永远不会返回相同项目两次的随机字典搜索

Question

当要求程序从键值条目字典中选择一个随机条目时，有没有一种方法，一旦每个条目都被选中，一旦程序将让用户知道所有条目都已被选中，并且然后停止，最终只允许每个条目被选择一次，如果只有 3 个条目，程序将只运行 3 次，如果有 100 个条目，它将运行 100 次等等？ python 编码新手，请多多包涵。

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.')

Answer 1

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

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)

Answer 2

使用集合来收集您想要的密钥。 向要求定义的函数提供 set 和 dict，以便它可以向其添加新键 - 不要使用全局变量。 循环直到无法再猜测 (len(already)==len(data)) 并处理这种情况，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：创建一个随机列表并使用它：

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.')