如何使用randomint python获取唯一数字？

Question

我正在创建一个'Euromillions彩票发生器'只是为了好玩,我不断打印出相同的数字.我该如何制作它以便我得到随机数字并且永远不会弹出相同的数字:

from random import randint

numbers = randint(1,50)

stars = randint(1,11)


print "Your lucky numbers are: ", numbers, numbers, numbers, numbers, numbers

print "Your lucky stars are: " , stars, stars

输出只是:

>>> Your lucky numbers are:  41 41 41 41 41
>>> Your lucky stars are:  8 8
>>> Good bye!

我怎样才能解决这个问题？

问候

Answer 1

您正在生成一个数字,然后多次打印出来.

生成几个数字:

print "Your lucky numbers are: ", randint(1,50), randint(1,50), randint(1,50), randint(1,50), randint(1,50)

或生成一个列表:

numbers = [randint(1,50) for _ in range(5)]
print "Your lucky numbers are: ", ' '.join(numbers)

或者更好的是,生成所有允许的数字(range()然后使用从中选择一个样本:

possible_numbers = range(1, 51)  # end point is not included
numbers = random.sample(possible_numbers, 5)
print "Your lucky numbers are: ", ' '.join(map(str, numbers))

现在numbers保证由完全唯一的数字组成.

该numbers变量不会奇迹般地更新每次打印它,因为它仅仅是指时间的结果的random.randint(1, 50).

Answer 2

设置一组数字,然后随机播放并切片.

from random import shuffle
numbers  = list(range(1,51))
shuffle(numbers)
draw = numbers[:6]
print(draw)