floating point rounding

[hg]

Hi,

Here is my issue:

f = 1.5 * 0.01
f

>0.014999999999999999

'%f' % f

>>'0.015000'

But I really want to get 0.02 as a result ... is there a way out ?

Thanks,

hg

[hg]

hg wrote:

Hi,

Here is my issue:

f = 1.5 * 0.01
f

>>0.014999999999999999

'%f' % f

>>>'0.015000'

But I really want to get 0.02 as a result ... is there a way out ?

Thanks,

hg

round

[John Henry]

On Mar 9, 5:45 am, hg <[e-mail]:

hg wrote:

Hi,

Here is my issue:

f = 1.5 * 0.01
f

>0.014999999999999999

'%f' % f

>>'0.015000'

But I really want to get 0.02 as a result ... is there a way out ?

Thanks,

hg

round

Or more precisely:

round(0.014999999999999999,2)

if that's what you wish to do.

[hg]

John Henry wrote:

On Mar 9, 5:45 am, hg <[e-mail]:

>hg wrote:

Hi,

Here is my issue:

f = 1.5 * 0.01
f
0.014999999999999999
'%f' % f
'0.015000'

But I really want to get 0.02 as a result ... is there a way out ?

Thanks,

hg

round

Or more precisely:

round(0.014999999999999999,2)

if that's what you wish to do.

Indeed.

hg

[greg]

John Henry wrote:

Or more precisely:

round(0.014999999999999999,2)

No, that *won't* solve the problem. Using a slightly
different example,

>>x = 1.5 * 0.1
x

0.15000000000000002

>>round(x, 2)

0.14999999999999999

The problem is that floats are stored internally in
binary, not decimal, and numbers like 0.1 and 0.01
have no exact representation as a binary float.
Using round() doesn't help, because the result is
still a binary float, and the result you're after
still can't be represented.

The best you can do is to *display* it rounded
to the number of digits you want using formatting,
e.g.

>>"%.2f" % x

'0.15'

Alternatively, use the Decimal module, which stores
numbers as decimal and does arithmetic in ways that
will match your calculator. It's slower, though.

--
Greg