I am trying to obtain the left inverse of a non-square matrix in python using either numpy or scipy. How can I translate the following Matlab code to Python?
>> A = [0,1; 0,1; 1,0]
A =
0 1
0 1
1 0
>> y = [2;2;1]
y =
2
2
1
>> A\y
ans =
1.0000
2.0000
Is there a numpy or scipy equivalent of the left inverse \
operator in Matlab?
We use numpy. linalg. inv() function to calculate the inverse of a matrix. The inverse of a matrix is such that if it is multiplied by the original matrix, it results in identity matrix.
Python provides a very easy method to calculate the inverse of a matrix. The function numpy. linalg. inv() which is available in the python NumPy module is used to compute the inverse of a matrix.
Compute the (Moore-Penrose) pseudo-inverse of a Hermitian matrix. The pseudo-inverse of a matrix A, denoted , is defined as: “the matrix that 'solves' [the least-squares problem] A x = b ,” i.e., if is said solution, then is that matrix such that x ¯ = A + b .
To Compute the (Moore-Penrose) pseudo-inverse of a matrix, use the numpy. linalg. pinv() method in Python. Calculate the generalized inverse of a matrix using its singular-value decomposition (SVD) and including all large singular values.
Use linalg.lstsq(A,y)
since A
is not square. See here for details. You can use linalg.solve(A,y)
if A
is square, but not in your case.
Here is a method that will work with sparse matrices (which from your comments is what you want) which uses the leastsq function from the optimize package
from numpy import *
from scipy.sparse import csr_matrix
from scipy.optimize import leastsq
from numpy.random import rand
A=csr_matrix([[0.,1.],[0.,1.],[1.,0.]])
b=array([[2.],[2.],[1.]])
def myfunc(x):
x.shape = (2,1)
return (A*x - b)[:,0]
print leastsq(myfunc,rand(2))[0]
generates
[ 1. 2.]
It is kind of ugly because of how I had to get the shapes to match up according to what leastsq wanted. Maybe someone else knows how to make this a little more tidy.
I have also tried to get something to work with the functions in scipy.sparse.linalg by using the LinearOperators, but to no avail. The problem is that all of those functions are made to handle square functions only. If anyone finds a way to do it that way, I would like to know as well.
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