Vectorial division
If $\boldsymbol{D}$ is a matrix. One can easily compute
r1=[(D[j+1]-D[j]) for j in range(D.nrows()-1)]
but is there a simple way to compute :
r1=[(D[j+1]-D[j])/D[j] for j in range(D.nrows()-1)]
2 Answers
Numpy arrays admit elementwise operations. So you could convert the matrix to a Numpy array, compute the new matrix R and then come back to a SageMath matrix:
import numpy
R = D.numpy()
R = numpy.diff(R, axis=0)/R[0:-1,:]
R = matrix(R)
You can also opt for a pure Python approach:
nr, nc = D.nrows()-1, D.ncols()
R = matrix(nr, nc, [(D[i+1,j]-D[i,j])/D[i,j]
for i in range(nr) for j in range(nc)])
The first method is faster.
Comments
Both
r1=[vector((D[j+1][i]-D[j][i])/D[j][i] for i in range(D[j].degree())) for j in range(D.nrows()-1)]
and
r1=[vector((D[j+1, i]-D[j, i])/D[j, i] for i in range(D[j].degree())) for j in range(D.nrows()-1)]
work for me.
(Elementwise vector division is not a standard mathematical operation, so it makes sense that just dividing by D[j] is not implemented in SageMath, nor should it be, since the software is aimed at mathematical use.)
Comments
Try it: D = ... some matrix ..., then v = D[0]. Then do v.degree? to see Return the degree of this vector, which is simply the number of entries. You could replace D[j].degree() by D.ncols().
Your Answer
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Asked: 2020-06-23 01:09:31 +0200
Seen: 331 times
Last updated: Jun 23 '20
What does it mean to divide a vector by a vector?
I think he refers to elementwise division.