百分比pythonic方式

Question

我目前正在尝试创建一个具有一个项目（代表游戏内）的随机机会的方法，例如一个项目的20％，或50％的百分比，甚至1％。

from random import randint 

def Break(): 
    print(".") 

def Treasure(): 

    Gen = randint(1,100) 
    if Gen <= 10: 
     print("10%") 
     Break() 

    Gen = randint(1,100) 
    if Gen <= 10: 
     print("10%") 
     Break() 

    Gen = randint(1,100) 
    if Gen <= 10: 
     print("10%") 
     Break() 

    Gen = randint(1,100) 
    if Gen <= 10: 
     print("10%") 
     Break() 

    Gen = randint(1,100) 
    if Gen <= 10: 
     print("10%") 
     Break() 

此代码方法这类作品的，问题是，即使选择从这里百分比和休息（）被调用，还有的可能：我曾尝试与如下所示的代码做这个它有超过一个百分比列出 因此，如果它说80％，然后Break（） 它也可以列出50％，然后Break（）一段时间。

所以我的问题是，我怎么能做到以下几点：

• 比较有效的代码
• 使百分比只能拨打ONE再突破（）
• 使得百分比肯定的方式是什么他们说，例如50％的一半机会。

测试代码：（编辑）

if gen <= 10: 
    print("10%") 
elif gen <= 20: 
    print("20%") 
elif gen <= 30: 
    print("30%") 
elif gen <= 40: 
    print("30%") 
elif gen <= 50: 
    print("50%") 
elif gen <= 60: 
    print("60%") 
elif gen <= 70: 
    print("70%") 
elif gen <= 80: 
    print("80%") 
elif gen <= 90: 
    print("90%") 

如果这没有使90％的最有可能的，反过来10％的最少？

Answer 1

几件事情第一：

• 资本事宜; break和Break是两个不同的东西

• break是一个关键字，而不是一个函数;称之为break而不是break()

• break是退出循环，但不在循环中;也许你的意思是return？

你是级联的情况下抛出的百分比注销的方式：

gen = randint(1,100) 
if gen <= 10: 
    return "10% (#1)" 

gen = randint(1,100) 
if gen <= 10: 
    return "10% (#2)"  # 10% of 90% == 9% 

gen = randint(1,100) 
if gen <= 10: 
    return "10% (#3)"  # 10% of 81% == 8.1% 

对于需要累计测试正确的价值观，像

gen = randint(1, 100) 
if gen <= 10: 
    return "10% (#1)" 
elif gen <= 20: 
    return "10% (#2)" 
elif gen <= 30: 
    return "10% (#3)" 

或许，这将节省大量的加重包装这一类，就像

from bisect import bisect # fast binary search 
from random import random 

class RandomItemGenerator: 
    def __init__(self, items=None, probs=None): 
     # start with no items 
     self.items  = [] 
     self.total  = 0. 
     self.breakpoints = [] 
     # add any initial items 
     if items is not None: 
      for item, prob in zip(items, probs): 
       self.add_item(item, prob) 

    def add_item(self, item, prob): 
     self.items.append(item) 
     self.total += prob 
     self.breakpoints.append(self.total) 

    def __call__(self): 
     if self.items: 
      value = random() * self.total 
      index = bisect(self.breakpoints, value) 
      return self.items[index] 
     else: 
      raise ValueError("you haven't got any items yet") 

那么你的代码看起来像

make_treasure = RandomItemGenerator() 
make_treasure.add_item("sword", 80) 
make_treasure.add_item("sword +1", 19) 
make_treasure.add_item("ring +2", 1) 

，我们可以测试像

from collections import Counter 

# generate 10,000 items 
test = Counter(make_treasure() for i in range(10000)) 

# check number of each item generated 
print(test.most_common()) 

这给像

[('sword', 8002), ('sword +1', 1887), ('ring +2', 111)] 

---

编辑：一个例子返回功能：

def sword_fn(): 
    # your stuff goes here 

def dagger_fn(): 
    # your stuff goes here 

def wand_fn(): 
    # your stuff goes here 

make_fn = RandomItemGenerator() 
make_fn.add_item(sword_fn, 80) 
make_fn.add_item(dagger_fn, 19) 
make_fn.add_item(wand_fn, 1) 

which = make_fn()  # pick a function 
which()    # run the function 

我最后的评论中使用的等同的简写，就像

make_fn = RandomItemGenerator([sword_fn, dagger_fn, wand_fn], [80, 19, 1]) 

这不正是同样的事情上面的四条线，我在那里创建实例，然后添加项目。

Answer 2

你可以尝试这样的事情：

def Break(): 
    print(".") 

def FindAnotherName(): 
    Gen = randint(1,100) 
    if Gen <= 10: 
     print("10%") 
     Break() 
     return True 
    else: 
     return False 

def Treasure(): 

    i = 0 
    finished = False 
    while (i < 5 && finished == False): 
     finished = FinAnotherName() 
     i = i + 1