Re: [Numpy-discussion] Round away from zero (towards +/- infinity)

2014-10-04 Thread Charles R Harris 
On Fri, Oct 3, 2014 at 11:28 AM, T J tjhn...@gmail.com wrote:

 It does, but it is not portable. That's why I was hoping NumPy might think
 about supporting more rounding algorithms.

 On Thu, Oct 2, 2014 at 10:00 PM, John Zwinck jzwi...@gmail.com wrote:

 On 3 Oct 2014 07:09, T J tjhn...@gmail.com wrote:
 
  Any bites on this?
 
  On Wed, Sep 24, 2014 at 12:23 PM, T J tjhn...@gmail.com wrote:
  Python's round function goes away from zero, so I am looking for the
 NumPy equivalent (and using vectorize() seems undesirable). In this sense,
 it seems that having a ufunc for this type of rounding could be helpful.
 
  Aside: Is there interest in a more general around() that allows users
 to specify alternative tie-breaking rules, with the default staying 'round
 half to nearest even'? [1]
  ---
  [1]
 http://stackoverflow.com/questions/16000574/tie-breaking-of-round-with-numpy

 I like the solution given in that Stack Overflow post, namely using
 ctypes to call fesetround(). Does that work for you?




In [4]: def roundout(x):
   ...: return trunc(x + copysign(.5, x))
   ...:

Will do what you want, if not quite as nicely as a ufunc. Won't work as is
for complex, but that could be handled with a view.

Chuck
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Re: [Numpy-discussion] Round away from zero (towards +/- infinity)

2014-10-03 Thread T J 
It does, but it is not portable. That's why I was hoping NumPy might think
about supporting more rounding algorithms.

On Thu, Oct 2, 2014 at 10:00 PM, John Zwinck jzwi...@gmail.com wrote:

 On 3 Oct 2014 07:09, T J tjhn...@gmail.com wrote:
 
  Any bites on this?
 
  On Wed, Sep 24, 2014 at 12:23 PM, T J tjhn...@gmail.com wrote:
  Python's round function goes away from zero, so I am looking for the
 NumPy equivalent (and using vectorize() seems undesirable). In this sense,
 it seems that having a ufunc for this type of rounding could be helpful.
 
  Aside: Is there interest in a more general around() that allows users
 to specify alternative tie-breaking rules, with the default staying 'round
 half to nearest even'? [1]
  ---
  [1]
 http://stackoverflow.com/questions/16000574/tie-breaking-of-round-with-numpy

 I like the solution given in that Stack Overflow post, namely using ctypes
 to call fesetround(). Does that work for you?

 ___
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 NumPy-Discussion@scipy.org
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Re: [Numpy-discussion] Round away from zero (towards +/- infinity)

2014-10-02 Thread T J 
Any bites on this?

On Wed, Sep 24, 2014 at 12:23 PM, T J tjhn...@gmail.com wrote:

 Is there a ufunc for rounding away from zero? Or do I need to do

 x2 = sign(x) * ceil(abs(x))

 whenever I want to round away from zero? Maybe the following is better?

 x_ceil = ceil(x)
 x_floor = floor(x)
 x2 = where(x = 0, x_ceil, x_floor)

 Python's round function goes away from zero, so I am looking for the NumPy
 equivalent (and using vectorize() seems undesirable). In this sense, it
 seems that having a ufunc for this type of rounding could be helpful.

 Aside: Is there interest in a more general around() that allows users to
 specify alternative tie-breaking rules, with the default staying 'round
 half to nearest even'? [1]

 Also, what is the difference between NumPy's fix() and trunc() functions?
 It seems like they achieve the same goal. trunc() was added in 1.3.0. So is
 fix() just legacy?

 ---
 [1]
 http://stackoverflow.com/questions/16000574/tie-breaking-of-round-with-numpy


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Re: [Numpy-discussion] Round away from zero (towards +/- infinity)

2014-10-02 Thread John Zwinck 
On 3 Oct 2014 07:09, T J tjhn...@gmail.com wrote:

 Any bites on this?

 On Wed, Sep 24, 2014 at 12:23 PM, T J tjhn...@gmail.com wrote:
 Python's round function goes away from zero, so I am looking for the
NumPy equivalent (and using vectorize() seems undesirable). In this sense,
it seems that having a ufunc for this type of rounding could be helpful.

 Aside: Is there interest in a more general around() that allows users to
specify alternative tie-breaking rules, with the default staying 'round
half to nearest even'? [1]
 ---
 [1]
http://stackoverflow.com/questions/16000574/tie-breaking-of-round-with-numpy

I like the solution given in that Stack Overflow post, namely using ctypes
to call fesetround(). Does that work for you?
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[Numpy-discussion] Round away from zero (towards +/- infinity)

2014-09-24 Thread T J 
Is there a ufunc for rounding away from zero? Or do I need to do

x2 = sign(x) * ceil(abs(x))

whenever I want to round away from zero? Maybe the following is better?

x_ceil = ceil(x)
x_floor = floor(x)
x2 = where(x = 0, x_ceil, x_floor)

Python's round function goes away from zero, so I am looking for the NumPy
equivalent (and using vectorize() seems undesirable). In this sense, it
seems that having a ufunc for this type of rounding could be helpful.

Aside: Is there interest in a more general around() that allows users to
specify alternative tie-breaking rules, with the default staying 'round
half to nearest even'? [1]

Also, what is the difference between NumPy's fix() and trunc() functions?
It seems like they achieve the same goal. trunc() was added in 1.3.0. So is
fix() just legacy?

---
[1]
http://stackoverflow.com/questions/16000574/tie-breaking-of-round-with-numpy
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