# working with VERY large 'float' and 'complex' types

Discussion in 'Python' started by Todd Steury, Sep 14, 2005.

1. ### Todd SteuryGuest

Greetings Python'ers:

I'm just an amature who occasionally uses Python for complex mathematical
models. The current model I'm working with occasionally generates really
large numbers that are either "float" or "complex" types. These numbers are
so large that I either get an overflow error, or some funky code like #INF
or 1.#INDj. However I really need these numbers to be calculated (although
precision isn't key). Is there a way to get python to increase the size
limit of float and complex numbers? I should mention that I'm using a lot of
pre-made modules and functions like math.exp() and scipy.special.erf() that
don't seem to be able to use available types like "Decimal" or "FixedPoint"
(especially since these don't seem to handle complex numbers).

Since I learn best by example, how could one solve the following problem:

from math import exp
>>>x=1000.
>>>z=exp(x)

so that z returns an actual value

Todd

Todd Steury, Sep 14, 2005

2. ### Paul RubinGuest

"Todd Steury" <> writes:
> or 1.#INDj. However I really need these numbers to be calculated (although
> precision isn't key). Is there a way to get python to increase the size
> limit of float and complex numbers?

Python just uses machine doubles by default. You might be able to edit
the source so it uses, say, 80-bit extended floats if you're on a machine
that supports them (like the x86). Those do support larger exponents.

You could also use something like gmpy, which supports arbitrary size
and arbitrary precision numbers. Of course it's slow by comparison.

> Since I learn best by example, how could one solve the following problem:
>
> from math import exp
> >>>x=1000.
> >>>z=exp(x)

>
> so that z returns an actual value

You could rearrange your formulas to not need such big numbers:

x = 1000.
log10_z = x / math.log(10)
c,m = divmod(log10_z, 1.)
print 'z = %.5fE%d' % (10.**c, m)

Paul Rubin, Sep 14, 2005

3. ### Fernando PerezGuest

Todd Steury wrote:

> Greetings Python'ers:
>
> I'm just an amature who occasionally uses Python for complex mathematical
> models. The current model I'm working with occasionally generates really
> large numbers that are either "float" or "complex" types. These numbers are
> so large that I either get an overflow error, or some funky code like #INF
> or 1.#INDj. However I really need these numbers to be calculated (although
> precision isn't key). Is there a way to get python to increase the size
> limit of float and complex numbers? I should mention that I'm using a lot of
> pre-made modules and functions like math.exp() and scipy.special.erf() that
> don't seem to be able to use available types like "Decimal" or "FixedPoint"
> (especially since these don't seem to handle complex numbers).

Python floats are C doubles underneath, so you're stuck. You need extended
numeric types. Decimal is slow as molasses, and was not designed for
mathematical work (rather for finance-type fixed-point work). Use this

http://calcrpnpy.sourceforge.net/clnumManual.html

Cheers,

f

Fernando Perez, Sep 14, 2005
4. ### Terry HancockGuest

> "Todd Steury" <> writes:
> > or 1.#INDj. However I really need these numbers to be calculated (although
> > precision isn't key). Is there a way to get python to increase the size
> > limit of float and complex numbers?

This is really a natural problem with such calculations.

On Wednesday 14 September 2005 02:30 pm, Paul Rubin wrote:
> You could rearrange your formulas to not need such big numbers:
>
> x = 1000.
> log10_z = x / math.log(10)
> c,m = divmod(log10_z, 1.)
> print 'z = %.5fE%d' % (10.**c, m)

I highly recommend you use this kind of solution. Solve the problem
with algebra, not with a new library. Most of the time, large numbers
can be avoided (particularly if you are not overly concerned with
precision), simply by dividing out large constant factors and the
like. Logs work better for this problem, as Paul points out.

--
Terry Hancock ( hancock at anansispaceworks.com )
Anansi Spaceworks http://www.anansispaceworks.com

Terry Hancock, Sep 15, 2005
5. ### Mikael OlofssonGuest

Paul Rubin calculates exp(1000.0):
> You could rearrange your formulas to not need such big numbers:
>
> x = 1000.
> log10_z = x / math.log(10)
> c,m = divmod(log10_z, 1.)
> print 'z = %.5fE%d' % (10.**c, m)

Nice approach. We should never forget that we do have mathematical
skills beside the computers. But, shouldn't you switch c and m in the
last row?

/MiO

Mikael Olofsson, Sep 15, 2005
6. ### Paul RubinGuest

Mikael Olofsson <> writes:
> > print 'z = %.5fE%d' % (10.**c, m)

>
> Nice approach. We should never forget that we do have mathematical
> skills beside the computers. But, shouldn't you switch c and m in the
> last row?

Yeah, doh. c=characteristic, m=mantissa.

Paul Rubin, Sep 15, 2005