Python 从不两次返回同一项的随机字典搜索

当要求程序从键值条目字典中选择一个随机条目时，是否有一种方法，一旦每个条目都被选中，程序会让用户知道所有条目都被选中，然后停止，最终只允许每个条目被选中一次，如果只有3个条目，程序将只运行3次，如果有100个条目，程序将运行100次，依此类推？python编码新手，请耐心听我说

from random import *
def show_keys():
    """ Show the user a random key and ask them
        to define it. Show the definition
        when the user presses return.    
    """
    random_key = choice(list(keys))
    print('Define: ', random_key)
    input('Press return to see the definition')
    print(keys [random_key])




# Set up the keys


keys = {'key1':'definition1',
        'key2':'definition2',
        'key3':'definition3'}



# The loop

exit = False
while not exit:
user_input = input('Enter s to show a key, or q to quit: ')
if user_input == 'q':
    exit = True
elif user_input == 's':
    show_keys()

else:
    print('You need to enter either q or s.')

---

您可以从字典中获取键，从该列表中选择随机元素，获取字典条目，然后从列表中删除键

import random

dictionary = {
    'one': '1',
    'two': '2',
    'three': '3'
}

# getting list of keys from dictionary
keys = list(dictionary.keys())
print(keys)

# picking random elemt from list
elem = random.choice(keys)
print(elem, dictionary.get(elem))

# remove an element from a list
keys.remove(elem)
print(keys)

---

使用集合收集您需要的钥匙。 向要求定义的函数提供集合和dict，以便它可以向其中添加新键-不要使用全局变量。 循环直到不再可能猜测（

len（已经）==len（数据）

）并处理该情况，或者直到用户想要退出：

from random import choice   # do not import ALL - just what you need

def show_keys(d,a):  # provide data and already used key set
    """ Show the user a random key and ask them
        to define it. Show the definition
        when the user presses return.    
        d : your dict of keys and definitions
        a : a set of already used keys
    """
    # choice only for those keys that are not yet used
    random_key = choice( [k for k in d if k not in a] )

    print('Define: ', random_key)
    input('Press return to see the definition')
    print(d [random_key])

    # add used key to a - you could ask (y/n) if the definition was correct
    # and only then add the key - if wrong it has a chance of being asked again
    a.add(random_key)

# Set up the data with keys and defs

data = {'key1':'definition1',
        'key2':'definition2',
        'key3':'definition3'}

# set up already seen keys 
already = set()

# The loop 
while True:  # loop until break

    # nothing more to guess
    if len(already)==len(data):
        y = input("Game over. Play again? [y, ]").lower()
        if y=="y":
            # reset set of seen stuff to empty set()
            already = set()
        else:
            break

    user_input = input('Enter s to show a key, or q to quit: ')
    if user_input == 'q':
        break
    elif user_input == 's': 
        # provide data and already to the function, don't use globals
        show_keys(data,already)
    else:
        print('You need to enter either q or s.')

---

或者：创建一个无序列表并使用：

from random import shuffle

data = {'key1':'definition1',
        'key2':'definition2',
        'key3':'definition3'}

already = set()

shuffled = list(data.items())
shuffle(shuffled)

# The loop 
while shuffled:
    user_input = input('Enter s to show a key, or q to quit: ')
    if user_input == 'q':
        break
    elif user_input == 's':
        key,value = shuffled.pop()
        print('Define: ', key)
        input('Press return to see the definition')
        print(value) 
    else:
        print('You need to enter either q or s.')

---

将所有键放入一个列表，random.shuffle（）列表中，弹出最后一个键，直到其为空。您可以获取所有键，将它们放入一个列表中，对列表进行洗牌，然后按顺序从列表中选择它们。您应该在问题中添加一些代码。人们更有可能以这种方式做出反应。@PatrickArtner做

随机选择会更有效。选择（…）

@Err取决于具体情况。假设您想用3到1000个键“清空”dict：将所有键的列表洗牌一次，然后从列表中弹出，直到清空，这样成本更低，然后执行choice（）1000次和list.remove（）1000次。如果您可以从dict本身删除密钥，那么使用dict中的选择并在之后删除密钥会更好，因为不需要列表-列表的提示是：dict未更改，您可以通过重新填充dict来“重新运行”游戏list@Redunited上面的评论不在这里。只需编辑您的问题并粘贴代码。如果你弄错了，有人会帮你编辑，但是代码应该在三个``块中（下面的键是

esc

）