Created
January 29, 2015 06:53
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Bilinear Interpolation
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for (int i = 0; i < output.rows; i++) | |
{ | |
uchar *out_data = output.ptr<uchar>(i); | |
for (int j = 0; j < output.cols; j++) | |
{ | |
double ox; | |
double oy; | |
//Calculate the old point(ox, oy) according to j, i | |
//Thus, find a function f:(j, i)->(ox, oy) to get the result like (ox, oy)=f(j, i) | |
int x1 = ox < 0 ? 0 : ox; | |
int y1 = oy < 0 ? 0 : oy; | |
int x2 = x1 + 1 >= WIDTH ? x1 : x1 + 1; | |
int y2 = y1 + 1 >= HEIGHT ? y1 : y1 + 1; | |
double u = ox - x1; | |
double v = oy - y1; | |
const uchar *src_data1 = src.ptr<uchar>(y1); | |
const uchar *src_data2 = src.ptr<uchar>(y2); | |
out_data[3 * j] = | |
(1 - u) * (1 - v) * src_data1[3 * x1] + | |
u * (1 - v) * src_data1[3 * x2] + | |
(1 - u) * v * src_data2[3 * x1] + | |
u * v * src_data2[3 * x2]; | |
out_data[3 * j + 1] = | |
(1 - u) * (1 - v) * src_data1[3 * x1 + 1] + | |
u * (1 - v) * src_data1[3 * x2 + 1] + | |
(1 - u) * v * src_data2[3 * x1 + 1] + | |
u * v * src_data2[3 * x2 + 1]; | |
out_data[3 * j + 2] = | |
(1 - u) * (1 - v) * src_data1[3 * x1 + 2] + | |
u * (1 - v) * src_data1[3 * x2 + 2] + | |
(1 - u) * v * src_data2[3 * x1 + 2] + | |
u * v * src_data2[3 * x2 + 2]; | |
} | |
} |
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