'Extract upper or lower triangular part of a numpy matrix
I have a matrix A and I want 2 matrices U and L such that U contains the upper triangular elements of A (all elements above and not including diagonal) and similarly for L(all elements below and not including diagonal). Is there a numpy method to do this?
e.g
A = array([[ 4., 9., -3.],
[ 2., 4., -2.],
[-2., -3., 7.]])
U = array([[ 0., 9., -3.],
[ 0., 0., -2.],
[ 0., 0., 0.]])
L = array([[ 0., 0., 0.],
[ 2., 0., 0.],
[-2., -3., 0.]])
Solution 1:[1]
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 2:[2]
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]])
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
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| Solution 1 | |
| Solution 2 |
