[Numpy-discussion] Random int64 and float64 numbers
josef.pktd at gmail.com
josef.pktd at gmail.com
Thu Nov 5 21:53:41 EST 2009
On Thu, Nov 5, 2009 at 9:23 PM, Charles R Harris
<charlesr.harris at gmail.com> wrote:
> On Thu, Nov 5, 2009 at 7:04 PM, <josef.pktd at gmail.com> wrote:
>> On Thu, Nov 5, 2009 at 6:36 PM, Charles R Harris
>> <charlesr.harris at gmail.com> wrote:
>> > On Thu, Nov 5, 2009 at 4:26 PM, David Warde-Farley <dwf at cs.toronto.edu>
>> > wrote:
>> >> On 5-Nov-09, at 4:54 PM, David Goldsmith wrote:
>> >> > Interesting thread, which leaves me wondering two things: is it
>> >> > documented
>> >> > somewhere (e.g., at the IEEE site) precisely how many *decimal*
>> >> > mantissae
>> >> > are representable using the 64-bit IEEE standard for float
>> >> > representation
>> >> > (if that makes sense);
>> >> IEEE-754 says nothing about decimal representations aside from how to
>> >> round when converting to and from strings. You have to provide/accept
>> >> *at least* 9 decimal digits in the significand for single-precision
>> >> and 17 for double-precision (section 5.6). AFAIK implementations will
>> >> vary in how they handle cases where a binary significand would yield
>> >> more digits than that.
>> > I believe that was the argument for the extended precision formats. The
>> > givien number of decimal digits is sufficient to recover the same float
>> > that
>> > produced them if a slightly higher precision is used in the conversion.
>> > Chuck
>> >From the discussion for the floating point representation, it seems that
>> a uniform random number generator would have a very coarse grid
>> in the range for example -1e30 to +1e30 compared to interval -0.5,0.5.
>> How many points can be represented by a float in [-0.5,0.5] compared
>> to [1e30, 1e30+1.]?
>> If I interpret this correctly, then there are as many floating point
>> in [0,1] as in [1,inf), or am I misinterpreting this.
>> So how does a PRNG handle a huge interval of uniform numbers?
> There are several implementations, but the ones I'm familiar with reduce to
> scaling. If the rng produces random unsigned integers, then the range of
> integers is scaled to the interval [0,1). The variations involve explicit
> scaling (portable) or bit twiddling of the IEEE formats. In straight forward
> scaling some ranges of the random integers may not map 1-1, so the unused
> bits are masked off first; if you want doubles you only need 52 bits, etc.
> For bit twiddling there is an implicit 1 in the mantissa, so the basic range
> works out to [1,2), but that can be fixed by subtracting 1 from the result.
> Handling larger ranges than [0,1) just involves another scaling.
So, since this is then a discrete distribution, what is the number of points
in the support? (My guess would be 2**52, but I don't know much about
This would be the largest set of integers that could be generated
without gaps in the distribution, and would determine the grid size
for floating point random variables. (?)
for Davids example:
low, high = -1e307, 1e307
np.random.uniform(low, high, 100) # much more reasonable
this would imply a grid size of
or something similar. (floating points are not very dense in the real line.)
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