How can I calculate the nearest positive semi-definite matrix?

Question

I\'m coming to Python from R and trying to reproduce a number of things that I\'m used to doing in R using Python. The Matrix library for R has a very nifty function called

Answer 1

I would submit a non-iterative approach. This is slightly modified from Rebonato and Jackel (1999) (page 7-9). Iterative approaches can take a long time to process on matrices of more than a few hundred variables.

import numpy as np

def nearPSD(A,epsilon=0):
   n = A.shape[0]
   eigval, eigvec = np.linalg.eig(A)
   val = np.matrix(np.maximum(eigval,epsilon))
   vec = np.matrix(eigvec)
   T = 1/(np.multiply(vec,vec) * val.T)
   T = np.matrix(np.sqrt(np.diag(np.array(T).reshape((n)) )))
   B = T * vec * np.diag(np.array(np.sqrt(val)).reshape((n)))
   out = B*B.T
   return(out)

Code is modified from a discussion of this topic here around nonPD/PSD matrices in R.

Answer 2

This is perhaps a silly extension to DomPazz answer to consider both correlation and covariance matrices. It also has an early termination if you are dealing with a large number of matrices.

def near_psd(x, epsilon=0):
    '''
    Calculates the nearest postive semi-definite matrix for a correlation/covariance matrix

    Parameters
    ----------
    x : array_like
      Covariance/correlation matrix
    epsilon : float
      Eigenvalue limit (usually set to zero to ensure positive definiteness)

    Returns
    -------
    near_cov : array_like
      closest positive definite covariance/correlation matrix

    Notes
    -----
    Document source
    http://www.quarchome.org/correlationmatrix.pdf

    '''

    if min(np.linalg.eigvals(x)) > epsilon:
        return x

    # Removing scaling factor of covariance matrix
    n = x.shape[0]
    var_list = np.array([np.sqrt(x[i,i]) for i in xrange(n)])
    y = np.array([[x[i, j]/(var_list[i]*var_list[j]) for i in xrange(n)] for j in xrange(n)])

    # getting the nearest correlation matrix
    eigval, eigvec = np.linalg.eig(y)
    val = np.matrix(np.maximum(eigval, epsilon))
    vec = np.matrix(eigvec)
    T = 1/(np.multiply(vec, vec) * val.T)
    T = np.matrix(np.sqrt(np.diag(np.array(T).reshape((n)) )))
    B = T * vec * np.diag(np.array(np.sqrt(val)).reshape((n)))
    near_corr = B*B.T    

    # returning the scaling factors
    near_cov = np.array([[near_corr[i, j]*(var_list[i]*var_list[j]) for i in xrange(n)] for j in xrange(n)])
    return near_cov

Answer 3

I know this thread is old, but the solutions provided here were not satisfactory for my covariance matrices: the transformed matrices always looked quite different from the original ones (for the cases I tested at least). So, I'm leaving here a very straightforward answer, based on the solution provided in this answer:

import numpy as np

def get_near_psd(A):
    C = (A + A.T)/2
    eigval, eigvec = np.linalg.eig(C)
    eigval[eigval < 0] = 0

    return eigvec.dot(np.diag(eigval)).dot(eigvec.T)

The idea is simple: I compute the symmetric matrix, then do an eigen decomposition to get the eigenvalues and eigenvectors. I zero out all negative eigenvalues and construct back the matrix, which will now be positive semi-definite.

For the sake of completness, I leave a simple code to check whether a matrix is positive semi-definite using numpy (basically checking whether all eigenvalues are non-negative):

def is_pos_semidef(x):
    return np.all(np.linalg.eigvals(x) >= 0)

Answer 4

I don't think there is a library which returns the matrix you want, but here is a "just for fun" coding of neareast positive semi-definite matrix algorithm from Higham (2000)

import numpy as np,numpy.linalg

def _getAplus(A):
    eigval, eigvec = np.linalg.eig(A)
    Q = np.matrix(eigvec)
    xdiag = np.matrix(np.diag(np.maximum(eigval, 0)))
    return Q*xdiag*Q.T

def _getPs(A, W=None):
    W05 = np.matrix(W**.5)
    return  W05.I * _getAplus(W05 * A * W05) * W05.I

def _getPu(A, W=None):
    Aret = np.array(A.copy())
    Aret[W > 0] = np.array(W)[W > 0]
    return np.matrix(Aret)

def nearPD(A, nit=10):
    n = A.shape[0]
    W = np.identity(n) 
# W is the matrix used for the norm (assumed to be Identity matrix here)
# the algorithm should work for any diagonal W
    deltaS = 0
    Yk = A.copy()
    for k in range(nit):
        Rk = Yk - deltaS
        Xk = _getPs(Rk, W=W)
        deltaS = Xk - Rk
        Yk = _getPu(Xk, W=W)
    return Yk

When tested on the example from the paper, it returns the correct answer

print nearPD(np.matrix([[2,-1,0,0],[-1,2,-1,0],[0,-1,2,-1],[0,0,-1,2]]),nit=10)
[[ 1.         -0.80842467  0.19157533  0.10677227]
 [-0.80842467  1.         -0.65626745  0.19157533]
 [ 0.19157533 -0.65626745  1.         -0.80842467]
 [ 0.10677227  0.19157533 -0.80842467  1.        ]]