Mark Dickinson added the comment:
> I think it is better to make a tip in the Python tutorial.
I'd recommend opening a separate issue (or pull request, if you're feeling
adventurous) for that; this issue is old and closed, and it's unlikely many
will be following it.
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Huan Wang added the comment:
Hi Mark,
Thank you for your reply.
I went over again the answer from Zachary Ware published on 2015-08-08 09:36. I
got the point that it is better to use string type of number.
>>> from decimal import Decimal, ROUND_HALF_UP
>>> Decimal("1.45")
Decimal('1.45')
Mark Dickinson added the comment:
Huan,
This isn't a bug: see the earlier comments from Zachary Ware on this issue for
explanations. When you compute `rounded(1.45, 0.1)`, you convert the *float*
1.45 to a Decimal instance. Thanks to the What You See Is Not What You Get
nature of binary
Huan Wang added the comment:
Hello,
I was confused by the decimal module. The problem is that I want to
from decimal import Decimal, ROUND_HALF_UP
def rounded(number, n):
''' Round the digits after the n_th decimal point by using
decimal module in python.
For example:
Zachary Ware added the comment:
I think the key point that you're missing (and which I could have made clearer
in my previous message) is that `Decimal(2.675) != Decimal('2.675')`. In the
first case, a Decimal instance is created from a float, and 2.675 cannot be
represented perfectly in
umedoblock added the comment:
last compared results are different.
should be bug or at least think that how to get a same result
about D(round(df2, 2)) == D(round(ds2, 2))
from decimal import Decimal as D
f1 = 1.65
s1 = str(f1)
df1 = D(f1)
ds1 = D(s1)
f2 = 2.675
s2 = str(f2)
df2 = D(f2)
umedoblock added the comment:
In this case.
round(1.65, 1) == 1.7
False
round(2.675, 2) == 2.68
False
I never say anything.
Because I understand what you said.
But I use the decimal module.
please pay attention to use decimal module.
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Python
umedoblock added the comment:
I have a headache.
because python reports many error after I patched below patches.
--- Lib/test/test_decimal.py.orig 2015-08-08 17:41:01.986316738 +0900
+++ Lib/test/test_decimal.py2015-08-08 17:41:05.470316878 +0900
@@ -1935,6 +1935,7 @@
Merlijn van Deen added the comment:
As Zachary explained, the behavior is correct. There are three issues in play
here.
1) The rounding method. With the ROUND_HALF_EVEN rounding mode, .5 is rounded
to the nearest *even* number, so 1.65 is rounded to 1.6, while 1.75 is rounded
to 1.8.
2)
umedoblock added the comment:
excuse me.
I understand ROUND_HALF_EVEN meaning.
I think that __round__() function work ROUND_HALF_UP.
so sorry.
I don't have exactly knowledge about ROUND_HALF_EVEN.
I misunderstand about ROUND_HALF_EVEN.
I have thought ROUND_HALF_EVEN means ROUND_HALF_UP.
SO
Zachary Ware added the comment:
I'm glad you understand it now :)
--
___
Python tracker rep...@bugs.python.org
http://bugs.python.org/issue24827
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Python-bugs-list
New submission from umedoblock:
round(1.65, 1) return 1.6 with decimal.
I feel bug adobe result.
not bug ?
import decimal
d1 = decimal.Decimal(1.65)
d2 = decimal.Decimal(10 ** -2) * 5
d1
Decimal('1.65')
d2
Decimal('0.05000104083408559')
d1 + d2
Zachary Ware added the comment:
The rounding mode of the default context is ROUND_HALF_EVEN[1]:
import decimal
decimal.getcontext()
Context(prec=28, rounding=ROUND_HALF_EVEN, Emin=-99, Emax=99,
capitals=1, clamp=0, flags=[], traps=[InvalidOperation, DivisionByZero,
Overflow])
For
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