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]

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]

Stargaming wrote:

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

Thank you!

[Paul Rubin]

brad <by*******@gmail.comwrites:

18,446,744,073,709,551,616

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

print 2**64

[brad]

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]

brad <by*******@gmail.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(2,10000)
raises an exception.

[Gary Herron]

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

18446744073699999744.000

>>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]

On Jul 26, 3:59 pm, Paul Rubin <http://[e-mail]:

brad <[e-mail]:

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(2,10000)
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))

99999999999999991611392L

[Paul Rubin]

Dan Bishop <da*****@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(2,10000)
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))

99999999999999991611392L

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

717897987691852578422784L

>>3**50 # right

717897987691852588770249L