Extract upper or lower triangular part of a numpy matrix

Solution 1:

Try numpy.triu (triangle-upper) and numpy.tril (triangle-lower).

Solution 2:

To extract the upper triangle values to a flat vector, you can do something like the following:

import numpy as np

a = np.array([[1,2,3],[4,5,6],[7,8,9]])
print(a)

#array([[1, 2, 3],
#       [4, 5, 6],
#       [7, 8, 9]])

a[np.triu_indices(3)]
#or
list(a[np.triu_indices(3)])

#array([1, 2, 3, 5, 6, 9])

Similarly, for the lower triangle, use np.tril.

IMPORTANT

If you want to extract the values that are above the diagonal (or below) then use the k argument. This is usually used when the matrix is symmetric.

import numpy as np

a = np.array([[1,2,3],[4,5,6],[7,8,9]])

#array([[1, 2, 3],
#       [4, 5, 6],
#       [7, 8, 9]])

a[np.triu_indices(3, k = 1)]

# this returns the following
array([2, 3, 6])

EDIT (on 11.11.2019):

To put back the extracted vector into a 2D symmetric array, one can follow my answer here: https://stackoverflow.com/a/58806626/5025009

Solution 3:

Use the Array Creation Routines of numpy.triu and numpy.tril to return a copy of a matrix with the elements above or below the k-th diagonal zeroed.

    >>> a = np.array([[1,2,3],[4,5,6],[7,8,9]])
    >>> a
    array([[1, 2, 3],
           [4, 5, 6],
           [7, 8, 9]])

    >>> tri_upper_diag = np.triu(a, k=0)
    >>> tri_upper_diag
    array([[1, 2, 3],
           [0, 5, 6],
           [0, 0, 9]])

    >>> tri_upper_no_diag = np.triu(a, k=1)
    >>> tri_upper_no_diag
    array([[0, 2, 3],
           [0, 0, 6],
           [0, 0, 0]])

    >>> tri_lower_diag = np.tril(a, k=0)
    >>> tri_lower_diag
    array([[1, 0, 0],
           [4, 5, 0],
           [7, 8, 9]])

    >>> tri_lower_no_diag = np.tril(a, k=-1)
    >>> tri_lower_no_diag
    array([[0, 0, 0],
           [4, 0, 0],
           [7, 8, 0]])