如何根据y分配一个值
我的这段代码写起来非常烦人，但可以完成工作，我想知道是否有某种函数或类可以帮助我。
这是我的功能：
func add_health():
if wave in range(10,20):
enemy_health = 2
elif wave in range(20,30):
enemy_health = 3.5
elif wave in range(30,40):
enemy_health = 5
elif wave in range(40,50):
enemy_health = 6.5
elif wave in range(50,59):
enemy_health = 8
elif wave in range(60,69):
enemy_health = 9.5
else:
enemy_health = 1
I have this code that is very annoying to write but does the job, I was wondering if there was a function or class of some sort to help me.
Here's my function:
func add_health():
if wave in range(10,20):
enemy_health = 2
elif wave in range(20,30):
enemy_health = 3.5
elif wave in range(30,40):
enemy_health = 5
elif wave in range(40,50):
enemy_health = 6.5
elif wave in range(50,59):
enemy_health = 8
elif wave in range(60,69):
enemy_health = 9.5
else:
enemy_health = 1
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评论（1）
如果您的
wave变量是整数，则很简单1。如果不是，您应该能够将其转换为 int 以获得相同的效果。将其作为函数的输入使这变得简单（因为您可以强制转换）。看起来你每 10 波都会将
enemy_health值增加 1.5（至少到 70 波时会恢复为 1）。因此，只需将wave除以 10，然后乘以 1.5，并将其添加到偏移量即可。您可能需要对基本情况进行一些特殊处理（wave<10 或 >=70）。如果不考虑这些边缘情况，您的函数将变为
1 将两个整数相除得到一个整数，并丢弃任何余数。所以 53/10 == 5
If your
wavevariable is an integer, this is simple1. If it is not, you should be able to cast it to an int for the same effect. Making it an input to the function makes this simple (since you can force the conversion).It looks like you increase the
enemy_healthvalue 1.5 per 10 wave (at least until 70, where you revert to 1). So simply dividewaveby 10 and then multiple by 1.5, and add it to an offset. You may need some special handling for base cases (wave<10 or >=70).Without considering those edge cases, your function becomes
1 Dividing two integers results in an integer, discarding any remainder. So 53/10 == 5