本文整理汇总了C++中Mat::data方法的典型用法代码示例。如果您正苦于以下问题：C++ Mat::data方法的具体用法？C++ Mat::data怎么用？C++ Mat::data使用的例子？那么, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类Mat的用法示例。
在下文中一共展示了Mat::data方法的4个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的C++代码示例。
示例1: lu
static void lu(const Mat<double> &X, Mat<double> &L, Mat<double> &U,
Vec<int> &p)
{
assert(X.ys() == X.xs());
int u, k, i, j, n = X.ys();
double Umax;
// temporary matrix
U = X;
p.setSize(n);
L.setSize(n, n);
for(k = 0; k < n - 1; k++) {
// determine u. Alt. u=max_index(abs(U(k,n-1,k,k)));
u = k;
Umax = fabs(U(k, k));
for(i = k + 1; i < n; i++) {
if(fabs(U(k, i)) > Umax) {
Umax=fabs(U(k, i));
u=i;
}
}
U.swapRows(k, u);
p(k) = u;
if(U(k, k) != 0.) {
//U(k+1,n-1,k,k)/=U(k,k);
for(i = k + 1; i < n; i++)
U(k, i) /= U(k, k);
// Should be: U(k+1,n-1,k+1,n-1)-=U(k+1,n-1,k,k)*U(k,k,k+1,n-1);
// but this is too slow.
// Instead work directly on the matrix data-structure.
double *iPos = U.data() + (k + 1) * U.xs();
double *kPos = U.data() + k * U.xs();
for(i = k + 1; i < n; i++) {
for(j = k + 1; j < n; j++) {
*(iPos + j) -= *(iPos + k) * *(kPos + j);
}
iPos += U.xs();
}
}
}
p(n - 1) = n - 1;
// Set L and reset all lower elements of U.
// set all lower triangular elements to zero
for(i = 0; i < n; i++) {
L(i, i) = 1.;
for(j = i + 1; j < n; j++) {
L(i, j) = U(i, j);
U(i, j) = 0;
L(j, i) = 0;
}
}
}示例2: forwardSubstitution
static void forwardSubstitution(const Mat<double> &L, const Vec<double> &b,
Vec<double> &x)
{
assert(L.ys() == L.xs() && L.xs() == b.size() && b.size() == x.size());
int n = L.ys(), i, j, iPos;
double temp;
x(0) = b(0) / L(0, 0);
for(i = 1; i < n; i++) {
// Should be: x(i)=((b(i)-L(i,i,0,i-1)*x(0,i-1))/L(i,i))(0);
// but this is to slow.
iPos = i * L.xs();
temp = 0;
for(j = 0; j < i; j++) {
temp += L.data()[iPos + j] * x(j);
}
x(i) = (b(i) - temp) / L.data()[iPos + i];
}
}示例3: backwardSubstitution
static void backwardSubstitution(const Mat<double> &U, const Vec<double> &b,
Vec<double> &x)
{
assert(U.ys() == U.xs() && U.xs() == b.size() && b.size() == x.size());
int n = U.ys(), i, j, iPos;
double temp;
x(n - 1) = b(n - 1) / U(n - 1, n - 1);
if(std::isnan(x(n - 1)))
x(n - 1) = 0.;
for(i = n - 2; i >= 0; i--) {
// Should be: x(i)=((b(i)-U(i,i,i+1,n-1)*x(i+1,n-1))/U(i,i))(0);
// but this is too slow.
temp = 0;
iPos = i * U.xs();
for(j = i + 1; j < n; j++) {
temp += U.data()[iPos + j] * x(j);
}
x(i) = (b(i) - temp) / U.data()[iPos + i];
if(std::isnan(x(i)))
x(i) = 0.;
}
}示例4: any_of
IGL_INLINE bool igl::any_of(const Mat & S)
{
return std::any_of(S.data(),S.data()+S.size(),[](bool s) {
return s;
});
}