Re: how to convert from Decimal('1.23456789') to Decimal('1.234')
On Mon, 23 Mar 2009 17:19:48 -0700, Mensanator wrote: And it also gives different results to my function: my function rounds to places decimal places, yours to i digits. Very different things. Yeah, I know all about that. I work in Environmental Remediation. That's real science, where rounding to decimal places is strictly forbidden, significant digits must be preserved. That means rounding to digits. Okay, so in other words you solved your problem rather than the OP's problem. People other than the OP read these threads. Do you want to give people the impression that quantize is the only option? The only option for what? Solving the OP's problem, or your problem, or some other problem? How many different problems are we going to solve in this thread? At the very least, when taking a detour to solve a different problem, say so, and point out why your problem is a more interesting/useful problem to solve. Do you know what kind of hoops I have to jump through to get Access or Excel to round properly when doing unit conversion? I feel your pain. So surely you don't want to give the impression that all Decimal is good for is emulating Excel? That's a non sequitor. It does not follow that just because I sympathize with your difficulties with Excel, that I therefore want to give the impression that Decimal is only good for emulating Excel. Surely you're not so maive that you think dividing by 1000 simply moves the decimal point three places? Of course it does, if you're using real numbers. Computers have real numbers? You don't have to use computers to divide by 1000. As difficult as it is to believe, throughout history people have done mathematical calculations without the use of computing devices. If you're using floats, no, not quite, there are rounding issues involved, and underflow. And that's why we have Decimal, right? I'm sure that the Decimal standard wasn't invented just so people can divide by 1000. -- Steven -- http://mail.python.org/mailman/listinfo/python-list
how to convert from Decimal('1.23456789') to Decimal('1.234')
I only need the 3 digits after '.' Is there any way other than converting from/to string? -- http://mail.python.org/mailman/listinfo/python-list
Re: how to convert from Decimal('1.23456789') to Decimal('1.234')
On Mar 23, 4:40 pm, valpa valpass...@gmail.com wrote:
I only need the 3 digits after '.'
Is there any way other than converting from/to string?
I'm not sure if this is the canonical way but it works:
d = Decimal('1.23456789')
three_places = Decimal('0.001') # or anything that has the exponent depth
you want
d.quantize(three_places, 'ROUND_DOWN')
Decimal('1.234')
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Re: how to convert from Decimal('1.23456789') to Decimal('1.234')
In that case, I usually use # when rounding is proper, s = '1.23456789' print round(float(s)) or # when cut out is proper, from math import floor print floor(float(s)*1000)/1000 Hyunchul valpa wrote: I only need the 3 digits after '.' Is there any way other than converting from/to string? -- http://mail.python.org/mailman/listinfo/python-list -- http://mail.python.org/mailman/listinfo/python-list
Re: how to convert from Decimal('1.23456789') to Decimal('1.234')
On Sun, 22 Mar 2009 23:40:38 -0700, valpa wrote:
I only need the 3 digits after '.'
Is there any way other than converting from/to string?
You should Read the Fine Manual:
http://docs.python.org/library/decimal.html
[quote]
The quantize() method rounds a number to a fixed exponent. This method is
useful for monetary applications that often round results to a fixed
number of places:
Decimal('7.325').quantize(Decimal('.01'), rounding=ROUND_DOWN)
Decimal('7.32')
Decimal('7.325').quantize(Decimal('1.'), rounding=ROUND_UP)
Decimal('8')
[end quote]
In my opinion, that's hideously ugly, but you can create a helper
function very easily:
def round(dec, places, rounding=decimal.ROUND_HALF_UP):
return dec.quantize(decimal.Decimal(str(10**-places)), rounding)
--
Steven
--
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Re: how to convert from Decimal('1.23456789') to Decimal('1.234')
On Mar 23, 7:01 am, alex23 wuwe...@gmail.com wrote:
On Mar 23, 4:40 pm, valpa valpass...@gmail.com wrote:
I only need the 3 digits after '.'
Is there any way other than converting from/to string?
I'm not sure if this is the canonical way but it works:
d = Decimal('1.23456789')
three_places = Decimal('0.001') # or anything that has the exponent depth
you want
d.quantize(three_places, 'ROUND_DOWN')
Decimal('1.234')
Yes, that's the official 'right way'. There's
also a _rescale method that does this:
Decimal('1.23456789')._rescale(-3, 'ROUND_HALF_EVEN')
Decimal('1.235')
... but as the leading underscore indicates, it's
private and undocumented, so you shouldn't rely on it
not to change or disappear in a future version.
The two methods are subtly different, in that the
quantize method respects the current context, while
the _rescale method ignores it. For example:
getcontext().prec = 3
Decimal('1.23456789').quantize(Decimal('0.001'))
Traceback (most recent call last):
File stdin, line 1, in module
File /Users/dickinsm/python_source/trunk/Lib/decimal.py, line
2364, in quantize
'quantize result has too many digits for current context')
File /Users/dickinsm/python_source/trunk/Lib/decimal.py, line
3735, in _raise_error
raise error(explanation)
decimal.InvalidOperation: quantize result has too many digits for
current context
Decimal('1.23456789')._rescale(-3, 'ROUND_DOWN')
Decimal('1.234')
[61114 refs]
Mark
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Re: how to convert from Decimal('1.23456789') to Decimal('1.234')
On Mar 23, 6:40 am, valpa valpass...@gmail.com wrote:
I only need the 3 digits after '.'
Is there any way other than converting from/to string?
And in Python 3.0, just use the built-in round function:
from decimal import Decimal
round(Decimal('1.23456789'), 3)
Decimal('1.235')
This uses the rounding specified by the context, so if
you want a rounding mode other than the default
ROUND_HALF_EVEN, just set the context rounding
appropriately:
from decimal import getcontext, ROUND_DOWN
getcontext().rounding = ROUND_DOWN
round(Decimal('1.23456789'), 3)
Decimal('1.234')
Mark
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Re: how to convert from Decimal('1.23456789') to Decimal('1.234')
On Mar 23, 2:24�am, Steven D'Aprano
ste...@remove.this.cybersource.com.au wrote:
On Sun, 22 Mar 2009 23:40:38 -0700, valpa wrote:
I only need the 3 digits after '.'
Is there any way other than converting from/to string?
You should Read the Fine Manual:
http://docs.python.org/library/decimal.html
[quote]
The quantize() method rounds a number to a fixed exponent. This method is
useful for monetary applications that often round results to a fixed
number of places:
Decimal('7.325').quantize(Decimal('.01'), rounding=ROUND_DOWN)
Decimal('7.32')
Decimal('7.325').quantize(Decimal('1.'), rounding=ROUND_UP)
Decimal('8')
[end quote]
In my opinion, that's hideously ugly,
In looking at this for the first time, it struck
me as funny why the first argument to quantize
even requires a number since you can pass a
Conext as an argument. But if you do, precision
is ignored. Quantize isn't the way to do that.
but you can create a helper
function very easily:
def round(dec, places, rounding=decimal.ROUND_HALF_UP):
� � return dec.quantize(decimal.Decimal(str(10**-places)), rounding)
Still ugly. I would do this:
a = Decimal('1.23456789')
for i in xrange(1,6):
print Context.create_decimal(Context(i,ROUND_DOWN),a)
1
1.2
1.23
1.234
1.2345
print a
1.23456789
--
Steven
--
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Re: how to convert from Decimal('1.23456789') to Decimal('1.234')
On Mon, 23 Mar 2009 01:45:53 -0700, Mensanator wrote:
but you can create a helper
function very easily:
def round(dec, places, rounding=decimal.ROUND_HALF_UP): � � return
dec.quantize(decimal.Decimal(str(10**-places)), rounding)
Still ugly. I would do this:
a = Decimal('1.23456789')
for i in xrange(1,6):
print Context.create_decimal(Context(i,ROUND_DOWN),a)
Well, that's hardly any less ugly.
And it also gives different results to my function: my function rounds to
places decimal places, yours to i digits. Very different things.
--
Steven
--
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Re: how to convert from Decimal('1.23456789') to Decimal('1.234')
On Mar 23, 5:48 am, Steven D'Aprano st...@remove-this-
cybersource.com.au wrote:
On Mon, 23 Mar 2009 01:45:53 -0700, Mensanator wrote:
but you can create a helper
function very easily:
def round(dec, places, rounding=decimal.ROUND_HALF_UP): return
dec.quantize(decimal.Decimal(str(10**-places)), rounding)
Still ugly. I would do this:
a = Decimal('1.23456789')
for i in xrange(1,6):
print Context.create_decimal(Context(i,ROUND_DOWN),a)
Well, that's hardly any less ugly.
I wouldn't say so since there are no strings attached. Didn't the
OP specifically ask for a solution that didn't involve strings?
And it also gives different results to my function: my function rounds to
places decimal places, yours to i digits. Very different things.
Yeah, I know all about that. I work in Environmental Remediation.
That's real science, where rounding to decimal places is strictly
forbidden, significant digits must be preserved. That means rounding
to digits. Do you know what kind of hoops I have to jump through to
get Access or Excel to round properly when doing unit conversion?
Surely you're not so maive that you think dividing by 1000 simply
moves
the decimal point three places?
--
Steven
--
http://mail.python.org/mailman/listinfo/python-list
Re: how to convert from Decimal('1.23456789') to Decimal('1.234')
On Mon, 23 Mar 2009 10:06:23 -0700, Mensanator wrote:
On Mar 23, 5:48 am, Steven D'Aprano st...@remove-this-
cybersource.com.au wrote:
On Mon, 23 Mar 2009 01:45:53 -0700, Mensanator wrote:
but you can create a helper
function very easily:
def round(dec, places, rounding=decimal.ROUND_HALF_UP): return
dec.quantize(decimal.Decimal(str(10**-places)), rounding)
Still ugly. I would do this:
a = Decimal('1.23456789')
for i in xrange(1,6):
print Context.create_decimal(Context(i,ROUND_DOWN),a)
Well, that's hardly any less ugly.
I wouldn't say so since there are no strings attached. Didn't the OP
specifically ask for a solution that didn't involve strings?
No, the OP asked for a solution that didn't involve converting the
decimal number to a string first.
Besides, I don't believe you can specify the rounding mode unless you use
strings *wink*
type(ROUND_DOWN)
type 'str'
And it also gives different results to my function: my function rounds
to places decimal places, yours to i digits. Very different things.
Yeah, I know all about that. I work in Environmental Remediation. That's
real science, where rounding to decimal places is strictly forbidden,
significant digits must be preserved. That means rounding to digits.
Okay, so in other words you solved your problem rather than the OP's
problem.
Do
you know what kind of hoops I have to jump through to get Access or
Excel to round properly when doing unit conversion?
I feel your pain.
Surely you're not so maive that you think dividing by 1000 simply moves
the decimal point three places?
Of course it does, if you're using real numbers. If you're using floats,
no, not quite, there are rounding issues involved, and underflow.
I'm not sure why you raise this. Is this a general rant, or do you have a
specific criticism?
--
Steven
--
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Re: how to convert from Decimal('1.23456789') to Decimal('1.234')
On Mar 23, 5:42 pm, Steven D'Aprano st...@remove-this-
cybersource.com.au wrote:
On Mon, 23 Mar 2009 10:06:23 -0700, Mensanator wrote:
On Mar 23, 5:48 am, Steven D'Aprano st...@remove-this-
cybersource.com.au wrote:
On Mon, 23 Mar 2009 01:45:53 -0700, Mensanator wrote:
but you can create a helper
function very easily:
def round(dec, places, rounding=decimal.ROUND_HALF_UP): return
dec.quantize(decimal.Decimal(str(10**-places)), rounding)
Still ugly. I would do this:
a = Decimal('1.23456789')
for i in xrange(1,6):
print Context.create_decimal(Context(i,ROUND_DOWN),a)
Well, that's hardly any less ugly.
I wouldn't say so since there are no strings attached. Didn't the OP
specifically ask for a solution that didn't involve strings?
No, the OP asked for a solution that didn't involve converting the
decimal number to a string first.
Besides, I don't believe you can specify the rounding mode unless you use
strings *wink*
type(ROUND_DOWN)
type 'str'
And it also gives different results to my function: my function rounds
to places decimal places, yours to i digits. Very different things.
Yeah, I know all about that. I work in Environmental Remediation. That's
real science, where rounding to decimal places is strictly forbidden,
significant digits must be preserved. That means rounding to digits.
Okay, so in other words you solved your problem rather than the OP's
problem.
People other than the OP read these threads. Do you want to
give people the impression that quantize is the only option?
Do
you know what kind of hoops I have to jump through to get Access or
Excel to round properly when doing unit conversion?
I feel your pain.
So surely you don't want to give the impression that
all Decimal is good for is emulating Excel?
Surely you're not so maive that you think dividing by 1000 simply moves
the decimal point three places?
Of course it does, if you're using real numbers.
Computers have real numbers?
If you're using floats,
no, not quite, there are rounding issues involved, and underflow.
And that's why we have Decimal, right?
I'm not sure why you raise this. Is this a general rant, or do you have a
specific criticism?
At first, I could not understand the need for nor the syntax of
quantize. It seemed very clunky having to specify a template and
a rounding, especially since it also accepts a Context object.
And upon seeing Context.create_decimal, I thought why would
I ever use quantize when I can do it the 'right' way?
But that was when I first saw it, so the rant isn't about
which way to do something, but pointing out there may be
other solutions that are more appropriate to the problem.
Speaking of doing things the 'right' way, I seem to have
blundered my way into the answer I was looking for. According
to the documention
quote
create_decimal(num)
Creates a new Decimal instance from num but using self as context.
Unlike the Decimal constructor, the context precision, rounding
method, flags, and traps are applied to the conversion.
/quote
and that's exactly what I want. I may have one context for
calculation and another for printing, so I want to be able
to programatically impose my printing context on a number
without affecting it.
Of course it didn't work.
Context.create_decimal('1.234456789')
Traceback (most recent call last):
File pyshell#13, line 1, in module
Context.create_decimal('1.234456789')
TypeError: unbound method create_decimal() must be called with Context
instance as first argument (got str instance instead)
Ok, the Traceback says I need a Context instance, so I tried
Context.create_decimal(Context(4),'1.234456789')
Decimal('1.234')
So I thought ok, but what's this self referred to in the docs
and why does it say create_decimal(num) if I have to have a
Context instance as the first parameter?
But wait, the Traceback also says unbound method. Was I supposed
to have parentheses after Context?
Context().create_decimal('1.234456789')
Decimal('1.234456789')
Duh.
The self must refer to the default Context that's in place.
Providing I use the bound method.
If I had done it the right way from the start I would have gotten
a precision of 28 (unless I mucked around with the default Context)
and probably would never have figured out that with an unbound
method I could impose a Context on a number on the fly.
Man, that's subtle.
--
Steven
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