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 randomdef 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。
