i have an array y with shape (n,), I want to compute the inner product matrix, which is a n * n matrix However, when I tried to do it in Python
np.dot(y , y) I got the answer n, this is not what I am looking for I have also tried:
np.dot(np.transpose(y),y) np.dot(y, np.transpose(y)) I always get the same answer n
I think you are looking for:
np.multiply.outer(y,y) or equally:
y = y[None,:] y.T@y example:
y = np.array([1,2,3])[None,:] output:
#[[1 2 3] # [2 4 6] # [3 6 9]] You can try to reshape y from shape (70,) to (70,1) before multiplying the 2 matrices.
# Reshape y = y.reshape(70,1) # Either below code would work y*y.T np.matmul(y,y.T) One-liner?
np.dot(a[:, None], a[None, :]) transpose doesn't work on 1-D arrays, because you need atleast two axes to 'swap' them. This solution adds a new axis to the array; in the first argument, it looks like a column vector and has two axes; in the second argument it still looks like a row vector but has two axes.
Looks like what you need is the @ matrix multiplication operator. dot method is only to compute dot product between vectors, what you want is matrix multiplication.
>>> a = np.random.rand(70, 1) >>> (a @ a.T).shape (70, 70) Above answer is incorrect. dot does the same things if the array is 2D. See the docs here.
np.dotcomputes the dot product of two arrays. Specifically,
- 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.
Simplest way to do what you want is to convert the vector to a matrix first using np.matrix and then using the @. Although, dot can also be used @ is better because conventionally dot is used for vectors and @ for matrices.
>>> a = np.random.rand(70) (70,) >>> a.shape >>> a = np.matrix(a).T >>> a.shape (70, 1) >>> (a @ a.T).shape (70, 70) 
shapeofx.Tis the same as that ofx. This is because x is a one-dimensional array. You want to use 2-d arrays. Try usingnp.reshapeto make x 2 dimensional.dotis the inner product. The equivalent ofnp.sum(y*y), a scalar. To get a (70,70) you want the 'outer' product.