Python |避免从列表中随机选择的先前值

Question

基本上我想要我的代码做的是打印出 5 wrong() 函数，而不是连续 2 个相同的文本。当然，我不希望它是靠运气。 :)

虽然不用担心它打印出 5 wrong()s 的部分，但我只是想确定我是否至少使用此函数两次我 100% 确定以前的值不会和下一个一样。

例如我想避免的：

Wrong!
Wrong!

虽然这样还是可以的：

Wrong!
Incorrect!
Wrong!

我的代码：

import random

def wrong():
    wrong_stats=["\n Wrong!","\n Tough Luck!","\n Better Luck Next Time!","\n Not there yet!","\n Incorrect!"]
    rand = random.choice(wrong_stats)
    rand3 = random.choice(wrong_stats)
    norep(rand,rand3,wrong_stats)

def norep(rand,rand3,wrong_stats):
    if rand == rand3:
        same = True
        while same:
            rand = random.choice(wrong_stats)
            if rand != rand3:
                print(rand)
                break

    elif rand != rand3:
        print(rand)

wrong()
wrong()
wrong()
wrong()
wrong()

Answer 1

将之前的值存入一个全局变量，并从列表中随机选择（不包括之前的值）：

import random

wrong_stats=["\n Wrong!","\n Tough Luck!","\n Better Luck Next Time!","\n Not there yet!","\n Incorrect!"]
prev = ""

def wrong():
  global prev
  if prev == "":
    prev = random.choice(wrong_stats)
  else:
    prev = random.choice(wrong_stats[:wrong_stats.index(prev)] + wrong_stats[wrong_stats.index(prev)+1:])
  print prev
if __name__ == "__main__":
  wrong()
  wrong()
  wrong()
  wrong()
  wrong()

Answer 2

全局变量是一种不好的做法。

您应该将最后打印的值传递给 wrong，然后在除该值之外的所有值之间进行选择。像这样：

import random

def wrong(last):
    chosen = random.choice([stat for stat in WRONG_STATS if stat != last])
    print(chosen)
    return chosen

if "__main__" == __name__:
    last = None
    for i in xrange(5):
        last = wrong(last)

Answer 3

使用random.shuffle:

from random import shuffle
>>> shuffle(wrong_stats)
>>> wrong_stats
['\n Better Luck Next Time!', '\n Wrong!', '\n Incorrect!', '\n Tough Luck!', '\n Not there yet!']
>>> shuffle(wrong_stats)
>>> wrong_stats
['\n Incorrect!', '\n Tough Luck!', '\n Better Luck Next Time!', '\n Not there yet!', '\n Wrong!']
>>> shuffle(wrong_stats)
>>> wrong_stats
['\n Incorrect!', '\n Wrong!', '\n Better Luck Next Time!', '\n Not there yet!', '\n Tough Luck!']
>>> shuffle(wrong_stats)
>>> wrong_stats
['\n Wrong!', '\n Incorrect!', '\n Better Luck Next Time!', '\n Tough Luck!', '\n Not there yet!']

Answer 4

这是另一个可能会给你一个想法的例子：

import random 

country = ["Spain", "Sweden", "Netherlands", "Germany"]
lastcountry = ''
i = 0
while i <= 5:
    country = (random.choice(country))
    if (country != lastcountry):
        i = i + 1
        print (country)
    lastcountry = country

它会记住最后选择的选项。
如果它进行了新的选择，它会查看它是否与之前的选择不同。

Answer 5

您需要跟踪它返回的最后一个值；你可以

为此使用一个全局模块（在实践中通常很混乱），
或将其变成 class （有点冗长）、
或者在外部保持跟踪，每次都传入（笨重乏味），

但我认为最好的方法是将 wrong 函数改为 generator：这样您就可以跟踪生成器执行状态中的最后一个返回值，并在下次避免它，而不必担心在任何地方的外部代码中。

def wrong():
    wrong_stats = ["Wrong!","Tough Luck!","Better Luck Next Time!","Not there yet!","Incorrect!"]
    previous_value = None
    while True:
        value = random.choice(wrong_stats)
        if value != previous_value:
            yield value
            previous_value = value

和用法：

w = wrong()
for i in range(5):
    print(next(w))

# Tough Luck!
# Incorrect!
# Not there yet!
# Tough Luck!
# Better Luck Next Time!

您可以继续使用生成器调用 next，它将生成无限数量的字符串，而不会重复之前的值。

Python |避免从列表中随机选择的先前值

Python | Avoid previous value from random selection from list

python

random