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More options Mar 5 2012, 2:14 pm
From: Neal Becker <ndbeck...@gmail.com>
Date: Mon, 05 Mar 2012 14:14:02 -0500
Local: Mon, Mar 5 2012 2:14 pm
Subject: [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?

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More options Mar 5 2012, 2:19 pm
From: Keith Goodman <kwgood...@gmail.com>
Date: Mon, 5 Mar 2012 11:19:44 -0800
Local: Mon, Mar 5 2012 2:19 pm
Subject: 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
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More options Mar 5 2012, 2:24 pm
From: Neal Becker <ndbeck...@gmail.com>
Date: Mon, 05 Mar 2012 14:24:07 -0500
Local: Mon, Mar 5 2012 2:24 pm
Subject: 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

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More options Mar 5 2012, 2:28 pm
From: Zachary Pincus <zachary.pin...@yale.edu>
Date: Mon, 5 Mar 2012 14:28:41 -0500
Local: Mon, Mar 5 2012 2:28 pm
Subject: Re: [Numpy-discussion] all elements equal
exact: numpy.all(a == a[0])
inexact: numpy.allclose(a, a[0])

On Mar 5, 2012, at 2:19 PM, Keith Goodman wrote:

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More options Mar 5 2012, 2:29 pm
From: Keith Goodman <kwgood...@gmail.com>
Date: Mon, 5 Mar 2012 11:29:13 -0800
Local: Mon, Mar 5 2012 2:29 pm
Subject: Re: [Numpy-discussion] all elements equal

Yeah, it is slow:

I[1] a = np.ones(100000)
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()
10000 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(100000)
I[10] np.allclose(a, a[0])
O[10] True
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More options Mar 5 2012, 2:32 pm
From: John Hunter <jdh2...@gmail.com>
Date: Mon, 5 Mar 2012 13:32:40 -0600
Local: Mon, Mar 5 2012 2:32 pm
Subject: 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(100000)
> 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.

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More options Mar 5 2012, 2:33 pm
From: Olivier Delalleau <sh...@keba.be>
Date: Mon, 5 Mar 2012 14:33:23 -0500
Local: Mon, Mar 5 2012 2:33 pm
Subject: Re: [Numpy-discussion] all elements equal

Le 5 mars 2012 14:29, Keith Goodman <kwgood...@gmail.com> a écrit :

Looks like the following is even faster:
np.max(a) == np.min(a)

-=- Olivier

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More options Mar 5 2012, 2:36 pm
From: josef.p...@gmail.com
Date: Mon, 5 Mar 2012 14:36:44 -0500
Local: Mon, Mar 5 2012 2:36 pm
Subject: Re: [Numpy-discussion] all elements equal

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

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More options Mar 5 2012, 2:44 pm
From: Keith Goodman <kwgood...@gmail.com>
Date: Mon, 5 Mar 2012 11:44:38 -0800
Local: Mon, Mar 5 2012 2:44 pm
Subject: 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(100000)
I[2] timeit (a == a[0]).all()
1000 loops, best of 3: 203 us per loop
I[3] timeit a.min() == a.max()
10000 loops, best of 3: 106 us per loop
I[4] timeit np.ptp(a)
10000 loops, best of 3: 106 us per loop

I[5] a[1] = 9
I[6] timeit (a == a[0]).all()
10000 loops, best of 3: 89.7 us per loop
I[7] timeit a.min() == a.max()
10000 loops, best of 3: 102 us per loop
I[8] timeit np.ptp(a)
10000 loops, best of 3: 103 us per loop
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More options Mar 5 2012, 2:52 pm
From: Benjamin Root <ben.r...@ou.edu>
Date: Mon, 5 Mar 2012 13:52:04 -0600
Local: Mon, Mar 5 2012 2:52 pm
Subject: 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.

Ben Root

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More options Mar 5 2012, 2:56 pm
From: Keith Goodman <kwgood...@gmail.com>
Date: Mon, 5 Mar 2012 11:56:56 -0800
Local: Mon, Mar 5 2012 2:56 pm
Subject: 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(100000)
I[3] timeit a.min() == a.max()
10000 loops, best of 3: 106 us per loop
I[4] timeit allequal(a)
10000 loops, best of 3: 68.9 us per loop

I[5] a[1] = 9
I[6] timeit a.min() == a.max()
10000 loops, best of 3: 102 us per loop
I[7] timeit allequal(a)
1000000 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
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More options Mar 5 2012, 3:01 pm
From: Brett Olsen <brett.ol...@gmail.com>
Date: Mon, 5 Mar 2012 14:01:52 -0600
Local: Mon, Mar 5 2012 3:01 pm
Subject: 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
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More options Mar 5 2012, 3:06 pm
From: Neal Becker <ndbeck...@gmail.com>
Date: Mon, 05 Mar 2012 15:06:23 -0500
Local: Mon, Mar 5 2012 3:06 pm
Subject: Re: [Numpy-discussion] all elements equal

But doesn't this one fail on empty array?

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More options Mar 5 2012, 3:12 pm
From: Keith Goodman <kwgood...@gmail.com>
Date: Mon, 5 Mar 2012 12:12:41 -0800
Local: Mon, Mar 5 2012 3:12 pm
Subject: 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
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More options Mar 5 2012, 3:30 pm
From: Keith Goodman <kwgood...@gmail.com>
Date: Mon, 5 Mar 2012 12:30:05 -0800
Local: Mon, Mar 5 2012 3:30 pm
Subject: Re: [Numpy-discussion] all elements equal

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
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