随机列表选择：如何确保同一项目不会连续重复两次，一个接一个？

Question

import random

welcomes = ["Hello","Hi","What's up","YO", "Piss off"]

chosen = random.choice(welcomes)

print("You have entered the welcome cave ----- {} -----".format(chosen))

我如何确保Hello 不会连续重复两次？以后再重复就好了，不要马上重复。

Answer 1

使用random.sample 而不是random.choice。 Find online demo

import random

welcomes = ["Hello","Hi","What's up","YO", "Piss off"]

chosen = random.sample(welcomes,2)

for item in chosen:
  print("You have entered the welcome cave ----- {} -----".format(item))

Answer 2

如果你想生成一个很长的问候流具有属性：没有连续的问候是相同的（在线演示：last version）：

import random

def random_hello():
    welcomes = ["Hello", "Hi", "What's up", "YO", "Piss off"]
    last_hello = None
    while 1:
        random.shuffle(welcomes)
        if welcomes[0] == last_hello:
            continue
        for item in welcomes:
            yield item
        last_hello = welcomes[-1]


hellower = iter(random_hello())
for _ in range(1000):
    print(next(hellower))

或者当您担心确定性时间时，交换元素（与第一个）：

if welcomes[0] == last_hello:
    welcomes[0], welcomes[1] = welcomes[1], welcomes[0]

或随机：

if welcomes[0] == last_hello:
    swap_with = random.randrange(1, len(welcomes))
    welcomes[0], welcomes[swap_with] = welcomes[swap_with], welcomes[0]

Answer 3

像其他建议的答案一样，使用random.sample 仅在您知道只需要一定数量的项目（例如2）时才有用。确保随机性和不重复的最佳方法是使用random.shuffle：

import random
welcomes = ["Hello","Hi","What's up","YO", "Piss off"]
random.shuffle(welcomes)

这很好地就地打乱列表，然后您可以开始将pop 项目从列表中移开，直到完成：

while len(welcomes)>0:
    print("You have entered the welcome cave ----- {} -----".format(welcomes.pop())

这适用于任何长度的列表，您可以使用此过程直到完成整个列表。如果您想让它永远持续下去，而不仅仅是直到列表结束，您还可以在整个过程中添加另一个循环。

Answer 4

你可以通过命中与错过的方法来做到这一点：

import random

class RandomChoiceNoImmediateRepeat(object):
    def __init__(self, lst):
        self.lst = lst
        self.last = None
    def choice(self):
        if self.last is None:
            self.last = random.choice(self.lst)
            return self.last
        else:
            nxt = random.choice(self.lst)
            # make a new choice as long as it's equal to the last.
            while nxt == self.last:   
                nxt = random.choice(self.lst)
            # Replace the last and return the choice
            self.last = nxt
            return nxt

可以使用random.choices 和权重（需要 python-3.6）对其进行优化，但该方法应该适用于所有 python 版本：

>>> welcomes = ["Hello","Hi","What's up","YO", "Piss off"]
>>> gen = RandomChoiceNoImmediateRepeat(welcomes)
>>> gen.choice()
'YO'

---

或者，如果您不喜欢 hit&miss，您也可以在 0 和列表长度之间绘制一个随机索引 - 2，如果它等于或高于前一个，则添加 1。这样可以确保不会发生重复，并且只需要一次调用 random 即可获得下一个选择：

import random

class RandomChoiceNoImmediateRepeat(object):
    def __init__(self, lst):
        self.lst = lst
        self.lastidx = None

    def choice(self):
        if self.lastidx is None:
            nxtidx = random.randrange(0, len(self.lst))
        else:
            nxtidx = random.randrange(0, len(self.lst)-1)
            if nxtidx >= self.lastidx:
                nxtidx += 1
        self.lastidx = nxtidx
        return self.lst[nxtidx]

Answer 5

我的看法：我们创建了两个相同的列表。在循环中，我们从一个列表中弹出一个值，如果该列表的长度小于原始列表 - 1，我们将列表重置为其原始状态：

import random

origin = ["Hello","Hi","What's up","YO", "Piss off"]
welcomes = origin.copy()

for i in range(5):
    if len(welcomes) < len(origin) - 1:
        welcomes = origin.copy()
    random.shuffle(welcomes) # shuffle
    chosen = welcomes.pop() # pop one value
    print("You have entered the welcome cave ----- {} -----".format(chosen))

例如输出 5 个循环：

You have entered the welcome cave ----- Piss off -----
You have entered the welcome cave ----- YO -----
You have entered the welcome cave ----- Piss off -----
You have entered the welcome cave ----- YO -----
You have entered the welcome cave ----- What's up -----

Answer 6

在= 输出中放入[a = list.consumableList] 而不仅仅是列表

（将括号中的小写列表替换为您的列表名称）