Sparse Matrix Representations | Set 3 ( CSR )

07 Jul, 2022
If most of the elements in the matrix are zero then the matrix is called a sparse matrix. It is wasteful to store the zero elements in the matrix since they do not affect the results of our computation. This is why we implement these matrices in more efficient representations than the standard 2D Array. Using more efficient representations we can cut down space and time complexities of operations significantly.We have discussed at 4 different representations in following articles : 
Sparse Matrix Representation | Set 1 Sparse Matrix Representation | Set 2 .
Sparse Matrix Representation | Set 1 
Sparse Matrix Representation | Set 2 .
In this article, we will discuss another representation of the Sparse Matrix which is commonly referred as the Yale Format.The CSR (Compressed Sparse Row) or the Yale Format is similar to the Array Representation (discussed in Set 1) of Sparse Matrix. We represent a matrix M (m * n), by three 1-D arrays or vectors called as A, IA, JA. Let NNZ denote the number of non-zero elements in M and note that 0-based indexing is used.  
The A vector is of size NNZ and it stores the values of the non-zero elements of the matrix. The values appear in the order of traversing the matrix row-by-row
The IA vector is of size m+1 stores the cumulative number of non-zero elements upto ( not including) the i-th row. It is defined by the recursive relation : IA[0] = 0IA[i] = IA[i-1] + no of non-zero elements in the (i-1) th row of the Matrix
IA[0] = 0
IA[i] = IA[i-1] + no of non-zero elements in the (i-1) th row of the Matrix
The JA vector stores the column index of each element in the A vector. Thus it is of size NNZ as well.
To find the no of non-zero elements in say row i, we perform IA[i+1] – IA[i]. Notice how this representation is different to the array based implementation where the second vector stores the row indices of non-zero elements.The following examples show how these matrixes are represented.Examples:  
Input : 0  0  0  0
        5  8  0  0
        0  0  3  0
        0  6  0  0

Solution: When the matrix is read row by 
          row, the A vector is [ 5 8 3 6]
          The JA vector stores column indices
          of elements in A hence, JA = [ 0 1 2 
           1]. IA[0] = 0. IA[1] = IA[0] + no  
          of non-zero elements in row 0 
          i.e 0 + 0 = 0.
          Similarly,
          IA[2] = IA[1] + 2 = 2
          IA[3] = IA[2] + 1 = 3  
          IA[4] = IA[3]+1 = 4
          Therefore IA = [0 0 2 3 4]
          The trick is remember that IA[i]
          stores NNZ upto and not-including 
          i row.

Input : 10  20  0  0  0  0
         0  30  0  4  0  0
         0   0 50 60 70  0
         0   0  0  0  0 80

Output :  A = [10 20 30 4 50 60 70 80],
         IA = [0 2 4 7 8]
         JA = [0  1 1 3 2 3 4 5]
Algorithm  
SPARSIFY (MATRIX)
Step 1: Set M to number of rows in MATRIX
Step 2: Set N to number of columns in MATRIX
Step 3: I = 0, NNZ = 0. Declare A, JA, and IA. 
        Set IA[0] to 0
Step 4: for I = 0 ... N-1
Step 5: for J = 0 ... N-1
Step 5: If MATRIX [I][J] is not zero
           Add MATRIX[I][J] to A
           Add J to JA
           NNZ = NNZ + 1
        [End of IF]
Step 6: [ End of J loop ]
        Add NNZ to IA
        [ End of I loop ]
Step 7: Print vectors A, IA, JA
Step 8: END
 
CPP
Java
C#
// CPP program to find sparse matrix rep-// resentation using CSR#include <algorithm>#include <iostream>#include <vector>using namespace std; typedef std::vector<int> vi; typedef vector<vector<int> > matrix; // Utility Function to print a Matrixvoid printMatrix(const matrix& M){    int m = M.size();    int n = M[0].size();    for (int i = 0; i < m; i++) {        for (int j = 0; j < n; j++)            cout << M[i][j] << " ";               cout << endl;    }} // Utility Function to print A, IA, JA vectors// with some decoration.void printVector(const vi& V, char* msg){     cout << msg << "[ ";    for_each(V.begin(), V.end(), [](int a) {        cout << a << " ";    });    cout << "]" << endl;} // Generate the three vectors A, IA, JAvoid sparesify(const matrix& M){    int m = M.size();    int n = M[0].size(), i, j;    vi A;    vi IA = { 0 }; // IA matrix has N+1 rows    vi JA;    int NNZ = 0;     for (i = 0; i < m; i++) {        for (j = 0; j < n; j++) {            if (M[i][j] != 0) {                A.push_back(M[i][j]);                JA.push_back(j);                 // Count Number of Non Zero                // Elements in row i                NNZ++;            }        }        IA.push_back(NNZ);    }     printMatrix(M);    printVector(A, (char*)"A = ");    printVector(IA, (char*)"IA = ");    printVector(JA, (char*)"JA = ");} // Driver codeint main(){    matrix M = {        { 0, 0, 0, 0, 1 },        { 5, 8, 0, 0, 0 },        { 0, 0, 3, 0, 0 },        { 0, 6, 0, 0, 1 },    };     sparesify(M);     return 0;}
import java.util.*; // Java program to find sparse matrix// resentation using CSR public class GFG {    // Utility Function to print a Matrix    private static void printMatrix(int[][] M)    {        int m = M.length;        int n = (M.length == 0 ? 0 : M[0].length);        for (int i = 0; i < m; i++) {            for (int j = 0; j < n; j++) {                System.out.print(M[i][j] + " ");            }            System.out.println();        }    }     // Utility Function to print A, IA, JA vectors    // with some decoration.    private static void printVector(ArrayList<Integer> V,                                    String msg)    {         System.out.print(msg + "[ ");        for (var a : V) {            System.out.print(a + " ");        }        System.out.println("]");    }     // Generate the three vectors A, IA, JA    private static void sparesify(int[][] M)    {        int m = M.length;        int n = (M.length == 0 ? 0 : M[0].length), i, j;        ArrayList<Integer> A = new ArrayList<Integer>();        ArrayList<Integer> IA            = new ArrayList<Integer>(); // IA matrix has N+1                                        // rows        ArrayList<Integer> JA = new ArrayList<Integer>();        int NNZ = 0;         for (i = 0; i < m; i++) {            for (j = 0; j < n; j++) {                if (M[i][j] != 0) {                    A.add(M[i][j]);                    JA.add(j);                     // Count Number of Non Zero                    // Elements in row i                    NNZ++;                }            }            IA.add(NNZ);        }         printMatrix(M);        printVector(A, "A = ");        printVector(IA, "IA = ");        printVector(JA, "JA = ");    }     // Driver code    public static void main(String[] args)    {        int[][] M = { { 0, 0, 0, 0, 1 },                      { 5, 8, 0, 0, 0 },                      { 0, 0, 3, 0, 0 },                      { 0, 6, 0, 0, 1 } };         // Function call        sparesify(M);    }} // This code is contributed by Aarti_Rathi
// C# program to find sparse matrix// resentation using CSR using System;using System.Collections.Generic; class GFG {    // Utility Function to print a Matrix    static void printMatrix(int[, ] M)    {        int m = M.GetLength(0);        int n = M.GetLength(1);        for (int i = 0; i < m; i++) {            for (int j = 0; j < n; j++)                Console.Write(M[i, j] + " ");            Console.WriteLine();        }    }     // Utility Function to print A, IA, JA vectors    // with some decoration.    static void printVector(List<int> V, string msg)    {         Console.Write(msg + "[ ");        foreach(var a in V) { Console.Write(a + " "); }        Console.WriteLine("]");    }     // Generate the three vectors A, IA, JA    static void sparesify(int[, ] M)    {        int m = M.GetLength(0);        int n = M.GetLength(1), i, j;        List<int> A = new List<int>();        List<int> IA            = new List<int>(); // IA matrix has N+1 rows        List<int> JA = new List<int>();        int NNZ = 0;         for (i = 0; i < m; i++) {            for (j = 0; j < n; j++) {                if (M[i, j] != 0) {                    A.Add(M[i, j]);                    JA.Add(j);                     // Count Number of Non Zero                    // Elements in row i                    NNZ++;                }            }            IA.Add(NNZ);        }         printMatrix(M);        printVector(A, "A = ");        printVector(IA, "IA = ");        printVector(JA, "JA = ");    }     // Driver code    public static void Main()    {        int[, ] M = {            { 0, 0, 0, 0, 1 },            { 5, 8, 0, 0, 0 },            { 0, 0, 3, 0, 0 },            { 0, 6, 0, 0, 1 },        };         // Function call        sparesify(M);    }} // This code is contributed by Aarti_Rathi
Output:  
0 0 0 0 1 
5 8 0 0 0 
0 0 3 0 0 
0 6 0 0 1 
A = [ 1 5 8 3 6 1 ]
IA = [ 0 1 3 4 6 ]
JA = [ 4 0 1 2 1 4 ]
Time Complexity : O(n x m)
Auxiliary Space: O(n + m)Notes  
The sparsity of the matrix = ( Total No of Elements – Number of Non Zero Elements) / ( Total No of Elements) or (1 – NNZ/mn ) or ( 1 – size(A)/mn ) .
The direct array based representation required memory 3 * NNZ while CSR requires ( 2*NNZ + m + 1) memory.
CSR matrices are memory efficient as long as .
Similar to CSR there exits CSC which stands for Compressed Sparse Columns. It is the column analogue for CSR.
The ‘New’ Yale format further compresses the A and JA vectors into 1 vector.
References https://en.wikipedia.org/wiki/Sparse_matrix
 This article is contributed by Aarti_Rathi. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
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