question about math module notation

[brad]

How does one make the math module spit out actual values without using
engineer or scientific notation?

I get this from <code>print math.pow(2,64)</code>:
1.84467440737e+ 19

I want this:
18,446,744,073, 709,551,616

I'm lazy... I don't want to convert it manually :)

[Stargaming]

Re: question about math module notation

On Thu, 26 Jul 2007 15:54:11 -0400, brad wrote:

How does one make the math module spit out actual values without using
engineer or scientific notation?
>
I get this from <code>print math.pow(2,64)</code>: 1.84467440737e+ 19
>
I want this:
18,446,744,073, 709,551,616
>
I'm lazy... I don't want to convert it manually :)

Explicitly converting it to `int` works for me. (Without the 3-digit-
block notation, of course.)

[brad]

Re: question about math module notation

Stargaming wrote:

Explicitly converting it to `int` works for me. (Without the 3-digit-
block notation, of course.)

Thank you!

[Paul Rubin]

Re: question about math module notation

brad <byte8bits@gmai l.comwrites:

18,446,744,073, 709,551,616
>
I'm lazy... I don't want to convert it manually :)

print 2**64

[brad]

Re: question about math module notation

Paul Rubin wrote:

print 2**64

Ah yes, that works too... thanks. I've settled on doing it this way:

print int(math.pow(2, 64))

I like the added parenthesis :)

[Paul Rubin]

Re: question about math module notation

brad <byte8bits@gmai l.comwrites:

Ah yes, that works too... thanks. I've settled on doing it this way:
print int(math.pow(2, 64))
I like the added parenthesis :)

I was surprised to find that gives an exact (integer, not
floating-point) answer. Still, I think it's better to say 2**64
which also works for (e.g.) 2**10000 where math.pow(2,1000 0)
raises an exception.

[Gary Herron]

Re: question about math module notation

Stargaming wrote:

On Thu, 26 Jul 2007 15:54:11 -0400, brad wrote:
>
>

>How does one make the math module spit out actual values without using
>engineer or scientific notation?
>>
>I get this from <code>print math.pow(2,64)</code>: 1.84467440737e+ 19
>>
>I want this:
>18,446,744,073 ,709,551,616
>>
>I'm lazy... I don't want to convert it manually :)
>>

>
Explicitly converting it to `int` works for me. (Without the 3-digit-
block notation, of course.)
>

It's got nothing to do with the math module. Any floating point number,
whether produced by the math module or not, can be converted to a
printable representation using the % operator with a format string:

>>x = 1.84467440737e+ 19
>>x

1.84467440737e+ 19

>>print '%25.3f' % x

184467440736999 99744.000

>>print '%12.5e' % x

1.84467e+19

>>print '%12.5g' % x

1.8447e+19

See http://docs.python.org/lib/typesseq-strings.html for all he details.

Gary Herron

[Dan Bishop]

Re: question about math module notation

On Jul 26, 3:59 pm, Paul Rubin <http://phr...@NOSPAM.i nvalidwrote:

brad <byte8b...@gmai l.comwrites:

Ah yes, that works too... thanks. I've settled on doing it this way:
print int(math.pow(2, 64))
I like the added parenthesis :)

>
I was surprised to find that gives an exact (integer, not
floating-point) answer. Still, I think it's better to say 2**64
which also works for (e.g.) 2**10000 where math.pow(2,1000 0)
raises an exception.

It *is* binary floating point. Powers of 2 are exactly
representable. Of course, it doesn't give an exact answer in general.

>>int(math.pow( 10, 23))

999999999999999 91611392L

[Paul Rubin]

Re: question about math module notation

Dan Bishop <danb_83@yahoo. comwrites:

I was surprised to find that gives an exact (integer, not
floating-point) answer. Still, I think it's better to say 2**64
which also works for (e.g.) 2**10000 where math.pow(2,1000 0)
raises an exception.

>
It *is* binary floating point. Powers of 2 are exactly
representable. Of course, it doesn't give an exact answer in general.
>

>int(math.pow(1 0, 23))

999999999999999 91611392L

Oh yikes, good point. math.pow(2,64) is really not appropriate for
what the OP wanted, I'd say. If you want integer exponentiation, then
write it that way. Don't do a floating point exponentiation that just
happens to not lose precision for the specific example.

>>int(math.pow( 3,50)) # wrong

717897987691852 578422784L

>>3**50 # right

717897987691852 588770249L