Python：如何生成12位随机数？

Question

在Python中，如何生成12位随机数？有没有可以指定类似范围的函数random.range(12)？

import random
random.randint()

输出应为包含0到9范围内的12位数字的字符串（允许使用前导零）。

Answer 1

一种简单的方法怎么了？

>>> import random
>>> random.randint(100000000000,999999999999)
544234865004L

而且，如果您希望以前导零开头，则需要一个字符串。

>>> "%0.12d" % random.randint(0,999999999999)
'023432326286'

编辑：

我自己对这个问题的解决方案是这样的：

import random

def rand_x_digit_num(x, leading_zeroes=True):
    """Return an X digit number, leading_zeroes returns a string, otherwise int"""
    if not leading_zeroes:
        # wrap with str() for uniform results
        return random.randint(10**(x-1), 10**x-1)  
    else:
        if x > 6000:
            return ''.join([str(random.randint(0, 9)) for i in xrange(x)])
        else:
            return '{0:0{x}d}'.format(random.randint(0, 10**x-1), x=x)

测试结果：

>>> rand_x_digit_num(5)
'97225'
>>> rand_x_digit_num(5, False)
15470
>>> rand_x_digit_num(10)
'8273890244'
>>> rand_x_digit_num(10)
'0019234207'
>>> rand_x_digit_num(10, False)
9140630927L

速度计时方法：

def timer(x):
        s1 = datetime.now()
        a = ''.join([str(random.randint(0, 9)) for i in xrange(x)])
        e1 = datetime.now()
        s2 = datetime.now()
        b = str("%0." + str(x) + "d") % random.randint(0, 10**x-1)
        e2 = datetime.now()
        print "a took %s, b took %s" % (e1-s1, e2-s2)

速度测试结果：

>>> timer(1000)
a took 0:00:00.002000, b took 0:00:00
>>> timer(10000)
a took 0:00:00.021000, b took 0:00:00.064000
>>> timer(100000)
a took 0:00:00.409000, b took 0:00:04.643000
>>> timer(6000)
a took 0:00:00.013000, b took 0:00:00.012000
>>> timer(2000)
a took 0:00:00.004000, b took 0:00:00.001000

它告诉我们的是：

对于长度在6000个字符以下的任何数字，我的方法都更快-
有时要快得多，但是对于较大的数字，arshajii建议的方法看起来更好。