Python: How to generate a 12-digit random number?
Whats wrong with a straightforward approach?
>>> import random
>>> random.randint(100000000000,999999999999)
544234865004L
And if you want it with leading zeros, you need a string.
>>> "%0.12d" % random.randint(0,999999999999)
'023432326286'
Edit:
My own solution to this problem would be something like this:
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)
Testing Results:
>>> 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
Timing methods for speed:
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)
Speed test results:
>>> 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
What it tells us:
For any digit under around 6000 characters in length my method is faster - sometimes MUCH faster, but for larger numbers the method suggested by arshajii looks better.
Since leading zeros are allowed (by your comment), you could also use:
int(''.join(str(random.randint(0,9)) for _ in xrange(12)))
EDIT: Of course, if you want a string, you can just leave out the int
part:
''.join(str(random.randint(0,9)) for _ in xrange(12))
This seems like the most straightforward way to do it in my opinion.
There are many ways to do that:
import random
rnumber1 = random.randint(10**11, 10**12-1) # randint includes endpoint
rnumber2 = random.randrange(10**11, 10**12) # randrange does not
# useful if you want to generate some random string from your choice of characters
digits = "123456789"
digits_with_zero = digits + "0"
rnumber3 = random.choice(digits) + ''.join(random.choice(digits_with_zero) for _ in range(11))
Do random.randrange(10**11, 10**12)
. It works like randint
meets range
From the documentation:
randrange(self, start, stop=None, step=1, int=<type 'int'>, default=None, maxwidth=9007199254740992L) method of random.Random instance
Choose a random item from range(start, stop[, step]).
This fixes the problem with randint() which includes the
endpoint; in Python this is usually not what you want.
Do not supply the 'int', 'default', and 'maxwidth' arguments.
This is effectively like doing random.choice(range(10**11, 10**12))
or random.randint(10**1, 10**12-1)
. Since it conforms to the same syntax as range()
, it's a lot more intuitive and cleaner than these two alternatives
If leading zeros are allowed:
"%012d" %random.randrange(10**12)