Precise calculations

[Ladvánszky Károly]

Could anybody help me on how to do precise calculations with Python the way
hand-held calculators do?

A small example to show the impreciseness coming from the floating point
arithmetics:

(100000000000000.4*3)*100 gives 30000000000000124.0

Thanks for any help,

Károly

[Ben Finney]

On Fri, 07 Nov 2003 13:17:44 GMT, Ladvánszky Károly wrote:

A small example to show the impreciseness coming from the floating
point arithmetics:
Issues with representing floating point numbers in binary computers are
discussed here with respect to Python:

<http://www.python.org/doc/current/tut/node14.html>
Could anybody help me on how to do precise calculations with Python
the way hand-held calculators do?

Use rounding, e.g. as implemented by the str() function. From the above
URL:

"Python's builtin str() function produces only 12 significant
digits, and you may wish to use that instead. It's unusual for
eval(str(x)) to reproduce x, but the output may be more pleasant to
look at:

print str(0.1)

0.1

It's important to realize that this is, in a real sense, an
illusion: the value in the machine is not exactly 1/10, you're
simply rounding the display of the true machine value."

--
\ "I know when I'm going to die, because my birth certificate has |
`\ an expiration date." -- Steven Wright |
_o__) |
Ben Finney <http://bignose.squidly.org/>

[Ben Finney]

On 8 Nov 2003 08:26:07 +1050, Ben Finney wrote:

On Fri, 07 Nov 2003 13:17:44 GMT, Ladvánszky Károly wrote:

Could anybody help me on how to do precise calculations with Python
the way hand-held calculators do?

Use rounding, e.g. as implemented by the str() function.

Apologies; str() doesn't do rounding per se, it shows the float values
with fewer significant digits without changing the internal value.

--
\ "Once consumers can no longer get free music, they will have to |
`\ buy the music in the formats we choose to put out." -- Steve |
_o__) Heckler, VP of Sony Music, 2001 |
Ben Finney <http://bignose.squidly.org/>

[Alex Martelli]

Ben Finney wrote:
...

Could anybody help me on how to do precise calculations with Python
the way hand-held calculators do?

Use rounding, e.g. as implemented by the str() function. From the above

Use str if you really want to do like hand-held calculators, i.e., hide some
digits. If you want really high precision in binary floating point, the
gmpy extension module may be useful; if in decimal floating point, there is
a class Decimal that's being worked on to eventually provide that, too.
Alex

[Bengt Richter]

On Fri, 07 Nov 2003 13:17:44 GMT, "Ladvánszky Károly" <[e-mail]> wrote:

Could anybody help me on how to do precise calculations with Python the way
hand-held calculators do?

A small example to show the impreciseness coming from the floating point
arithmetics:

(100000000000000.4*3)*100 gives 30000000000000124.0

By coincidence I just posted a module (very alpha code, see "prePEP: Decimal data type" thread)
that does exact decimal floating point, so you can do better than hand calculators ;-)

First, let's use it to show what the full floating point bits ('all') show when converted
to an exact representation:

from exactdec import ED
ED(100000000000000.4,'all') ED('100000000000000.40625')
^
+--the _actual_ floating point value represented
ED(100000000000000.4,'all')*3 ED('300000000000001.21875')
^
+-- an exact multiple by 3
ED(100000000000000.4*3,'all') ED('300000000000001.25')
^
+--the exact value of the inexact floating point product
ED(100000000000000.4*3,'all')*100 ED('30000000000000125')
^
+--times 100 exactly
ED(100000000000000.4*3*100,'all') ED('30000000000000124')
^
+--the result of the additional floating point multiply

Now doing the math starting with an exact string literal for your starting value:
ED('100000000000000.4') ED('100000000000000.4')
^
+--value is exact
ED('100000000000000.4')*3 ED('300000000000001.2')
^
+--product is exact (the right operand of 3 is converted like ED(3) which is exact
since integer is exact.
(ED('100000000000000.4')*3)*100 ED('30000000000000120')
^
+--similarly with the 100 factor, which is converted to exact before multiplying as well.

Now try this on your calculator (note the binomial thing):
b = ED('1001001001001001001001001001')
b ED('1.001001001001001001001001001e27') b*b ED('1.00200300400500600700800901000900800700600500 4003002001e54') b*b*b

ED('1.00300601001502102803604505506306907307507507 3069063055045036028021015010006003001e81')

;-)

Regards,
Bengt Richter