传递给 `np.ones` 的参数是什么，它在这里是如何工作的？

Question

I actually wanted to multiply the array into 3 and tried to use * 3 inside the shape. I then realized it has to be outside the np.ones like np.ones((1,2))*4 . But was wondering why this is producing exponential results. Can someone please explain to me the below behavior?

np.ones((1,2)*1) returns array([[1., 1.]])

np.ones((1,2)*2) returns

array([[[[1., 1.]],
        [[1., 1.]]]])

np.ones((1,2)*3) returns

array([[[[[[1., 1.]],
          [[1., 1.]]]],
        [[[[1., 1.]],
          [[1., 1.]]]]]])

Similarly, np.ones((1,2)*4) returns

array([[[[[[[[1., 1.]],
            [[1., 1.]]]],
          [[[[1., 1.]],
            [[1., 1.]]]]]],
        [[[[[[1., 1.]],
            [[1., 1.]]]],
          [[[[1., 1.]],
            [[1., 1.]]]]]]]])

Unfortunately, the documentation doesn't have any explanation on this.

Answer 1

np.ones accepts a shape parameter and returns an ND array according to your specification. For example, with np.ones((10,)) , you get a 1D array with 10 elements... np.ones((3, 5)) will give you a 2D array of size 3x5 with 3*5=15 elements,... and so on.

Now, you've done (for example) (1, 2) * 3 , which if you run in a python REPL will show

(1, 2) * 3
# (1, 2, 1, 2, 1, 2)

Passing this to np.ones will return a 6D array of shape (1, 2, 1, 2, 1, 2) with 8 elements.

np.ones((1, 2)*3)     
array([[[[[[1., 1.]],    
          [[1., 1.]]]],
        [[[[1., 1.]],    
          [[1., 1.]]]]]])

_.shape
# (1, 2, 1, 2, 1, 2)

And similar, for the others.