为什么numpy数组的元素与自己不同?
问题描述：
我如何解释这些的最后一行?
How do I explain the last line of these?
>>> a = 1
>>> a is a
True
>>> a = [1, 2, 3]
>>> a is a
True
>>> a = np.zeros(3)
>>> a
array([ 0., 0., 0.])
>>> a is a
True
>>> a[0] is a[0]
False
我一直以为一切都至少是本身"！
I always thought that everything is at least "is" that thing itself!
答
NumPy不会将数组元素存储为Python对象.如果您尝试访问单个元素，则NumPy必须创建一个新的包装器对象来表示该元素，并且每次访问该元素时，它必须每次执行.两次访问a[0]的包装对象是不同的对象，因此a[0] is a[0]返回False.
NumPy doesn't store array elements as Python objects. If you try to access an individual element, NumPy has to create a new wrapper object to represent the element, and it has to do this every time you access the element. The wrapper objects from two accesses to a[0] are different objects, so a[0] is a[0] returns False.