Iterating over symmetric matrix

Question

I have a symmetric matrix, and I want to loop over its elements. Currently, I'm using:

import torch

# sample symmetric matrix
W = torch.tensor([[0., 3., 0., 0., 0., 0., 0., 0., 0.],
        [3., 0., 1., 0., 0., 0., 0., 0., 0.],
        [0., 1., 0., 1., 1., 0., 0., 0., 0.],
        [0., 0., 1., 0., 0., 0., 0., 0., 0.],
        [0., 0., 1., 0., 0., 1., 0., 1., 0.],
        [0., 0., 0., 0., 1., 0., 1., 0., 0.],
        [0., 0., 0., 0., 0., 1., 0., 1., 0.],
        [0., 0., 0., 0., 1., 0., 1., 0., 2.],
        [0., 0., 0., 0., 0., 0., 0., 2., 0.]])

for ix, row in enumerate(W):
    for iy, element in enumerate(row):

        if iy <= ix:
            continue

        print(element)

I was wondering whether is this the most efficient way or if it can be written in a better way.

This matrix doesn't look symmetric. The first row starts with 3., 0., 1. but the first column starts with 3.0, 0., 0. — vnp
– vnp, Commented Jun 7, 2020 at 20:45
@JosefZ Because this symmetric matrix is representing an adjacency matrix for an undirected graph with no self-loops. Having no self-loops means that the node is not connected to itself, so the diagonal is zero anyways and does not need to be processed in the loop in my case. Nothing fancy... — Blade
– Blade, Commented Jun 30, 2020 at 11:11
While I get that you're very likely doing other things to this tensor, in this situation, there appears to be very little use for it vs. just a list of lists. — Chris
– Chris, Commented Oct 14 at 14:22

Toby Speight · Accepted Answer · 2025-10-14 12:51:46Z

2

It's not clear whether you need to iterate over these values just once, or in several different places in the code.

If the latter, then you probably want to encapsulate the logic in a reusable function, like this:

def top_right_triangle(tensor):
    for ix, row in enumerate(tensor):
        for iy, element in enumerate(row):
            if iy > ix:
                yield element

answered Oct 14 at 12:51

Toby Speight

88.7k14 gold badges104 silver badges327 bronze badges

1

\$\begingroup\$ Good news: such a function already exists. \$\endgroup\$

Reinderien
– Reinderien

2025-10-17 01:28:44 +00:00
Commented Oct 17 at 1:28

Add a comment |

toolic · Accepted Answer · 2025-10-14 10:21:03Z

Simpler

Overall, the code is easy to understand as it is written. However, there is no need for the continue statement if you change the if condition.

Layout

There is some inconsistent indentation.
The black program can be used to automatically adjust the indentation.

import torch

# sample symmetric matrix
W = torch.tensor(
    [
        [0.0, 3.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0],
        [3.0, 0.0, 1.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0],
        [0.0, 1.0, 0.0, 1.0, 1.0, 0.0, 0.0, 0.0, 0.0],
        [0.0, 0.0, 1.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0],
        [0.0, 0.0, 1.0, 0.0, 0.0, 1.0, 0.0, 1.0, 0.0],
        [0.0, 0.0, 0.0, 0.0, 1.0, 0.0, 1.0, 0.0, 0.0],
        [0.0, 0.0, 0.0, 0.0, 0.0, 1.0, 0.0, 1.0, 0.0],
        [0.0, 0.0, 0.0, 0.0, 1.0, 0.0, 1.0, 0.0, 2.0],
        [0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 2.0, 0.0],
    ]
)

for ix, row in enumerate(W):
    for iy, element in enumerate(row):
        if iy > ix:
            print(element)

Reinderien · Accepted Answer · 2025-10-17 01:27:34Z

I have a symmetric matrix, and I want to loop over its elements

Well... no. You only want to loop over its lower triangle (equivalent to its upper triangle), and there are more direct ways:

import numpy as np

w = np.array((
    (0, 3, 0, 0, 0, 0, 0, 0, 0),
    (3, 0, 1, 0, 0, 0, 0, 0, 0),
    (0, 1, 0, 1, 1, 0, 0, 0, 0),
    (0, 0, 1, 0, 0, 0, 0, 0, 0),
    (0, 0, 1, 0, 0, 1, 0, 1, 0),
    (0, 0, 0, 0, 1, 0, 1, 0, 0),
    (0, 0, 0, 0, 0, 1, 0, 1, 0),
    (0, 0, 0, 0, 1, 0, 1, 0, 2),
    (0, 0, 0, 0, 0, 0, 0, 2, 0),
))
i, j = np.triu_indices_from(w, k=1)
triu = w[i, j]
print(triu)

[3 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 1 0 1 0 1 0 0 1 0 2]

There is an equivalent method in PyTorch.

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