I have a matrix and I want to check if it is sparse or not.
Things I have tried:
isinstance method:
if isinstance[, scipy.sparse.csc.csc_matrix]:
This works fine if I know exactly which sparse class I want to check.
- getformat method: But it assumes that my matrix is sparse and give format
But I want a way to know if matrix is sparse or not, and should work irrespective of which sparse class.
Kindly help me.
asked Mar 8, 2016 at 18:06
scipy.sparse.issparse[my_matrix]
answered Mar 8, 2016 at 18:44
BoaBoa
2,5341 gold badge21 silver badges37 bronze badges
You can do sparsity = 1.0 - count_nonzero[X] / X.size
This works for any matrices.
answered Mar 16, 2020 at 3:50
Not the answer you're looking for? Browse other questions tagged python class matrix sparse-matrix or ask your own question.
View Discussion
Improve Article
Save Article
View Discussion
Improve Article
Save Article
A matrix is a two-dimensional data object having m rows and n columns, therefore a total of m*n values. If most of the values of a matrix are 0 then
we say that the matrix is sparse.
Consider a definition of Sparse where a matrix is considered sparse if the number of 0s is more than half of the elements in the matrix,
Examples:
Input : 1 0 3
0 0 4
6 0 0
Output : Yes
There are 5 zeros. This count
is more than half of matrix
size.
Input : 1 2 3
0 7 8
5 0 7
Output: No To check whether a matrix is a sparse matrix, we only need to check the total number of elements that are equal to zero. If this count is more than [m * n]/2, we return true.
Implementation:
C++
#include
using namespace std;
const int MAX = 100;
bool isSparse[int array[][MAX], int m, int n]
{
int counter = 0;
for [int i = 0; i < m; ++i]
for [int j = 0; j < n; ++j]
if [array[i][j] == 0]
++counter;
return [counter > [[m * n] / 2]];
}
int main[]
{
int array[][MAX] = { { 1, 0, 3 },
{ 0, 0, 4 },
{ 6, 0, 0 } };
int m = 3,
n = 3;
if [isSparse[array, m, n]]
cout
[[m * n] // 2]]
array = [ [ 1, 0, 3 ],
[ 0, 0, 4
],
[ 6, 0, 0 ] ]
m = 3
n = 3
if [isSparse[array, m, n]] :
print["Yes"]
else :
print["No"]
C#
using System;
class GFG {
static bool isSparse[int [,]array, int m,
int n]
{
int counter = 0;
for [int i = 0; i < m; ++i]
for [int j = 0; j < n; ++j]
if [array[i,j] == 0]
++counter;
return [counter > [[m * n] / 2]];
}
public static void Main[]
{
int [,]array = { { 1, 0, 3 },
{ 0, 0, 4 },
{ 6, 0, 0 } };
int m = 3,
n = 3;
if [isSparse[array, m, n]]
Console.WriteLine["Yes"];
else
Console.WriteLine["No"];
}
}
PHP
Javascript
let MAX = 100;
function isSparse[array, m, n]
{
let counter = 0;
for [let i = 0; i < m; ++i]
for [let j = 0; j < n; ++j]
if [array[i][j] == 0]
++counter;
return [counter > parseInt[[m * n] / 2], 10];
}
let array = [ [ 1, 0, 3 ],
[ 0, 0, 4 ],
[ 6, 0, 0 ] ];
let m = 3,
n = 3;
if [isSparse[array, m, n]]
document.write["Yes"];
else
document.write["No"];
Time Complexity: O[m*n]
Auxiliary Space: O[1]
This article is contributed by Aarti_Rathi and Vineet Joshi. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to . See your article appearing on the GeeksforGeeks main page and help other Geeks.