Re: [Numpy-discussion] Is this odd?
In python empty sequences are always equivalent to False, and non-empty to True. You can use this property or: if len(b) 0: . Nadav -Original Message- From: numpy-discussion-boun...@scipy.org on behalf of Shailendra Sent: Fri 02-Apr-10 06:07 To: numpy-discussion@scipy.org Subject: [Numpy-discussion] Is this odd? Hi All, Below is some array behaviour which i think is odd a=arange(10) a array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9]) b=nonzero(a0) b (array([], dtype=int32),) if not b[0]: ... print 'b[0] is false' ... b[0] is false Above case the b[0] is empty so it is fine it is considered false b=nonzero(a1) b (array([0]),) if not b[0]: ... print 'b[0] is false' ... b[0] is false Above case b[0] is a non-empty array. Why should this be consider false. b=nonzero(a8) b (array([9]),) if not b[0]: ... print 'b[0] is false' ... Above case b[0] is non-empty and should be consider true.Which it does. I don't understand why non-empty array should not be considered true irrespective to what value they have. Also, please suggest the best way to differentiate between an empty array and non-empty array( irrespective to what is inside array). Thanks, Shailendra ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion winmail.dat___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] Is this odd?
On 4/2/2010 8:29 AM, Nadav Horesh wrote: In python empty sequences are always equivalent to False, and non-empty to True. I think that was why the OP objected to this behavior: bool(np.array([0])) False Alan Isaac ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] Is this odd?
On Thu, Apr 1, 2010 at 22:07, Shailendra shailendra.vi...@gmail.com wrote: Hi All, Below is some array behaviour which i think is odd a=arange(10) a array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9]) b=nonzero(a0) b (array([], dtype=int32),) if not b[0]: ... print 'b[0] is false' ... b[0] is false Above case the b[0] is empty so it is fine it is considered false b=nonzero(a1) b (array([0]),) if not b[0]: ... print 'b[0] is false' ... b[0] is false Above case b[0] is a non-empty array. Why should this be consider false. b=nonzero(a8) b (array([9]),) if not b[0]: ... print 'b[0] is false' ... Above case b[0] is non-empty and should be consider true.Which it does. I don't understand why non-empty array should not be considered true irrespective to what value they have. We raise an exception for most arrays. We used to evaluate bool(arr) as True if any element in arr were nonzero. However, people routinely got confused and thought that the behavior was that it would return False if any element were zero. It's ambiguous, so we raise an exception to prevent errors and misconceptions. Empty arrays are very rare, so we decided that evaluating bool(arr) as True if there were any elements had very little use case and would still have the potential to confuse people. As shown below, there is a better, more explicit way to test for emptiness. However, we do allow single-element arrays to evaluate to True or False based on its single element because that is unambiguous. This is possibly a mistake, but it tends not to cause too many problems. I don't know why we also allow bool(array([])) to evaluate to False, too; we probably shouldn't. Also, please suggest the best way to differentiate between an empty array and non-empty array( irrespective to what is inside array). a.size 0 -- Robert Kern I have come to believe that the whole world is an enigma, a harmless enigma that is made terrible by our own mad attempt to interpret it as though it had an underlying truth. -- Umberto Eco ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] Is this odd?
On Fri, Apr 2, 2010 at 10:11 AM, Robert Kern robert.k...@gmail.com wrote: On Thu, Apr 1, 2010 at 22:07, Shailendra shailendra.vi...@gmail.com wrote: Hi All, Below is some array behaviour which i think is odd a=arange(10) a array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9]) b=nonzero(a0) b (array([], dtype=int32),) if not b[0]: ... print 'b[0] is false' ... b[0] is false Above case the b[0] is empty so it is fine it is considered false b=nonzero(a1) b (array([0]),) if not b[0]: ... print 'b[0] is false' ... b[0] is false Above case b[0] is a non-empty array. Why should this be consider false. b=nonzero(a8) b (array([9]),) if not b[0]: ... print 'b[0] is false' ... Above case b[0] is non-empty and should be consider true.Which it does. I don't understand why non-empty array should not be considered true irrespective to what value they have. We raise an exception for most arrays. We used to evaluate bool(arr) as True if any element in arr were nonzero. However, people routinely got confused and thought that the behavior was that it would return False if any element were zero. It's ambiguous, so we raise an exception to prevent errors and misconceptions. Empty arrays are very rare, so we decided that evaluating bool(arr) as True if there were any elements had very little use case and would still have the potential to confuse people. As shown below, there is a better, more explicit way to test for emptiness. However, we do allow single-element arrays to evaluate to True or False based on its single element because that is unambiguous. This is possibly a mistake, but it tends not to cause too many problems. I don't know why we also allow bool(array([])) to evaluate to False, too; we probably shouldn't. Also, please suggest the best way to differentiate between an empty array and non-empty array( irrespective to what is inside array). a.size 0 -- Robert Kern with examples: bool(0) False bool(1) True bool(np.array([0])) False bool(np.array([1])) True bool([0]) True bool([1]) True bool(np.array([[0]])) False bool(np.array([[1]])) True np.array([[1]]).size 1 Josef I have come to believe that the whole world is an enigma, a harmless enigma that is made terrible by our own mad attempt to interpret it as though it had an underlying truth. -- Umberto Eco ___ 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] Is this odd?
On Thu, Apr 1, 2010 at 10:07 PM, Shailendra shailendra.vi...@gmail.com wrote: Hi All, Below is some array behaviour which i think is odd a=arange(10) a array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9]) b=nonzero(a0) b (array([], dtype=int32),) if not b[0]: ... print 'b[0] is false' ... b[0] is false Above case the b[0] is empty so it is fine it is considered false b=nonzero(a1) b (array([0]),) if not b[0]: ... print 'b[0] is false' ... b[0] is false Above case b[0] is a non-empty array. Why should this be consider false. b=nonzero(a8) b (array([9]),) if not b[0]: ... print 'b[0] is false' ... Above case b[0] is non-empty and should be consider true.Which it does. I don't understand why non-empty array should not be considered true irrespective to what value they have. Also, please suggest the best way to differentiate between an empty array and non-empty array( irrespective to what is inside array). But by using: if not b[0]: You're not considering the array as a whole, you're looking at the first element, which is giving expected results. As I'm sure you're aware, however, you can't simply do: if not b: # Raises exception So what you need to do is: if b.any(): or: if b.all() Now for determining empty or not, you'll need to look at len(b) or b.shape Ryan -- Ryan May Graduate Research Assistant School of Meteorology University of Oklahoma ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] Is this odd?
On Fri, Apr 2, 2010 at 08:28, Ryan May rma...@gmail.com wrote: On Thu, Apr 1, 2010 at 10:07 PM, Shailendra shailendra.vi...@gmail.com wrote: Hi All, Below is some array behaviour which i think is odd a=arange(10) a array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9]) b=nonzero(a0) b (array([], dtype=int32),) if not b[0]: ... print 'b[0] is false' ... b[0] is false Above case the b[0] is empty so it is fine it is considered false b=nonzero(a1) b (array([0]),) if not b[0]: ... print 'b[0] is false' ... b[0] is false Above case b[0] is a non-empty array. Why should this be consider false. b=nonzero(a8) b (array([9]),) if not b[0]: ... print 'b[0] is false' ... Above case b[0] is non-empty and should be consider true.Which it does. I don't understand why non-empty array should not be considered true irrespective to what value they have. Also, please suggest the best way to differentiate between an empty array and non-empty array( irrespective to what is inside array). But by using: if not b[0]: You're not considering the array as a whole, you're looking at the first element, which is giving expected results. No, b is a tuple containing the array. b[0] is the array itself. -- Robert Kern I have come to believe that the whole world is an enigma, a harmless enigma that is made terrible by our own mad attempt to interpret it as though it had an underlying truth. -- Umberto Eco ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] Is this odd?
Thanks everyone for replies/suggestion. It is simple to avoid this problem. But my point that given the behavior of python this behavior seems inconsistent. There could other method provided which could evaluate bool value depending on values stored in the array. Thanks, Shailendra On Fri, Apr 2, 2010 at 10:28 AM, Ryan May rma...@gmail.com wrote: On Thu, Apr 1, 2010 at 10:07 PM, Shailendra shailendra.vi...@gmail.com wrote: Hi All, Below is some array behaviour which i think is odd a=arange(10) a array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9]) b=nonzero(a0) b (array([], dtype=int32),) if not b[0]: ... print 'b[0] is false' ... b[0] is false Above case the b[0] is empty so it is fine it is considered false b=nonzero(a1) b (array([0]),) if not b[0]: ... print 'b[0] is false' ... b[0] is false Above case b[0] is a non-empty array. Why should this be consider false. b=nonzero(a8) b (array([9]),) if not b[0]: ... print 'b[0] is false' ... Above case b[0] is non-empty and should be consider true.Which it does. I don't understand why non-empty array should not be considered true irrespective to what value they have. Also, please suggest the best way to differentiate between an empty array and non-empty array( irrespective to what is inside array). But by using: if not b[0]: You're not considering the array as a whole, you're looking at the first element, which is giving expected results. As I'm sure you're aware, however, you can't simply do: if not b: # Raises exception So what you need to do is: if b.any(): or: if b.all() Now for determining empty or not, you'll need to look at len(b) or b.shape Ryan -- Ryan May Graduate Research Assistant School of Meteorology University of Oklahoma ___ 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] Is this odd?
On Fri, Apr 2, 2010 at 8:31 AM, Robert Kern robert.k...@gmail.com wrote: On Fri, Apr 2, 2010 at 08:28, Ryan May rma...@gmail.com wrote: On Thu, Apr 1, 2010 at 10:07 PM, Shailendra shailendra.vi...@gmail.com wrote: Hi All, Below is some array behaviour which i think is odd a=arange(10) a array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9]) b=nonzero(a0) b (array([], dtype=int32),) if not b[0]: ... print 'b[0] is false' ... b[0] is false Above case the b[0] is empty so it is fine it is considered false b=nonzero(a1) b (array([0]),) if not b[0]: ... print 'b[0] is false' ... b[0] is false Above case b[0] is a non-empty array. Why should this be consider false. b=nonzero(a8) b (array([9]),) if not b[0]: ... print 'b[0] is false' ... Above case b[0] is non-empty and should be consider true.Which it does. I don't understand why non-empty array should not be considered true irrespective to what value they have. Also, please suggest the best way to differentiate between an empty array and non-empty array( irrespective to what is inside array). But by using: if not b[0]: You're not considering the array as a whole, you're looking at the first element, which is giving expected results. No, b is a tuple containing the array. b[0] is the array itself. Wow, that's what I get for trying to read code *before* coffee. On the plus side, I now know how nonzero() actually works. Ryan -- Ryan May Graduate Research Assistant School of Meteorology University of Oklahoma ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
[Numpy-discussion] Is this odd?
Hi All, Below is some array behaviour which i think is odd a=arange(10) a array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9]) b=nonzero(a0) b (array([], dtype=int32),) if not b[0]: ... print 'b[0] is false' ... b[0] is false Above case the b[0] is empty so it is fine it is considered false b=nonzero(a1) b (array([0]),) if not b[0]: ... print 'b[0] is false' ... b[0] is false Above case b[0] is a non-empty array. Why should this be consider false. b=nonzero(a8) b (array([9]),) if not b[0]: ... print 'b[0] is false' ... Above case b[0] is non-empty and should be consider true.Which it does. I don't understand why non-empty array should not be considered true irrespective to what value they have. Also, please suggest the best way to differentiate between an empty array and non-empty array( irrespective to what is inside array). Thanks, Shailendra ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
