def main():
x = [randint(1,100) for i in range(1,100)]
return x这将返回100个随机数btw 1和100。每次我调用这个函数,它都会返回一个不同的数字序列。我想要的是,每次都得到相同的数字序列。也许可以将结果保存到某个文件中?
发布于 2017-03-09 15:36:09
您可以提供一些固定的种子。
import random
def main():
random.seed(9001)
x = [random.randint(1,100) for i in range(1,100)]
return x有关种子的更多信息,请访问:random.seed(): What does it do?
发布于 2017-03-09 15:44:59
下面是一个模仿您的代码的基本示例
import random
s = random.getstate()
print([random.randint(1,100) for i in range(10)])
random.setstate(s)
print([random.randint(1,100) for i in range(10)])在这两个调用中,您都会得到相同的输出。关键是,您可以在任何时候检索并在以后重新分配rng的当前状态。
发布于 2017-03-09 15:45:05
In [19]: for i in range(10):
...: random.seed(10)
...: print [random.randint(1, 100) for j in range(5)]
...:
...:
[58, 43, 58, 21, 82]
[58, 43, 58, 21, 82]
[58, 43, 58, 21, 82]
[58, 43, 58, 21, 82]
[58, 43, 58, 21, 82]
[58, 43, 58, 21, 82]
[58, 43, 58, 21, 82]
[58, 43, 58, 21, 82]
[58, 43, 58, 21, 82]
[58, 43, 58, 21, 82]必须在对random的新调用之前调用random.seed函数。
In [20]: random.seed(10)
In [21]: for i in range(10):
...: print [random.randint(1,10) for j in range(10)]
...:
[5, 6, 3, 9, 9, 7, 2, 6, 4, 3]
[10, 10, 1, 9, 7, 4, 3, 7, 5, 7]
[7, 2, 8, 10, 10, 7, 1, 1, 2, 10]
[4, 4, 9, 4, 6, 5, 1, 6, 9, 2]
[3, 5, 1, 5, 9, 7, 6, 9, 2, 6]
[4, 7, 2, 8, 1, 2, 9, 10, 5, 5]
[3, 3, 7, 2, 2, 5, 2, 7, 9, 8]
[5, 4, 5, 1, 8, 4, 4, 1, 5, 6]
[4, 9, 7, 3, 6, 10, 6, 7, 1, 5]
[5, 5, 5, 6, 6, 5, 2, 5, 10, 5]https://stackoverflow.com/questions/42689301
复制相似问题