In a message of Sun, 22 Feb 2015 09:53:33 -0800, LJ writes: >Hi everyone. Quick question here. Lets suppose if have the following numpy >array: > >b=np.array([[0]*2]*3) > >and then: > >>>> id(b[0]) >45855552 >>>> id(b[1]) >45857512 >>>> id(b[2]) >45855552 > >Please correct me if I am wrong, but according to this b[2] and b[0] are the >same object. Now, > >>>> b[0] is b[2] >False
You are running into one of the peculiarities of the python representation of numbers. It can make things more efficient to represent all common numbers as 'there is only one' of them. So. Python 2.7.9 (default, Dec 11 2014, 08:58:12) [GCC 4.9.2] on linux2 Type "help", "copyright", "credits" or "license" for more information. >>> a = 1 >>> b = 1 >>> a is b True >>> a = 1001 >>> b = 1001 >>> a is b False -------- Don't rely on this. Other implementations are free to implement this however they like. -------- [PyPy 2.4.0 with GCC 4.9.2] on linux2 Type "help", "copyright", "credits" or "license" for more information. >>>> a = 1001 >>>> b = 1001 >>>> a is b True Laura -- https://mail.python.org/mailman/listinfo/python-list
