我是一名新手程序員，我試圖在我的書中為挑戰編寫代碼，其中包含一個循環，該循環隨機獲取數字和/或字母以宣布彩票中獎者。我正在嘗試編寫代碼：從元組中取出一個未被拾取的隨機對象 4 次將每個對象存儲在列表中打印清單from random import choice #Import choice() function from the random modulelottery_1 = (1,2,3,4,5,6,7,8,9,'a','b','c','d','e')lottery_winner = []for i in range(4): #Takes 4 random numbers and/or letters random = choice(lottery_1) if random not in lottery_winner: lottery_winner.append(pulled_number)print('$1M Winner\n')print(lottery_winner)有時它只選擇 2 個字符結果：$1M Winner[1, 'e']>>>為什么會這樣？我可以更改什么以使其選擇 4 個字符？
3 回答
慕蓋茨4494581
TA貢獻1850條經驗 獲得超11個贊
實際上，它是選擇四個字符。但是，當所選字符之一已在 中時lottery_winner，不會添加它。在這種情況下，您最終得到的總結果少于四個。
lenik 的回答是最實用的解決方案。但是，如果您對如何使用該choice函數進行操作的邏輯感到好奇，請記住，您需要在帽子出現重復時再次選擇，或者您需要從帽子中消除選項當你去時。
選項 #1，每當獲勝者重復時再試一次：
for i in range(4):
new_winner = False # Initially, we have not found a new winner yet.
while not new_winner: # Repeat the following as long as new_winner is false:
random = choice(lottery_1)
if random not in lottery_winner:
lottery_winner.append(pulled_number)
new_winner = True # We're done! The while loop won't run again.
# (The for loop will keep going, of course.)
選項 #2，每次都從列表中刪除獲勝者，這樣他們就不會被再次選中：
for i in range(4):
random = choice(lottery_1)
lottery_winner.append(pulled_number)
lottery_1.remove(pulled_number) # Now it can't get chosen again.
請注意，remove()刪除指定值的第一個實例，在值不唯一的情況下，這可能不會執行您想要的操作。
GCT1015
TA貢獻1827條經驗 獲得超4個贊
import random
lottery_1 = (1, 2, 3, 4, 5, 6, 7, 8, 9, 'a', 'b', 'c', 'd', 'e')
'''Instead of using the choice method which can randomly grab the same value,
i suggest that you use the sample method which ensures that you only get randomly unique values'''
# The k value represents the number of items that you want to get
lottery_winner = random.sample(lottery_1, k=4)
print('$1M Winner\n')
print(lottery_winner)
HUWWW
TA貢獻1874條經驗 獲得超12個贊
這對我有用：
>>> import random
>>> lottery_1 = (1,2,3,4,5,6,7,8,9,'a','b','c','d','e')
>>> random.sample(lottery_1, 4)
[1, 7, 'a', 'e']
>>>
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