how to convert coo format to csr ?

[liunsteingmail_com]

Hi all:

It's driving me crazy even after reading the examples and docs

here's the code:



#include

#include "mkl_spblas.h"

#include "mkl_types.h"

int main()

{

//*******************************************************************************

// Definition arrays for sparse matrix formats

//*******************************************************************************

#define M 4

#define N 4

#define LDA 4

#define NZMAX 8

#define NNZ 8

#define MBLK 2

#define NN 2

#define INFO 0

#define MN 16

#define IBASE1 1

#define IBASE2 1

#define LOCAT 2

#define IDIAG 3

#define NDIAG 4

#define INDIA 12

MKL_INT n=N, nzmax=NZMAX, nnz = NNZ, info=INFO;

MKL_INT ibase1 = IBASE1,ibase2 = IBASE2, locat = LOCAT;

double Acsr[NZMAX];

double Acoo[NZMAX]= {5.0, 8.0, 9.0, 2.0, 3.0, 6.0, 1.0, 4.0};

MKL_INT AI[M+1];

MKL_INT AJ[NZMAX] = {1, 2, 1, 2, 3, 4, 3, 4};

MKL_INT ir[NZMAX];

MKL_INT jc[NZMAX];

MKL_INT job[8];

locat=2;

ibase1=1;

ibase2=1;

job[1]=ibase1;

job[2]=ibase2;

job[3]=locat;

job[4]=nzmax;

job[0]=1;

job[5]=1;

mkl_dcsrcoo (job,&n, Acsr, AJ,AI,&nnz,Acoo, ir,jc,&info);

for(int i = 0; i< M+1;i++ )

{

printf("AI %d = %d\n",i,AI);

}

for(int i = 0; i< 8;i++ )

{

printf("Job %d = %d\n",i,job);

}

}

[Gennady_F_Intel]

it's not clear - what is the problem?

[liunsteingmail_com]

Quoting - Gennady Fedorov (Intel)

it's not clear - what is the problem?

sorry about that
let me repeat

I have a matrix A in coo one-base format. and i would like to convert it to csr one-based "3 array-variation" by using mkl_dcsrcoo

here is the code, could you please point out where the problem is ? since i dont get any result





int nb_values_A = 13;
int nb_lines_A = 5;
int nb_columns_A = 5;

double *values_coo_A;
values_coo_A = new double[nb_values_A];

MKL_INT *columns_coo_A;
columns_coo_A = new MKL_INT[nb_values_A];

MKL_INT *rows_coo_A;
rows_coo_A = new MKL_INT[nb_values_A];

values_coo_A[0] = 1.0;
values_coo_A[1] = -1.0;
values_coo_A[2] = -3.0;
values_coo_A[3] = -2.0;
values_coo_A[4] = 5.0;
values_coo_A[5] = 4.0;
values_coo_A[6] = 6.0;
values_coo_A[7] = 4.0;
values_coo_A[8] = -4.0;
values_coo_A[9] = 2.0;
values_coo_A[10] = 7.0;
values_coo_A[11] = 8.0;
values_coo_A[12] = -5.0;

columns_coo_A[0] = 1;
columns_coo_A[1] = 2;
columns_coo_A[2] = 4;
columns_coo_A[3] = 1;
columns_coo_A[4] = 2;
columns_coo_A[5] = 3;
columns_coo_A[6] = 4;
columns_coo_A[7] = 5;
columns_coo_A[8] = 1;
columns_coo_A[9] = 3;
columns_coo_A[10] = 4;
columns_coo_A[11] = 2;
columns_coo_A[12] = 5;

rows_coo_A[0] = 1;
rows_coo_A[1] = 1;
rows_coo_A[2] = 1;
rows_coo_A[3] = 2;
rows_coo_A[4] = 2;
rows_coo_A[5] = 3;
rows_coo_A[6] = 3;
rows_coo_A[7] = 3;
rows_coo_A[8] = 4;
rows_coo_A[9] = 4;
rows_coo_A[10] = 4;
rows_coo_A[11] = 5;
rows_coo_A[12] = 5;

MKL_INT job[8];


job[0] = 1; // the matrix in the coordinate format is converted to the CSR format.
job[1] = 1; // one-based indexing for the matrix in CSR format is used.
job[2] = 1; // one-based indexing for the matrix in coordinate format is used.
job[5] = 1; // only array ia is filled in for the output storage.


double *values_csr_A;
MKL_INT *columns_csr_A;

MKL_INT *rowIndex_csr_A;
rowIndex_csr_A = new MKL_INT[nb_lines_A+1];

MKL_INT nnz_coo;
MKL_INT info_coo;


// Convert COO format to CSR format
mkl_dcsrcoo (job,&nb_columns_A, values_csr_A,columns_csr_A,rowIndex_csr_A,&nnz_coo,values_coo_A,rows_coo_A,columns_coo_A,&info_coo);

int nb_values_csr_A = rowIndex_csr_A[nb_lines_A] -1;

values_csr_A = new double[nb_values_csr_A];
columns_csr_A = new MKL_INT[nb_values_csr_A];

job[5]=2;// only array ia is filled in for the output storage.

mkl_dcsrcoo (job,&nb_columns_A, values_csr_A,columns_csr_A,rowIndex_csr_A,&nnz_coo,values_coo_A,rows_coo_A,columns_coo_A,&info_coo);

[liunsteingmail_com]

especially where in the documents can we find about the size of job? in the example i find it is of size 8
and where in the documents ca we find about job(4), job(7), job(8)?
and also dimention of the matrix A "m" . what is it ?

[liunsteingmail_com]

Quoting - liunsteingmail.com

especially where in the documents can we find about the size of job? in the example i find it is of size 8
and where in the documents ca we find about job(4), job(7), job(8)?
and also dimention of the matrix A "m" . what is it ?

i managed to make it work now
sorry about the fuss i made

#include
#include "mkl_spblas.h"
#include "mkl_types.h"


#define NZMAX 8
#define INFO 0
#define NNZ 8
#define N 4
#define M 4
double Acoo[NZMAX] = {5.0, 8.0, 9.0, 2.0, 3.0, 6.0, 1.0, 4.0};
MKL_INT AJ[NZMAX] ;
MKL_INT AI[M+1];
MKL_INT ir[NZMAX] = {1, 1, 2, 2, 3, 3, 4, 4};;
MKL_INT jc[NZMAX] = {1, 2, 1, 2, 3, 4, 3, 4};
//double Acsr[NZMAX] = {5.0, 8.0, 9.0, 2.0, 3.0, 6.0, 1.0, 4.0};
double Acsr[NZMAX] ;
MKL_INT n=N, info=INFO, nnz = NNZ;

int main()
{
MKL_INT job[8];

job[1] = 1;
job[2] = 1;
//job[3] = 1; // why ?
job[4] = NZMAX;


job[0]=1;
job[5]=1;
mkl_dcsrcoo (job,&n, Acsr, AJ,AI,&nnz,Acoo, ir,jc,&info);
for(int i=0; i< 8; i++)
{
printf("AI%d = %dn",i,AI);
}
}