[Numpy-discussion] all elements equal
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
Re: [Numpy-discussion] all elements equal
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# All equal I[3] a = np.array([1,1,1,2]) I[4] np.unique(a).size O[4] 2 # All not equal ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] all elements equal
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] np.unique(a).size O[2] 1# All equal I[3] a = np.array([1,1,1,2]) I[4] np.unique(a).size O[4] 2 # All not equal I considered this - just not sure if it's the most efficient ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] all elements equal
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 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] a = np.array([1,1,1,2]) I[4] np.unique(a).size O[4] 2 # All not equal ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] all elements equal
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()) epsilon is an alternative that avoids this. Another good use case for a minmax func. ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] all elements equal
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 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] a = np.array([1,1,1,2]) I[4] np.unique(a).size O[4] 2 # All not equal I considered this - just not sure if it's the most efficient Yeah, it is slow: I[1] a = np.ones(10) I[2] timeit np.unique(a).size 1000 loops, best of 3: 1.56 ms per loop I[3] timeit (a == a[0]).all() 1000 loops, best of 3: 203 us per loop I think all() short-circuits for bool arrays: I[4] a[1] = 9 I[5] timeit (a == a[0]).all() 1 loops, best of 3: 89 us per loop You could avoid making the bool array by writing a function in cython. It could grab the first array element and then return False as soon as it finds an element that is not equal to it. And you could check for closeness. Or: 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 Looks like the following is even faster: np.max(a) == np.min(a) -=- Olivier ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] all elements equal
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 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 # All equal I[3] a = np.array([1,1,1,2]) I[4] np.unique(a).size O[4] 2 # All not equal I considered this - just not sure if it's the most efficient Yeah, it is slow: I[1] a = np.ones(10) I[2] timeit np.unique(a).size 1000 loops, best of 3: 1.56 ms per loop I[3] timeit (a == a[0]).all() 1000 loops, best of 3: 203 us per loop I think all() short-circuits for bool arrays: I[4] a[1] = 9 I[5] timeit (a == a[0]).all() 1 loops, best of 3: 89 us per loop You could avoid making the bool array by writing a function in cython. It could grab the first array element and then return False as soon as it finds an element that is not equal to it. And you could check for closeness. Or: 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 Looks like the following is even faster: np.max(a) == np.min(a) 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 Josef -=- Olivier ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] all elements equal
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] timeit a.min() == a.max() 1 loops, best of 3: 106 us per loop I[4] timeit np.ptp(a) 1 loops, best of 3: 106 us per loop I[5] a[1] = 9 I[6] timeit (a == a[0]).all() 1 loops, best of 3: 89.7 us per loop I[7] timeit a.min() == a.max() 1 loops, best of 3: 102 us per loop I[8] timeit np.ptp(a) 1 loops, best of 3: 103 us per loop ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] all elements equal
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) I[2] timeit (a == a[0]).all() 1000 loops, best of 3: 203 us per loop I[3] timeit a.min() == a.max() 1 loops, best of 3: 106 us per loop I[4] timeit np.ptp(a) 1 loops, best of 3: 106 us per loop I[5] a[1] = 9 I[6] timeit (a == a[0]).all() 1 loops, best of 3: 89.7 us per loop I[7] timeit a.min() == a.max() 1 loops, best of 3: 102 us per loop I[8] timeit np.ptp(a) 1 loops, best of 3: 103 us per loop 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. Ben Root ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] all elements equal
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 allclose: I[2] a = np.ones(10) I[3] timeit a.min() == a.max() 1 loops, best of 3: 106 us per loop I[4] timeit allequal(a) 1 loops, best of 3: 68.9 us per loop I[5] a[1] = 9 I[6] timeit a.min() == a.max() 1 loops, best of 3: 102 us per loop I[7] timeit allequal(a) 100 loops, best of 3: 269 ns per loop where @cython.boundscheck(False) @cython.wraparound(False) def allequal(np.ndarray[np.float64_t, ndim=1] a): cdef: np.float64_t a0 Py_ssize_t i, n=a.size a0 = a[0] for i in range(n): if a[i] != a0: return False return True ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] all elements equal
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. ~Brett ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] all elements equal
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 the speed of a simple cython allclose: I[2] a = np.ones(10) I[3] timeit a.min() == a.max() 1 loops, best of 3: 106 us per loop I[4] timeit allequal(a) 1 loops, best of 3: 68.9 us per loop I[5] a[1] = 9 I[6] timeit a.min() == a.max() 1 loops, best of 3: 102 us per loop I[7] timeit allequal(a) 100 loops, best of 3: 269 ns per loop where @cython.boundscheck(False) @cython.wraparound(False) def allequal(np.ndarray[np.float64_t, ndim=1] a): cdef: np.float64_t a0 Py_ssize_t i, n=a.size a0 = a[0] for i in range(n): if a[i] != a0: return False return True But doesn't this one fail on empty array? ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] all elements equal
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 allequal(np.ndarray[np.float64_t, ndim=1] a): cdef: np.float64_t a0 Py_ssize_t i, n=a.size if n == 0: return False # Or would you like True? a0 = a[0] for i in range(n): if a[i] != a0: return False return True ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] all elements equal
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: @cython.boundscheck(False) @cython.wraparound(False) def allequal(np.ndarray[np.float64_t, ndim=1] a): cdef: np.float64_t a0 Py_ssize_t i, n=a.size if n == 0: return False # Or would you like True? a0 = a[0] for i in range(n): if a[i] != a0: return False return True Sorry for all the posts. I'll go back to being quiet. Seems like np.allclose returns True for empty arrays: I[2] a = np.array([]) I[3] np.allclose(np.array([]), np.array([])) O[3] True The original allequal cython code did the same: I[4] allequal(a) O[4] True ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
