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pythonnumpymatplotlibscipysparse-matrix
Jing
Jing

Reputation: 945

python matplotlib plot sparse matrix pattern

Given a sparse binary matrix A (csr, coo, whatever) I want to make a plot such that I can see the position (i,j) = white in the figure if A(i,j) = 1, and (i,j) = black if A(i,j) = 0;

For a dense numpy array, matshow will do the job. However, the dimension of my sparse matrix (say 100000 x 1000000) is to big to be converted to a dense array. I wonder how could I plot the pattern in my sparse matrix.

Thanks

Upvotes: 10

Views: 22847

Answers (3)

Adam
Adam

Reputation: 17339

See matspy.

pip install matspy

Large matrices are no problem, a spy plot of tens of millions of nonzeros takes less than half a second. A small example:

from matspy import spy
import scipy

n = 9000
A = scipy.sparse.random(n, n, density=0.001) + scipy.sparse.eye(n)

spy(A)

matspy output

If you want to modify the plot further, use fig, ax = matspy.spy_to_mpl(A) instead.

Bigger example:

spy() took 0.42s to plot on my laptop:

mat = scipy.sparse.eye(10_000_000).tocsr()
spy(mat)

matspy eye spy

Upvotes: 1

Saullo G. P. Castro
Saullo G. P. Castro

Reputation: 58915

You can get a nice result using a coo_matrix, plot() and some adjustments:

import matplotlib.pyplot as plt
from scipy.sparse import coo_matrix

def plot_coo_matrix(m):
    if not isinstance(m, coo_matrix):
        m = coo_matrix(m)
    fig = plt.figure()
    ax = fig.add_subplot(111, facecolor='black')
    ax.plot(m.col, m.row, 's', color='white', ms=1)
    ax.set_xlim(0, m.shape[1])
    ax.set_ylim(0, m.shape[0])
    ax.set_aspect('equal')
    for spine in ax.spines.values():
        spine.set_visible(False)
    ax.invert_yaxis()
    ax.set_aspect('equal')
    ax.set_xticks([])
    ax.set_yticks([])
    return ax

Note that the y axis is inverted to put the first row at the top of the figure. One example:

import numpy as np
from scipy.sparse import coo_matrix

shape = (100000, 100000)
rows = np.int_(np.round_(shape[0]*np.random.random(1000)))
cols = np.int_(np.round_(shape[1]*np.random.random(1000)))
vals = np.ones_like(rows)

m = coo_matrix((vals, (rows, cols)), shape=shape)
ax = plot_coo_matrix(m)
ax.figure.show()

enter image description here

Upvotes: 20

Santi Peñate-Vera
Santi Peñate-Vera

Reputation: 1186

There is a function (spy) in Matplotlib for that:

scipy equivalent for MATLAB spy

http://matplotlib.org/examples/pylab_examples/spy_demos.html

Upvotes: 12

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