What is a simple, efficient way to determine if all elements in an array (in my
case, 1D) are equal? How about close?
___
NumPy-Discussion mailing list
NumPy-Discussion@scipy.org
http://mail.scipy.org/mailman/listinfo/numpy-discussion
On Mon, Mar 5, 2012 at 11:14 AM, Neal Becker ndbeck...@gmail.com 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
Keith Goodman wrote:
On Mon, Mar 5, 2012 at 11:14 AM, Neal Becker ndbeck...@gmail.com 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]
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 ndbeck...@gmail.com wrote:
What is a simple, efficient way to determine if all elements in an array (in
my
On Mon, Mar 5, 2012 at 1:29 PM, Keith Goodman kwgood...@gmail.com 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())
Le 5 mars 2012 14:29, Keith Goodman kwgood...@gmail.com a écrit :
On Mon, Mar 5, 2012 at 11:24 AM, Neal Becker ndbeck...@gmail.com wrote:
Keith Goodman wrote:
On Mon, Mar 5, 2012 at 11:14 AM, Neal Becker ndbeck...@gmail.com
wrote:
What is a simple, efficient way to determine if all
On Mon, Mar 5, 2012 at 2:33 PM, Olivier Delalleau sh...@keba.be wrote:
Le 5 mars 2012 14:29, Keith Goodman kwgood...@gmail.com a écrit :
On Mon, Mar 5, 2012 at 11:24 AM, Neal Becker ndbeck...@gmail.com wrote:
Keith Goodman wrote:
On Mon, Mar 5, 2012 at 11:14 AM, Neal Becker
On Mon, Mar 5, 2012 at 11:36 AM, josef.p...@gmail.com 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]
On Mon, Mar 5, 2012 at 1:44 PM, Keith Goodman kwgood...@gmail.com wrote:
On Mon, Mar 5, 2012 at 11:36 AM, josef.p...@gmail.com 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)
On Mon, Mar 5, 2012 at 11:52 AM, Benjamin Root ben.r...@ou.edu 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
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.
Keith Goodman wrote:
On Mon, Mar 5, 2012 at 11:52 AM, Benjamin Root ben.r...@ou.edu 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
On Mon, Mar 5, 2012 at 12:06 PM, Neal Becker ndbeck...@gmail.com 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
On Mon, Mar 5, 2012 at 12:12 PM, Keith Goodman kwgood...@gmail.com wrote:
On Mon, Mar 5, 2012 at 12:06 PM, Neal Becker ndbeck...@gmail.com 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:
14 matches
Mail list logo
