Question
Please elaborate the answer in Numpy array broadcasting rules in 2012, and clarify what trailing axes are, as I am not sure which "linked documentation page" the answer refers to. Perhaps it has changed in the last 8 years.
As axes in trailing axes is plural, at least two last axes sizes must match (except sigular)? If so why at least two?
The given answer was:
Well, the meaning of trailing axes is explained on the linked documentation page. If you have two arrays with different dimensions number, say one 1x2x3 and other 2x3, then you compare only the trailing common dimensions, in this case 2x3. But if both your arrays are two-dimensional, then their corresponding sizes have to be either equal or one of them has to be 1.
In your case you have a 2x2 and 4x2 and 4 != 2 and neither 4 or 2 equals 1, so this doesn't work.
The error and the question raised were:
A = np.array([[1,2],[3,4]])
B = np.array([[2,3],[4,6],[6,9],[8,12]])
print("A.shape {}".format(A.shape))
print("B.shape {}".format(B.shape))
A*B
---
A.shape (2, 2) # <---- The last axis size is 2 in both shapes.
B.shape (4, 2) # <---- Apparently this "2" is not the size of trailing axis/axes
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
<ipython-input-91-7a3f7e97944d> in <module>
3 print("A.shape {}".format(A.shape))
4 print("B.shape {}".format(B.shape))
----> 5 A*B
ValueError: operands could not be broadcast together with shapes (2,2) (4,2)
Since both A and B have two columns, I would have thought this would work.
So, I'm probably misunderstanding something here about the term "trailing axis",
and how it applies to N-dimensional arrays.
References
The Broadcasting Rule
In order to broadcast, the size of the trailing axes for both arrays in an operation must either be the same size or one of them must be one.
Update
Understanding based on the reply by @Akshay Sehgal. Consider 2 arrays A.shape = (4,5,1) and B.shape = (1,2).
A = np.arange(20).reshape((4, 5, 1))
B = np.arange(2).reshape((1,2))
print("A.shape {}".format(A.shape))
print("B.shape {}".format(B.shape))
---
A.shape (4, 5, 1)
B.shape (1, 2)
Firstly, look at axis=-1 and the shape 01 in A is broadcast from 01 to 02, because it is singular, to match that of B. Then shape 01 in B for axis=-2 is broadcast from 01 (singular) to 05 to match that of A. The result is shape (4, 5, 2).
print("A * B shape is {}".format((A*B).shape))
---
A * B shape is (4, 5, 2)
Based on the answer from @hpaulj, a way to simulate broadcasting.
print("A.shape {}".format(A.shape))
print("B.shape {}".format(B.shape))
---
A.shape (4, 5, 1)
B.shape (1, 2)
# Check ranks.
print("rank(A) {} rank(B) {}".format(A.ndim, B.ndim))
---
rank(A) 3 rank(B) 2
# Expand B because rank(B) < rank(A).
B = B[
None,
::
]
B.shape
---
(1, 1, 2)
A:(4,5,1)
↑ ↑ ↓
B:(1,1,2)
----------
C:(4,5,2)
