百分比pythonic方式

Question

我目前正在嘗試創建一個具有一個項目（代表遊戲內）的隨機機會的方法，例如一個項目的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() 

此代碼方法這類作品的，問題是，即使選擇從這裏百分比和休息（）被調用，還有的可能：我曾嘗試與如下所示的代碼做這個它有超過一個百分比列出 因此，如果它說80％，然後Break（） 它也可以列出50％，然後Break（）一段時間。

所以我的問題是，我怎麼能做到以下幾點：

• 比較有效的代碼
• 使百分比只能撥打ONE再突破（）
• 使得百分比肯定的方式是什麼他們說，例如50％的一半機會。

測試代碼：（編輯）

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%") 

如果這沒有使90％的最有可能的，反過來10％的最少？

Answer 1

幾件事情第一：

• 資本事宜; break和Break是兩個不同的東西

• break是一個關鍵字，而不是一個函數;稱之爲break而不是break()

• break是退出循環，但不在循環中;也許你的意思是return？

你是級聯的情況下拋出的百分比註銷的方式：

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% 

對於需要累計測試正確的價值觀，像

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

或許，這將節省大量的加重包裝這一類，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") 

那麼你的代碼看起來像

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

，我們可以測試像

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()) 

這給像

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

---

編輯：一個例子返回功能：

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 

我最後的評論中使用的等同的簡寫，就像

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

這不正是同樣的事情上面的四條線，我在那裏創建實例，然後添加項目。

Answer 2

你可以嘗試這樣的事情：

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