python - 将号码拆分为可变费率计费的频段，即 "tiered pricing"

Question

我有一个在纸面上很简单的问题，但我很难用代码解决。在我继续之前，这不是字符串拆分问题。

我有一个本质上是带状时间表应用程序。比例因客户而异，但一个例子是一个月的前三个小时为 80，接下来的三个小时为 70，其余为 50。我目前在代码中表示为:

scale = [80, 80, 80, 70, 70, 70, 50]

...但我也愿意接受那里的建议。

步骤的规模和数量是​​——而且必须是——可变的。对于某些客户，我的一些计费要简单得多，但我希望能够提供这种高使用率计划。

但是我如何计算工作小时数(例如 15.2)并计算他们应该支付多少？我如何将这个大数字分成乐队？正如我所说，纸上谈兵很容易，但随着我的客户越来越多，计划越来越复杂，这就变得很无聊了。以下是我将如何计算 15.2 小时:

3 hours at 80 = 240
3 hours at 70 = 210
9.2 hours at 50 = 460
total = 910

当我在做这件事的时候，我很感激对我试图描述的内容的专有名称的评论。还有 Oli，如果你在 2023 年回到这里，下次选择一个更简单的计费方案，伙计。

Answer 1

首先，我会把这个:

scale = [80, 80, 80, 70, 70, 70, 50]

进入这个:

import math

scale = {(0, 3): 80, (3, 6): 70, (6, math.inf): 50}

然后我的算法的其余部分如下:

# Total hours worked
hours_worked = 15.2

# Handle the decimal (if any) to begin with... First find the "max" rate
decimal_rate = next(rate for (lower, upper), rate in scale.items()
                        if lower <= hours_worked and upper >= hours_worked)

# Then calculate the last "sliver" of pay
decimal_end = hours_worked - int(hours_worked)
end_pay = decimal_end * decimal_rate

# Use an integer for ease of calculation
hours_worked = int(hours_worked)

hours_paid_for = 0

# Beginning total pay is just the decimal "ending"
total_pay = end_pay

while hours_paid_for < hours_worked:
    # Find the rate for the current bucket of hours
    rate_filter = (rate for (lower, upper), rate in scale.items() if lower <= hours_paid_for and hours_paid_for < upper)
    current_level = next(rate_filter)

    print('Hour: {}'.format(hours_paid_for))
    print('Pay rate: ${}'.format(current_level))

    total_pay += current_level

    hours_paid_for += 1

print('Total earned: ${}'.format(total_pay))

输出如下:

Hour: 0
Pay rate: $80
Hour: 1
Pay rate: $80
Hour: 2
Pay rate: $80
Hour: 3
Pay rate: $70
Hour: 4
Pay rate: $70
Hour: 5
Pay rate: $70
Hour: 6
Pay rate: $50
Hour: 7
Pay rate: $50
Hour: 8
Pay rate: $50
Hour: 9
Pay rate: $50
Hour: 10
Pay rate: $50
Hour: 11
Pay rate: $50
Hour: 12
Pay rate: $50
Hour: 13
Pay rate: $50
Hour: 14
Pay rate: $50
Total earned: $910.0

这里还有一个简洁的函数:

def calculate_pay(scale, hours_worked):

    # Handle the decimal (if any) to begin with... First find the "max" rate
    decimal_rate = next(rate for (lower, upper), rate in scale.items()
                            if lower <= hours_worked and upper >= hours_worked)

    # Then calculate the last "sliver" of pay
    decimal_end = hours_worked - int(hours_worked)
    end_pay = decimal_end * decimal_rate

    # Use an integer for ease of calculation
    hours_worked = int(hours_worked)

    # Hours already paid for (int)
    hours_paid_for = 0

    # Beginning 'total pay' can be the decimal end, if any
    total_pay = end_pay

    while hours_paid_for < hours_worked:
        # Find the rate for the current bucket of hours
        rate_filter = (rate for (lower, upper), rate in scale.items()
                        if lower <= hours_paid_for and hours_paid_for < upper)
        current_level = next(rate_filter)

        total_pay += current_level

        hours_paid_for += 1

    return total_pay