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If you can explain what's the big picture of the project, I may can help better. Sometimes, you can iterate over every even or odd row/col. This is possible if the accuracy is not a very big deal!

For instance, I would re-write your code to be:

vector <double> temp_row;
double temp;
vector <double> oup;
for (int i = 0; i < rows; i++) { //you can say i +=2 and say the output, if you like it it makes the iteration much faster.
    for (int j = 0; j < cols; j++) {
        temp = signal[i][j];
        temp_row.push_back(temp);
    }
    branch_lp_dn(name, temp_row, oup);
    temp_row.clear();

    for (int j = 0; j < (int) oup.size(); j++) {
        lp_dn1[i][j] = oup[j];
    }
}

If you can explain what's the big picture of the project, I may can help better. Sometimes, you can iterate over every even or odd row/col. This is possible if the accuracy is not a very big deal!

For instance, I would re-write your code to be:

vector <double> temp_row;
double temp;
vector <double> oup;
for (int i = 0; i < rows; i++) { //you can say i +=2 and say the output, if you like it it makes the iteration much faster.
    for (int j = 0; j < cols; j++) {
        temp = signal[i][j];
        temp_row.push_back(temp);
    }
    branch_lp_dn(name, temp_row, oup);
    temp_row.clear();

    for (int j = 0; j < (int) oup.size(); j++) {
        lp_dn1[i][j] = oup[j];
    }
}

If you can explain what's the big picture of the project, I may help better. Sometimes, you can iterate over every even or odd row/col. This is possible if the accuracy is not a very big deal!

For instance, I would re-write your code to be:

vector <double> temp_row;
double temp;
vector <double> oup;
for (int i = 0; i < rows; i++) { //you can say i +=2 and say the output, if you like it it makes the iteration much faster.
    for (int j = 0; j < cols; j++) {
        temp = signal[i][j];
        temp_row.push_back(temp);
    }
    branch_lp_dn(name, temp_row, oup);
    temp_row.clear();

    for (int j = 0; j < (int) oup.size(); j++) {
        lp_dn1[i][j] = oup[j];
    }
}