Advanced random number generator?

[shing]

I'm making an advanced random number generator, and I want to be able to give it a set amount of digits and from there it can use the digits 1-9 for the result. I might also want to make a logfile.

For example:

I want it to generate 5 numbers using the digits 1,2,3,4,5,6,7,8,9.

How must I go about this?

[ghostdog74]

I'm making an advanced random number generator, and I want to be able to give it a set amount of digits and from there it can use the digits 1-9 for the result. I might also want to make a logfile.

For example:

I want it to generate 5 numbers using the digits 1,2,3,4,5,6,7,8,9.

How must I go about this?

you can use the random module. Look here and here

[bartonc]

I'm making an advanced random number generator, and I want to be able to give it a set amount of digits and from there it can use the digits 1-9 for the result. I might also want to make a logfile.

For example:

I want it to generate 5 numbers using the digits 1,2,3,4,5,6,7,8,9.

How must I go about this?

>>> import random as rand
>>> help(rand)
Help on module random:

NAME
random - Random variable generators.

FILE
d:\python24\lib\random.py
>>> rand.sample([1,2,3,4,5,6,7,8,9], 5)
[8, 4, 5, 7, 6]
>>>

[bvdet]

I'm making an advanced random number generator, and I want to be able to give it a set amount of digits and from there it can use the digits 1-9 for the result. I might also want to make a logfile.

For example:

I want it to generate 5 numbers using the digits 1,2,3,4,5,6,7,8,9.

How must I go about this?

>>> import random
>>> seq = [1,2,3,4,5,6,7,8,9]
>>> n = 7
>>> int(''.join([str(random.choice(seq)) for i in range(n)]))
3683319
>>> 

[bartonc]

>>> import random
>>> seq = [1,2,3,4,5,6,7,8,9]
>>> n = 7
>>> int(''.join([str(random.choice(seq)) for i in range(n)]))
3683319
>>> 

That's very nice. How 'bout

>>> int("".join(str(i) for i in rand.sample([1,2,3,4,5,6,7,8,9], 5)))

[bvdet]

That's very nice. How 'bout

>>> int("".join(str(i) for i in rand.sample([1,2,3,4,5,6,7,8,9], 5)))

I am stuck in 2.3, so no generator expressions - only list comprehensions. random.sample() works great except it will not reuse the numbers and you cannot ask for more numbers than are in the sample list.

[bvdet]

I am stuck in 2.3, so no generator expressions - only list comprehensions. random.sample() works great except it will not reuse the numbers and you cannot ask for more numbers than are in the sample list.

Here's what I had in mind:

>>> i = varRandom([0,1,2,3,4,5,6,7,8,9], 32)
>>> i
71725441469245314962196254882137L
>>> int_format(str(i))
'71,725,441,469,245,314,962,196,254,882,137'
>>> 

[bartonc]

Here's what I had in mind:

>>> i = varRandom([0,1,2,3,4,5,6,7,8,9], 32)
>>> i
71725441469245314962196254882137L
>>> int_format(str(i))
'71,725,441,469,245,314,962,196,254,882,137'
>>> 

So, what library (your own?) are those functions from?

[bvdet]

So, what library (your own?) are those functions from?

It's somethig I wrote:

import random
 
def int_format(s):
    if len(s)<=3:
        return s
    else:
        outList = [str(s[max(0,i-3):i]) for i in range(len(s), -1, -3)]
        outList.reverse()
        return ','.join([x for x in outList if len(x) > 0])
 
def varRandom(dList, n):
    num = ''.join([str(random.choice(dList)) for i in range(n)])
    while num.startswith('0'):
        num = num[1:]
        num += str(random.choice(dList))
    return int(num)
 
i = varRandom([0,1,2,3,4,5,6,7,8,9], 32)
print i
print int_format(str(i))
 
print int_format('12345678923')
print int_format('1234567892')
print int_format('123456789')
print int_format('12345678')
 
'''
>>> 46010601035631449157303537548986
46,010,601,035,631,449,157,303,537,548,986
12,345,678,923
1,234,567,892
123,456,789
12,345,678
>>>
'''

This function formats long integers using recursion, but I do not know where I got it nor to whom to give credit:

def int_format(s):
    if len(s) <= 3:
        return s
    return int_format(s[:-3]) + "," + s[-3:]

I wanted to write my own but could not match the elegance.

[bartonc]

It's somethig I wrote:

import random
 
def int_format(s):
    if len(s)<=3:
        return s
    else:
        outList = [str(s[max(0,i-3):i]) for i in range(len(s), -1, -3)]
        outList.reverse()
        return ','.join([x for x in outList if len(x) > 0])
 
def varRandom(dList, n):
    num = ''.join([str(random.choice(dList)) for i in range(n)])
    while num.startswith('0'):
        num = num[1:]
        num += str(random.choice(dList))
    return int(num)
 
i = varRandom([0,1,2,3,4,5,6,7,8,9], 32)
print i
print int_format(str(i))
 
print int_format('12345678923')
print int_format('1234567892')
print int_format('123456789')
print int_format('12345678')
 
'''
>>> 46010601035631449157303537548986
46,010,601,035,631,449,157,303,537,548,986
12,345,678,923
1,234,567,892
123,456,789
12,345,678
>>>
'''

This function formats long integers using recursion, but I do not know where I got it nor to whom to give credit:

def int_format(s):
    if len(s) <= 3:
        return s
    return int_format(s[:-3]) + "," + s[-3:]

I wanted to write my own but could not match the elegance.

A slight modification might be an improvement:

def int_format(s, sep=","):
    if len(s) <= 3:
        return s
    return int_format(s[:-3]) + sep + s[-3:]

for use in Europe where the comma is not so common.

BTW, how do you like your new status?

[bvdet]

A slight modification might be an improvement:

def int_format(s, sep=","):
    if len(s) <= 3:
        return s
    return int_format(s[:-3]) + sep + s[-3:]

for use in Europe where the comma is not so common.

BTW, how do you like your new status?

No pressure, huh? I noticed it today and assumed there was a problem with the website. I was thinking it should be TBE (training to be an expert).