numpy .dot using lists and arrays, whats the difference

Question

When multiplication of two similar matrices 1*2 like [1,2], [3,5] is carried out using numpy.dot, it gives a result, when in fact it should be giving a shape and dimension error like while multiplying two similar arrays. What is going on under the hood?

a=[1,2]
b=[6,3]
result=[np.dot(b, a)]
print(result)

O/P= 12

But,

a=[[1,2]]
b=[[6,3]]
result=[np.dot(b, a)]
print(result)

Error:

O/P= ValueError: shapes (1,2) and (1,2) not aligned: 2 (dim 1) != 1 (dim 0)

• hpaulj: To be picky, both of your inputs are lists. When converted to ndarray, which dot will do, they are 1d and 2d respectively.

Answer 1

As per the documentation here,

• If both a and b are 1-D arrays, it is inner product of vectors (without complex conjugation).

• If both a and b are 2-D arrays, it is matrix multiplication, but using matmul or a @ b is preferred.

Case 1: a and b are 1-D arrays, so result is 1*6+2*3=12.

Case 2: a and b are 2-D arrays, so we will do matrix product of these two. It raises ValueError since the last dimension of a is not the same size as the second-to-last dimension of b.

Answer 2

Adding on to Anubhav Singh's correct answer, note that a matrix product of a row vector with a column vector returns a 1-by-1 matrix whose sole entry is the dot product of the two vectors, so in this case,

In [32]: a = np.array([[1,2]])
In [33]: b = np.array([[6,3]])
In [34]: a @ b.T
Out[34]: array([[12]])
In [35]: np.dot(a, b.T)
Out[35]: array([[12]])
In [36]: np.dot(a[0], b[0])
Out[36]: 12

This is why np.dot behaves the way it does.