Johann Cohen-Tanugi wrote:

how about np.any(a!=b) ??

On 10/27/2010 12:25 PM, Neal Becker wrote:

...

Is there a way to get a short circuit != ?

That is, compare 2 arrays but stop as soon as the first element comparison fails?

I'm assuming that np.all (a != b) will _not_ do this, but will first compare all elements.

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I don't think that will do short ciruit, will it? I think that will compare each element, returning a bool array, then short-circuit eval that bool array.

On Wed, Oct 27, 2010 at 9:37 AM, Johann Cohen-Tanugi <cohen@lpta.in2p3.fr> wrote:

On 10/27/2010 03:31 PM, Neal Becker wrote:

...

Johann Cohen-Tanugi wrote:

...

how about np.any(a!=b)  ??

On 10/27/2010 12:25 PM, Neal Becker wrote:

...

Is there a way to get a short circuit != ?

That is, compare 2 arrays but stop as soon as the first element comparison fails?

I'm assuming that np.all (a != b) will _not_ do this, but will first compare all elements.

_______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion

I don't think that will do short ciruit, will it?  I think that will compare each element, returning a bool array, then short-circuit eval that bool array.

In [3]: a=np.array([2,3,

In [4]: b=np.array([2,5,

In [5]: np.any(a!=b) Out[5]: True

it does not return a bool array, it seems. I do not see how you would "broadcast" the notion of "any".... along axes maybe?

Not definitive by any means, but In [28]: a = np.arange(1,10000000) In [29]: b = np.arange(1,10000000) In [30]: timeit np.any(a!=b) 10 loops, best of 3: 37.8 ms per loop In [31]: a[0] = 10. In [32]: timeit np.any(a!=b) 10 loops, best of 3: 24.7 ms per loop In [33]: a[0] = 1 In [34]: a[-1] = 1 In [35]: timeit np.any(a!=b) 10 loops, best of 3: 37.7 ms per loop It seems to at least take less time when the difference is at the "beginning," though I'm sure there could be exceptions. Skipper

Pauli Virtanen wrote:

Wed, 27 Oct 2010 09:44:59 -0400, Skipper Seabold wrote: [clip]

...

In [35]: timeit np.any(a!=b) [clip] It seems to at least take less time when the difference is at the "beginning," though I'm sure there could be exceptions.

It performs all the comparisons to create a temporary boolean array. any() does return when it sees the first True, but this is not full short-circuiting.

I propose adding is_equal(a,b) function which does true short-ciruciting

This is silly: the structure of the python language prevents meaningful short-circuiting in the case of np.any(a!=b) While it's true that np.any itself may short-circuit, the 'a!=b' statement itself will be evaluated in its entirety before the result (a boolean array) is passed to np.any. This is just how python (like most other languages) works. This means that the big-O complexity isn't reduced any: all elements of a and b have to be examined. If any short-circuits, then at least all the elements don't need to be examined twice! If performance actually matters here (unlikely?), a quick bit of cython code could do this easily enough in the simple case where a and b are the same shape (so no broadcasting required). Note that the below hard-codes the dtype also. import cython cimport numpy import numpy TYPE_T = numpy.int64_t TYPE = numpy.int64 @cython.boundscheck(False) def equal(a, b): cdef: numpy.ndarray[TYPE, ndim=1, negative_indices=False] a_flat = a.astype(TYPE).flatten() numpy.ndarray[TYPE, ndim=1, negative_indices=False] b_flat = b.astype(TYPE).flatten() unsigned int i, l assert a_flat.shape[0] == b_flat.shape[0] l = a_flat.shape[0] for i in range(l): if a_flat[i] != b_flat[i]: return False return True Zach On Oct 27, 2010, at 9:44 AM, Skipper Seabold wrote:

On Wed, Oct 27, 2010 at 9:37 AM, Johann Cohen-Tanugi <cohen@lpta.in2p3.fr> wrote:

...

On 10/27/2010 03:31 PM, Neal Becker wrote:

...

Johann Cohen-Tanugi wrote:

...

how about np.any(a!=b) ??

On 10/27/2010 12:25 PM, Neal Becker wrote:

...

Is there a way to get a short circuit != ?

That is, compare 2 arrays but stop as soon as the first element comparison fails?

I'm assuming that np.all (a != b) will _not_ do this, but will first compare all elements.

_______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion

I don't think that will do short ciruit, will it? I think that will compare each element, returning a bool array, then short-circuit eval that bool array.

In [3]: a=np.array([2,3,

In [4]: b=np.array([2,5,

In [5]: np.any(a!=b) Out[5]: True

it does not return a bool array, it seems. I do not see how you would "broadcast" the notion of "any".... along axes maybe?

Not definitive by any means, but

In [28]: a = np.arange(1,10000000)

In [29]: b = np.arange(1,10000000)

In [30]: timeit np.any(a!=b) 10 loops, best of 3: 37.8 ms per loop

In [31]: a[0] = 10.

In [32]: timeit np.any(a!=b) 10 loops, best of 3: 24.7 ms per loop

In [33]: a[0] = 1

In [34]: a[-1] = 1

In [35]: timeit np.any(a!=b) 10 loops, best of 3: 37.7 ms per loop

It seems to at least take less time when the difference is at the "beginning," though I'm sure there could be exceptions.

Skipper _______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion

Alan G Isaac writes:

Maybe: any((ai != bi) for ai,bi in izip(a.flat,b.flat)) ?

I think this would indeed work, but it's not really friendly, while it should work out-of-the-box. Of course, we do not always want operations to perform lazily, so a generic approach might be on the lines of: any(a.lazy() != b) Invoking 'lazy' on an ndarray would return an object that instead of computing the given operations will just record them, and try to defer their evaluation as much as possible. E.g.,: class LazyNdarray: def __init__ (self, left, op = None, right = None): self._left = left self._op = op self._right = right # transitivize deferral on these operations def __neq__ (self, other): return LazyNdarray(self, "__neq__", other) # evaluate on these operations def __getitem__ (self, index): return self._evaluate(index) def __iter__ (self): for idx in range(len(self._left)): yield self._evaluate(idx) # do the evaluation def _evaluate (self, index) # still, this will not work unles no broadcast is necessary l = self._left[index] op = getattr(l, self._op) return op(right[index]) class ndarray: def lazy (self): return LazyNdarray(self) Or something like that. Lluis -- "And it's much the same thing with knowledge, for whenever you learn something new, the whole world becomes that much richer." -- The Princess of Pure Reason, as told by Norton Juster in The Phantom Tollbooth