如何使Python在序列中正常工作而不必每次重新启动？

我对python/pygame非常陌生，因此如果我没有使用正确的术语，请原谅我

我正在做一个游戏，玩家射击两个物体。当一个物体被击中时，生命值在时间上减少380%，在时间上减少1320%。我已经通过使用rand.choice和一个列表实现了这一点

我想重做它的设置方式，以便我确切地知道结果会是什么，以及以什么顺序

一位朋友建议使用一个长度范围，但他只知道如何使用不同的编码语言

这是我的代码：

hits = pg.sprite.groupcollide(self.enemies, self.shards, False, True)
for hit in hits:
   hit.health -= choice(RANDO)
   hit.vel = vec(0, 0)
   print('win')

RANDO的定义如下：

RANDO = [LOSE, LOSE, JACKPOT, LOSE, LOSE,
         LOSE, LOSE, LOSE, LOSE, JACKPOT]

损失=3，头奖=13

我希望每次射击敌人的结果都是我随机列出的。所以玩家只会在第3次和第10次中大奖。基本上，玩家每次射击时，所造成的伤害将根据列表中的下一步决定

我知道我必须摆脱rand.choice，我用它来让游戏攻击以一定的伤害和一定的百分比随机命中，但我不再需要它了

---

如果有人知道我应该使用什么或插入什么，我将不胜感激。我只需要它从列表中提取一个值，然后通过它来计算。

您可以保留一个计数器：

hit_count = 0
for hit in hits:
   hit.health -= RANDO[hit_count % len(RAND0)]
   hit_count += 1
   hit.vel = vec(0, 0)
   print('win')

如果列表中只有一个元素，只需根据需要

pop

for hit in hits:
   hit.health -= RANDO.pop()
   hit.vel = vec(0, 0)
   print('win')

---

功能检查：

LOSE = 3
JACKPOT = 13
RANDO = [LOSE, LOSE, JACKPOT, LOSE, LOSE,
         LOSE, LOSE, LOSE, LOSE, JACKPOT]

hit_count = 0
health = 50
for hit in range(10):
   health -= RANDO[hit_count % len(RANDO)]
   hit_count += 1
   print('new health', health)

输出：

new health 47
new health 44
new health 31        JACKPOT
new health 28
new health 25
new health 22
new health 19
new health 16
new health 13
new health 0         JACKPOT

new health 47
new health 44
new health 31
new health 28
new health 25
new health 22
new health 19
new health 16
new health 13
new health 0

我在第二个解决方案上的错误：

pop

从另一端起作用：

for hit in range(10):
   health -= RANDO.pop(0)
   print('new health', health)

输出：

new health 47
new health 44
new health 31        JACKPOT
new health 28
new health 25
new health 22
new health 19
new health 16
new health 13
new health 0         JACKPOT

new health 47
new health 44
new health 31
new health 28
new health 25
new health 22
new health 19
new health 16
new health 13
new health 0

---

您可以保留一个柜台：

hit_count = 0
for hit in hits:
   hit.health -= RANDO[hit_count % len(RAND0)]
   hit_count += 1
   hit.vel = vec(0, 0)
   print('win')

如果列表中只有一个元素，只需根据需要

pop

for hit in hits:
   hit.health -= RANDO.pop()
   hit.vel = vec(0, 0)
   print('win')

---

功能检查：

LOSE = 3
JACKPOT = 13
RANDO = [LOSE, LOSE, JACKPOT, LOSE, LOSE,
         LOSE, LOSE, LOSE, LOSE, JACKPOT]

hit_count = 0
health = 50
for hit in range(10):
   health -= RANDO[hit_count % len(RANDO)]
   hit_count += 1
   print('new health', health)

输出：

new health 47
new health 44
new health 31        JACKPOT
new health 28
new health 25
new health 22
new health 19
new health 16
new health 13
new health 0         JACKPOT

new health 47
new health 44
new health 31
new health 28
new health 25
new health 22
new health 19
new health 16
new health 13
new health 0

我在第二个解决方案上的错误：

pop

从另一端起作用：

for hit in range(10):
   health -= RANDO.pop(0)
   print('new health', health)

输出：

new health 47
new health 44
new health 31        JACKPOT
new health 28
new health 25
new health 22
new health 19
new health 16
new health 13
new health 0         JACKPOT

new health 47
new health 44
new health 31
new health 28
new health 25
new health 22
new health 19
new health 16
new health 13
new health 0

---

一种简单的方法是创建一个预期序列的列表（正如您已经使用

RANDO

所做的那样），然后使用来“循环”此序列：

λ python
Python 3.7.3 (v3.7.3:ef4ec6ed12, Mar 25 2019, 21:26:53) [MSC v.1916 32 bit (Intel)] on win32
Type "help", "copyright", "credits" or "license" for more information.
>>> RANDO = ['LOSE', 'LOSE', 'JACKPOT', 'LOSE', 'LOSE', 'LOSE', 'LOSE', 'LOSE', 'LOSE', 'LOSE', 'JACKPOT']
>>> from itertools import cycle
>>> RANDO_GEN = cycle(RANDO)
>>> for _ in range(25):
...     print(next(RANDO_GEN))
...
LOSE
LOSE
JACKPOT
LOSE
LOSE
LOSE
LOSE
LOSE
LOSE
LOSE
JACKPOT
LOSE
LOSE
JACKPOT
LOSE
LOSE
LOSE
LOSE
LOSE
LOSE
LOSE
JACKPOT
LOSE
LOSE
JACKPOT
>>>

cycle

返回一个生成器，使用

next

获取这个无限序列的下一个元素；因此，您只需将代码替换为：

LOSE = 3
JACKPOT = 13
RANDO = cycle([LOSE, LOSE, JACKPOT, LOSE, LOSE,
         LOSE, LOSE, LOSE, LOSE, JACKPOT])

...

for hit in hits:
    hit.health -= next(RANDO)
    hit.vel = vec(0, 0)
    print('win')

---

一种简单的方法是创建一个预期序列的列表（正如您已经使用

RANDO

所做的那样），然后使用来“循环”此序列：

λ python
Python 3.7.3 (v3.7.3:ef4ec6ed12, Mar 25 2019, 21:26:53) [MSC v.1916 32 bit (Intel)] on win32
Type "help", "copyright", "credits" or "license" for more information.
>>> RANDO = ['LOSE', 'LOSE', 'JACKPOT', 'LOSE', 'LOSE', 'LOSE', 'LOSE', 'LOSE', 'LOSE', 'LOSE', 'JACKPOT']
>>> from itertools import cycle
>>> RANDO_GEN = cycle(RANDO)
>>> for _ in range(25):
...     print(next(RANDO_GEN))
...
LOSE
LOSE
JACKPOT
LOSE
LOSE
LOSE
LOSE
LOSE
LOSE
LOSE
JACKPOT
LOSE
LOSE
JACKPOT
LOSE
LOSE
LOSE
LOSE
LOSE
LOSE
LOSE
JACKPOT
LOSE
LOSE
JACKPOT
>>>

cycle

返回一个生成器，使用

next

获取这个无限序列的下一个元素；因此，您只需将代码替换为：

LOSE = 3
JACKPOT = 13
RANDO = cycle([LOSE, LOSE, JACKPOT, LOSE, LOSE,
         LOSE, LOSE, LOSE, LOSE, JACKPOT])

...

for hit in hits:
    hit.health -= next(RANDO)
    hit.vel = vec(0, 0)
    print('win')

---

最终起作用的是使用了popleft元素

    '''for hit in hits:
          hit.health -= RANDO.pop()
          hit.vel = vec(0, 0)
          print('win')'''

在我希望它运行的列表之前添加deque

    '''LOSE = 3
       JACKPOT = 13
       RANDO = deque([LOSE, LOSE, JACKPOT, LOSE, LOSE,
                 LOSE, LOSE, LOSE, LOSE, JACKPOT])'''

---

现在，它按顺序一个接一个地浏览列表——在它从末尾开始向后运行之前。

最终运行的是使用了popleft元素

    '''for hit in hits:
          hit.health -= RANDO.pop()
          hit.vel = vec(0, 0)
          print('win')'''

在我希望它运行的列表之前添加deque

    '''LOSE = 3
       JACKPOT = 13
       RANDO = deque([LOSE, LOSE, JACKPOT, LOSE, LOSE,
                 LOSE, LOSE, LOSE, LOSE, JACKPOT])'''

---

现在，它按顺序一个接一个地浏览列表——在它从末尾开始并向后运行之前。

计数器在计时时没有选择JACKPOT值。RANDO.pop也没有按顺序做，但它似乎能满足我的需要。在第6次拍摄中，程序关闭时出现错误：indexError:pop from empty List计数器在拍摄时没有选择JACKPOT值。RANDO.pop也没有按顺序做，但它似乎能满足我的需要。在第6个镜头中，程序关闭时出现错误：indexer-ror:pop from empty list