On Mon, Dec 26, 2011 at 3:17 PM, Roy Smith wrote:
> Of course it is. Those things constitute doc bugs which need to get
> fixed. The fact that the specification is flawed does not change the
> fact that it *is* the specification.
Also, the specification is what can be trusted across multiple
im
In article <4ef7e337$0$29973$c3e8da3$54964...@news.astraweb.com>,
Steven D'Aprano wrote:
> And I'm afraid that you have missed my point. The above comment from the
> source is from a docstring: it *is* public, official documentation. See
> help(random.Random).
When you wrote, "the source expl
On Mon, Dec 26, 2011 at 2:00 PM, Steven D'Aprano
wrote:
> The implementation of getrandbits is not documented at all. You would
> have to read the C source of the _random module to find out how the
> default pseudo-random number generator produces random bits. But note the
> leading underscore: _r
On Sun, 25 Dec 2011 12:41:29 -0500, Roy Smith wrote:
> On Mon, 26 Dec 2011 03:11:56 +1100, Chris Angelico wrote:
>> > I prefer not to rely on the source. That tells me what happens, not
>> > what's guaranteed to happen.
>
> Steven D'Aprano wrote:
>> In this case, the source explicitly tells you
Roy Smith於 2011年12月26日星期一UTC+8上午1時41分29秒寫道：
> On Mon, 26 Dec 2011 03:11:56 +1100, Chris Angelico wrote:
> > > I prefer not to rely on the source. That tells me what happens, not
> > > what's guaranteed to happen.
>
> Steven D'Aprano wrote:
> > In this case, the source explicitly tells you that t
On 25 December 2011 17:32, Serhiy Storchaka wrote:
> 25.12.11 15:48, Steven D'Aprano написав(ла):
>
> On Sun, 25 Dec 2011 08:30:46 -0500, Roy Smith wrote:
>>
>>> I want to create a string of 20 random digits (I'm OK with leading
>>> zeros). The best I came up with is:
>>> ''.join(str(random.ran
On Mon, 26 Dec 2011 03:11:56 +1100, Chris Angelico wrote:
> > I prefer not to rely on the source. That tells me what happens, not
> > what's guaranteed to happen.
Steven D'Aprano wrote:
> In this case, the source explicitly tells you that the API includes
> support for arbitrary large ranges if
25.12.11 15:48, Steven D'Aprano написав(ла):
On Sun, 25 Dec 2011 08:30:46 -0500, Roy Smith wrote:
I want to create a string of 20 random digits (I'm OK with leading
zeros). The best I came up with is:
''.join(str(random.randint(0, 9)) for i in range(20))
Is there something better?
'%20d' % ran
On Mon, 26 Dec 2011 03:11:56 +1100, Chris Angelico wrote:
> On Mon, Dec 26, 2011 at 2:46 AM, Steven D'Aprano
> wrote:
>> Use the Source, Luke, er, Chris :)
>>
>> If I've read the source correctly, randint() will generate sufficient
>> bits of randomness to ensure that the entire int is random.
>>
On Mon, Dec 26, 2011 at 2:46 AM, Steven D'Aprano
wrote:
> Use the Source, Luke, er, Chris :)
>
> If I've read the source correctly, randint() will generate sufficient
> bits of randomness to ensure that the entire int is random.
>
> http://hg.python.org/cpython/file/default/Lib/random.py
I prefer
On Mon, 26 Dec 2011 00:54:40 +1100, Chris Angelico wrote:
> On Mon, Dec 26, 2011 at 12:48 AM, Steven D'Aprano
> wrote:
>> On Sun, 25 Dec 2011 08:30:46 -0500, Roy Smith wrote:
>>
>>> I want to create a string of 20 random digits (I'm OK with leading
>>> zeros). The best I came up with is:
>>>
>>>
On Mon, 26 Dec 2011 01:51:30 +1100, Chris Angelico wrote:
> On Mon, Dec 26, 2011 at 1:21 AM, Roy Smith wrote:
>> It turns out, I don't really need 20 digits. If I can count on
>>
> "%020d" % random.randint(0,999)
>>
>> to give me 15-ish digits, that's good enough for my needs and
Chris Angelico wrote:
> On Mon, Dec 26, 2011 at 12:30 AM, Roy Smith wrote:
>> I want to create a string of 20 random digits (I'm OK with leading
>> zeros). The best I came up with is:
>>
>> ''.join(str(random.randint(0, 9)) for i in range(20))
>>
>> Is there something better?
>
> The simple opt
On Mon, Dec 26, 2011 at 1:21 AM, Roy Smith wrote:
> It turns out, I don't really need 20 digits. If I can count on
>
"%020d" % random.randint(0,999)
>
> to give me 15-ish digits, that's good enough for my needs and I'll
> probably go with that. Thanks.
I'd say you can. The info
In article ,
Chris Angelico wrote:
> "%020d"%random.randint(0,)
> (the latter gives you a string, padded with leading zeroes). But I'm
> assuming that you discarded that option due to lack of entropy (ie you
> can't trust randint() over that huge a range).
Actually, the only
On Mon, Dec 26, 2011 at 12:48 AM, Steven D'Aprano
wrote:
> On Sun, 25 Dec 2011 08:30:46 -0500, Roy Smith wrote:
>
>> I want to create a string of 20 random digits (I'm OK with leading
>> zeros). The best I came up with is:
>>
>> ''.join(str(random.randint(0, 9)) for i in range(20))
>>
>> Is there
On Mon, Dec 26, 2011 at 12:50 AM, Chris Angelico wrote:
> The Python 2 docs [1]
Forgot to provide link.
Python 2: http://docs.python.org/library/random.html
And same in Python 3: http://docs.python.org/py3k/library/random.html
ChrisA
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On Mon, Dec 26, 2011 at 12:30 AM, Roy Smith wrote:
> I want to create a string of 20 random digits (I'm OK with leading
> zeros). The best I came up with is:
>
> ''.join(str(random.randint(0, 9)) for i in range(20))
>
> Is there something better?
The simple option is:
random.randint(0,99
On Sun, 25 Dec 2011 08:30:46 -0500, Roy Smith wrote:
> I want to create a string of 20 random digits (I'm OK with leading
> zeros). The best I came up with is:
>
> ''.join(str(random.randint(0, 9)) for i in range(20))
>
> Is there something better?
'%20d' % random.randint(0, 10**20-1)
--
St
I want to create a string of 20 random digits (I'm OK with leading
zeros). The best I came up with is:
''.join(str(random.randint(0, 9)) for i in range(20))
Is there something better?
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http://mail.python.org/mailman/listinfo/python-list
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