On Mon, Mar 5, 2012 at 12:12 PM, Keith Goodman wrote:
> On Mon, Mar 5, 2012 at 12:06 PM, Neal Becker wrote:
>> But doesn't this one fail on empty array?
>
> Yes. I'm optimizing for fun, not for corner cases. This should work
> for size zero and NaNs:
>
> @cython.boundscheck(False)
> @cython.wrapa
On Mon, Mar 5, 2012 at 12:06 PM, Neal Becker wrote:
> But doesn't this one fail on empty array?
Yes. I'm optimizing for fun, not for corner cases. This should work
for size zero and NaNs:
@cython.boundscheck(False)
@cython.wraparound(False)
def allequal(np.ndarray[np.float64_t, ndim=1] a):
c
Keith Goodman wrote:
> On Mon, Mar 5, 2012 at 11:52 AM, Benjamin Root wrote:
>> Another issue to watch out for is if the array is empty. Technically
>> speaking, that should be True, but some of the solutions offered so far
>> would fail in this case.
>
> Good point.
>
> For fun, here's the sp
> Another issue to watch out for is if the array is empty. Technically
> speaking, that should be True, but some of the solutions offered so far
> would fail in this case.
Similarly, NaNs or Infs could cause problems: they should signal as
False, but several of the solutions would return True.
On Mon, Mar 5, 2012 at 11:52 AM, Benjamin Root wrote:
> Another issue to watch out for is if the array is empty. Technically
> speaking, that should be True, but some of the solutions offered so far
> would fail in this case.
Good point.
For fun, here's the speed of a simple cython allclose:
I
On Mon, Mar 5, 2012 at 1:44 PM, Keith Goodman wrote:
> On Mon, Mar 5, 2012 at 11:36 AM, wrote:
> > How about numpy.ptp, to follow this line? I would expect it's single
> > pass, but wouldn't short circuit compared to cython of Keith
>
> I[1] a = np.ones(10)
> I[2] timeit (a == a[0]).all()
>
On Mon, Mar 5, 2012 at 11:36 AM, wrote:
> How about numpy.ptp, to follow this line? I would expect it's single
> pass, but wouldn't short circuit compared to cython of Keith
I[1] a = np.ones(10)
I[2] timeit (a == a[0]).all()
1000 loops, best of 3: 203 us per loop
I[3] timeit a.min() == a.max
On Mon, Mar 5, 2012 at 2:33 PM, Olivier Delalleau wrote:
> Le 5 mars 2012 14:29, Keith Goodman a écrit :
>
>> On Mon, Mar 5, 2012 at 11:24 AM, Neal Becker wrote:
>> > Keith Goodman wrote:
>> >
>> >> On Mon, Mar 5, 2012 at 11:14 AM, Neal Becker
>> >> wrote:
>> >>> What is a simple, efficient way
Le 5 mars 2012 14:29, Keith Goodman a écrit :
> On Mon, Mar 5, 2012 at 11:24 AM, Neal Becker wrote:
> > Keith Goodman wrote:
> >
> >> On Mon, Mar 5, 2012 at 11:14 AM, Neal Becker
> wrote:
> >>> What is a simple, efficient way to determine if all elements in an
> array (in
> >>> my case, 1D) are
On Mon, Mar 5, 2012 at 1:29 PM, Keith Goodman wrote:
>
> I[8] np.allclose(a, a[0])
> O[8] False
> I[9] a = np.ones(10)
> I[10] np.allclose(a, a[0])
> O[10] True
>
>
One disadvantage of using a[0] as a proxy is that the result depends on the
ordering of a
(a.max() - a.min()) < epsilon
is a
On Mon, Mar 5, 2012 at 11:24 AM, Neal Becker wrote:
> Keith Goodman wrote:
>
>> On Mon, Mar 5, 2012 at 11:14 AM, Neal Becker wrote:
>>> What is a simple, efficient way to determine if all elements in an array (in
>>> my case, 1D) are equal? How about close?
>>
>> For the exactly equal case, how
How about the following?
exact: numpy.all(a == a[0])
inexact: numpy.allclose(a, a[0])
On Mar 5, 2012, at 2:19 PM, Keith Goodman wrote:
> On Mon, Mar 5, 2012 at 11:14 AM, Neal Becker wrote:
>> What is a simple, efficient way to determine if all elements in an array (in
>> my
>> case, 1D) are equ
Keith Goodman wrote:
> On Mon, Mar 5, 2012 at 11:14 AM, Neal Becker wrote:
>> What is a simple, efficient way to determine if all elements in an array (in
>> my case, 1D) are equal? How about close?
>
> For the exactly equal case, how about:
>
> I[1] a = np.array([1,1,1,1])
> I[2] np.unique(a)
On Mon, Mar 5, 2012 at 11:14 AM, Neal Becker wrote:
> What is a simple, efficient way to determine if all elements in an array (in
> my
> case, 1D) are equal? How about close?
For the exactly equal case, how about:
I[1] a = np.array([1,1,1,1])
I[2] np.unique(a).size
O[2] 1# All equal
I[3]
What is a simple, efficient way to determine if all elements in an array (in my
case, 1D) are equal? How about close?
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