <= 1400:
    print("The motor tax for your vehicle is €200")
elif num >= 1401 <= 1600:
    print("The motor tax for your vehicle is €250")
elif num >= 1601 <= 1800:
    print("The motor tax for your vehicle is €300")
elif num >= 1801 <= 2000:
    print("The motor tax for your vehicle is €350")
else:
    print("The motor tax for your vehicle is €500")

我知道我在这里可能犯了一个愚蠢的错误，我只是希望有人可以指出我正确的方向。

我正在尝试让python为每个指定的引擎尺寸打印相应的数量。

每次运行它的金额大于1000时，它只会给我输出€175。

非常感谢！

Answer 1

这不是您想要的或期望的：

elif num >= 1001 <= 1200:

您应将其替换为：

elif num in range(1001, 1201):

注意：要检查<=，您需要将范围的上限增加1 ！！

否则，您可以将原来的内容写为：

elif 1001 <= num <= 1200:

Answer 2

欢迎堆栈溢出！

我已将PL200的答案更新为您的问题的答案。这是一个改进您的代码的建议：

# Motor tax is based on ranges
if num <= 1000:
    tax = 150
elif num <= 1200:
    tax = 175
elif num <= 1400:
    tax = 200
elif num <= 1600:
    tax = 250
elif num <= 1800:
    tax = 300
elif num <= 2000:
    tax = 350
else:
    tax = 500
print("The motor tax for your vehicle is €{}".format(tax))

由于if / elif链，实际上不需要两个不等式。

或者您甚至可以避免长时间使用if / elif：

def get_tax(num):
    """Return the vehicle tax based on ranges"""
    # Ranges are stored as "maximum size" / "tax" 
    tax_ranges = (
        (1000, 150),
        (1200, 175),
        (1400, 200),
        (1600, 250),
        (1800, 300),
        (2000, 350),
    )
    default_tax = 500

    for max_size, tax in tax_ranges:
        if num <= max_size:
            return tax
    return default_tax

num = int(input("Please enter engine size: "))
print(f"The motor tax for your vehicle is {get_tax(num)}")

注意：我在末尾添加了一个f-string以进行打印，但是如果您的Python版本早于3.6，则您的Python版本可能不支持它。在这种情况下，只需将{get_tax(num)}替换为"... {}".format(get_tax(num))，然后删除字符串前的f

python只激活第一个elif，忽略其他

2 个答案: