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How to get XY value from ct.
Ex: ct = 217, I want to get x="0.3127569", y= "0.32908".
I'm able to convert XY value into ct value using this below code.
float R1 = [hue[0] floatValue];
float S1 = [hue[1] floatValue];
float result = ((R1-0.332)/(S1-0.1858));
NSString *ctString = [NSString stringWithFormat:#"%f", ((-449*result*result*result)+(3525*result*result)-(6823.3*result)+(5520.33))];
float micro2 = (float) (1 / [ctString floatValue] * 1000000);
NSString *ctValue = [NSString stringWithFormat:#"%f", micro2];
ctValue = [NSString stringWithFormat:#"%d", [ctValue intValue]];
if ([ctValue integerValue] < 153) {
ctValue = [NSString stringWithFormat:#"%d", 153];
}
Now I want reverse value, which is from ct to XY.
On Phillips HUE
2000K maps to 500 and 6500K maps to 153 given in ct as color temperature but can be thought as actually being Mired.
Mired means micro reciprocal degree wikipedia.
ct is possibly used because it is not 100% Mired. Quite sure Phillips uses a lookup table as a lot CIE algorithms do because there are just 347 indexes in this range from 153 to 500.
The following is not a solution, it's just simple concept of a lookup table.
And as the CIE 1931 xy to CCT Formula by McCamy suggests found here it is possible to use a lookup table to find x and y as well.
A table can be found here but i am not sure if that is the right lookup table.
reminder so the following is not a solution, but to find an reverse algo the code may help.
typedef int Kelvin;
typedef float Mired;
Mired linearMiredByKelvin(Kelvin k) {
if (k==0) return 0;
return 1000000.0/k;
}
-(void)mired {
Mired miredMin = 2000.0/13.0; // 153,84 = reciprocal 6500K
Mired miredMax = 500.0; // 500,00 = reciprocal 2000K
Mired lookupMiredByKelvin[6501]; //max 6500 Kelvin + 1 safe index
//Kelvin lookupKelvinByMired[501]; //max 500 Mired + 1 safe index
// dummy stuff, empty unused table space
for (Kelvin k = 0; k < 2000; k++) {
lookupMiredByKelvin[k] = 0;
}
//for (Mired m = 0.0; m < 154.0; m++) {
// lookupKelvinByMired[(int)m] = 0;
//}
for (Kelvin k=2000; k<6501; k++) {
Mired linearMired = linearMiredByKelvin(k);
float dimm = (linearMired - miredMin) / ( miredMax - miredMin);
Kelvin ct = (Kelvin)(1000000.0/(dimm*miredMax - dimm*miredMin + miredMin));
lookupMiredByKelvin[k] = linearMiredByKelvin(ct);
if (k==2000 || k==2250 || k==2500 || k==2750 ||
k==3000 || k==3250 || k==3500 || k==3750 ||
k==4000 || k==4250 || k==4500 || k==4750 ||
k==5000 || k==5250 || k==5500 || k==5750 ||
k==6000 || k==6250 || k==6500 || k==6501 )
fprintf(stderr,"%d %f %f\n",ct, dimm, lookupMiredByKelvin[k]);
}
}
at least this is proof that x and y will not sit on a simple vector.
CCT means correlated colour temperature and like the implementation in the question shows can be calculated via n= (x-0.3320)/(0.1858-y); CCT = 437*n^3 + 3601*n^2 + 6861*n + 5517. (after McCamy)
but a cct=217 is out of range of above link'ed lookup table.
following the idea in this git-repo from colour-science
and ported to C it could look like..
void CCT_to_xy_CIE_D(float cct) {
//if (CCT < 4000 || CCT > 25000) fprintf(stderr, "Correlated colour temperature must be in domain, unpredictable results may occur! \n");
float x = calculateXviaCCT(cct);
float y = calculateYviaX(x);
NSLog(#"cct=%f x%f y%f",cct,x,y);
}
float calculateXviaCCT(float cct) {
float cct_3 = pow(cct, 3); //(cct*cct*cct);
float cct_2 = pow(cct, 2); //(cct*cct);
if (cct<=7000)
return -4.607 * pow(10, 9) / cct_3 + 2.9678 * pow(10, 6) / cct_2 + 0.09911 * pow(10, 3) / cct + 0.244063;
return -2.0064 * pow(10, 9) / cct_3 + 1.9018 * pow(10, 6) / cct_2 + 0.24748 * pow(10, 3) / cct + 0.23704;
}
float calculateYviaX(float x) {
return -3.000 * pow(x, 2) + 2.870 * x - 0.275;
}
CCT_to_xy_CIE_D(6504.38938305); //proof of concept
//cct=6504.389160 x0.312708 y0.329113
CCT_to_xy_CIE_D(217.0);
//cct=217.000000 x-387.131073 y-450722.750000
// so for sure Phillips hue temperature given in ct between 153-500 is not a good starting point
//but
CCT_to_xy_CIE_D(2000.0);
//cct=2000.000000 x0.459693 y0.410366
this seems to work fine with CCT between 2000 and 25000, but maybe confusing is CCT is given in Kelvin here.
EDIT
This has been through so many revisions and ideas. To keep it simple I edited most of that out and just give you the final result.
This fits your function perfectly except for a region in the middle (temp from 256 to 316) where it deviates a bit.
The problem with your function is that it has approximately infinite solutions, so to solve it nicely you need more constraints, but what? Ol Sen's reference https://www.waveformlighting.com/tech/calculate-color-temperature-cct-from-cie-1931-xy-coordinates discusses it in some detail and then mentions that you want a Duv to be zero. It also gives a way to calculate Duv and so I added that to my optimiser and voila!
Nice and smooth. The optimiser now solves for x and y that both satisfies your function and also minimises Duv.
To get it to work nicely I had to scale Duv quite a bit. That page mentions that Duv should be very small so I think this is a good thing. Also, as the temp increases the scaling should to help the optimiser.
Below prints from 153 to 500.
#import <Foundation/Foundation.h>
// Function taken from your code
// Simplified a bit
int ctFuncI ( float x, float y )
{
// float R1 = [hue[0] floatValue];
// float S1 = [hue[1] floatValue];
float result = (x-0.332)/(y-0.1858);
float cubic = - 449 * result * result * result + 3525 * result * result - 6823.3 * result + 5520.33;
float micro2 = 1 / cubic * 1000000;
int ct = ( int )( micro2 + 0.5 );
if ( ct < 153 )
{
ct = 153;
}
return ct;
}
// Need this
// Float version of your code
float ctFuncF ( float x, float y )
{
// float R1 = [hue[0] floatValue];
// float S1 = [hue[1] floatValue];
float result = (x-0.332)/(y-0.1858);
float cubic = - 449 * result * result * result + 3525 * result * result - 6823.3 * result + 5520.33;
return 1000000 / cubic;
}
// We need an additional constraint
// https://www.waveformlighting.com/tech/calculate-duv-from-cie-1931-xy-coordinates
// Given x, y calculate Duv
// We want this to be 0
float duv ( float x, float y )
{
float f = 1 / ( - 2 * x + 12 * y + 3 );
float u = 4 * x * f;
float v = 6 * y * f;
// I'm typing float but my heart yells double
float k6 = -0.00616793;
float k5 = 0.0893944;
float k4 = -0.5179722;
float k3 = 1.5317403;
float k2 = -2.4243787;
float k1 = 1.925865;
float k0 = -0.471106;
float du = u - 0.292;
float dv = v - 0.24;
float Lfp = sqrt ( du * du + dv * dv );
float a = acos( du / Lfp );
float Lbb = k6 * pow ( a, 6 ) + k5 * pow( a, 5 ) + k4 * pow( a, 4 ) + k3 * pow( a, 3 ) + k2 * pow(a,2) + k1 * a + k0;
return Lfp - Lbb;
}
// Solver!
// Returns iterations
int ctSolve ( int ct, float * x, float * y )
{
int iter = 0;
float dx = 0.001;
float dy = 0.001;
// Error
// Note we scale duv a bit
// Seems the higher the temp, the higher scale we require
// Also note the jump at 255 ...
float s = 1000 * ( ct > 255 ? 10 : 1 );
float d = fabs( ctFuncF ( * x, * y ) - ct ) + s * fabs( duv ( * x, * y ) );
// Approx
while ( d > 0.5 && iter < 250 )
{
iter ++;
dx *= fabs( ctFuncF ( * x + dx, * y ) - ct ) + s * fabs( duv ( * x + dx, * y ) ) < d ? 1.2 : - 0.5;
dy *= fabs( ctFuncF ( * x, * y + dy ) - ct ) + s * fabs( duv ( * x, * y + dy ) ) < d ? 1.2 : - 0.5;
* x += dx;
* y += dy;
d = fabs( ctFuncF ( * x, * y ) - ct ) + s * fabs( duv ( * x, * y ) );
}
return iter;
}
// Tester
int main(int argc, const char * argv[]) {
#autoreleasepool
{
// insert code here...
NSLog(#"Hello, World!");
float x, y;
int sume = 0;
int sumi = 0;
for ( int ct = 153; ct <= 500; ct ++ )
{
// Initial guess
x = 0.4;
y = 0.4;
// Approx
int iter = ctSolve ( ct, & x, & y );
// CT and error
int ctEst = ctFuncI ( x, y );
int e = ct - ctEst;
// Diagnostics
sume += abs ( e );
sumi += iter;
// Print out results
NSLog ( #"want ct = %d x = %f y = %f got ct %d in %d iter error %d", ct, x, y, ctEst, iter, e );
}
NSLog ( #"Sum of abs errors %d iterations %d", sume, sumi );
}
return 0;
}
To use it, do as below.
// To call it, init x and y to some guess
float x = 0.4;
float y = 0.4;
// Then call solver with your temp
int ct = 217;
ctSolve( ct, & x, & y ); // Note you pass references to x and y
// Done, answer now in x and y
a bit more compact answer and functions to convert back and forth..
beware there are rounding issues because McCamy's formula relies and mathematical assumptions. And so the backward calculation does also.
if you want to find more results search directly for "n= (x-0.3320)/(0.1858-y); CCT = 437*n^3 + 3601*n^2 + 6861*n + 5517" there are plenty of different methods to convert back and forth.
so here Phillips-Hue #[#x,#y] to Phillips-ct,Phillips-ct to CCT, CCT to x,y
void CCT_to_xy_CIE_D(float cct) {
//if (CCT < 4000 || CCT > 25000) fprintf(stderr, "Correlated colour temperature must be in domain, unpredictable results may occur! \n");
float x = calculateXviaCCT(cct);
float y = calculateYviaX(x);
fprintf(stderr,"cct=%f x%f y%f",cct,x,y);
}
float calculateXviaCCT(float cct) {
float cct_3 = pow(cct, 3); //(cct*cct*cct);
float cct_2 = pow(cct, 2); //(cct*cct);
if (cct<=7000.0)
return -4.607 * pow(10, 9) / cct_3 + 2.9678 * pow(10, 6) / cct_2 + 0.09911 * pow(10, 3) / cct + 0.244063;
return -2.0064 * pow(10, 9) / cct_3 + 1.9018 * pow(10, 6) / cct_2 + 0.24748 * pow(10, 3) / cct + 0.23704;
}
float calculateYviaX(float x) {
return -3.000 * x*x + 2.870 * x - 0.275;
}
int calculate_PhillipsHueCT_withCCT(float cct) {
if (cct>6500.0) return 2000.0/13.0;
if (cct<2000.0) return 500.0;
//return (float) (1 / cct * 1000000); // same as..
return 1000000 / cct;
}
float calculate_CCT_withPhillipsHueCT(float ct) {
if (ct == 0.0) return 0.0;
return 1000000 / ct;
}
float calculate_CCT_withHueXY(NSArray *hue) {
float x = [hue[0] floatValue]; //R1
float y = [hue[1] floatValue]; //S1
//x = 0.312708; y = 0.329113;
float n = (x-0.3320)/(0.1858-y);
float cct = 437.0*n*n*n + 3601.0*n*n + 6861.0*n + 5517.0;
return cct;
}
// MC Camy formula n=(x-0.3320)/(0.1858-y); cct = 437*n^3 + 3601*n^2 + 6861*n + 5517;
-(void)testPhillipsHueCt_backAndForth {
NSArray *hue = #[#(0.312708),#(0.329113)];
float cct = calculate_CCT_withHueXY(hue);
float ct = calculate_PhillipsHueCT_withCCT(cct);
NSLog(#"ct %f",ct);
CCT_to_xy_CIE_D(cct); // check
CCT_to_xy_CIE_D(6504.38938305); //proof of concept
CCT_to_xy_CIE_D(2000.0);
CCT_to_xy_CIE_D(calculate_CCT_withPhillipsHueCT(217.0));
}
The XYZ color space encompasses all possible colors, not just those which can be generated by a particular device like a monitor. Not all XYZ triplets represent a color that is physically possible. Is there a way, given an XYZ triplet, to determine if it represents a real color?
I wanted to generate a CIE 1931 chromaticity diagram (seen bellow) for myself, but wasn't sure how to go about it. It's easy to, for example, take all combinations of sRGB triplets and then transform them into the xy coordinates of the chromaticity diagram and then plot them. You cannot use this same approach in the XYZ color space though since not all combinations are valid colors. So far the best I have come up with is a stochastic approach, where I generate a random spectral distribution by summing a random number of random Gaussians, then converting it to XYZ using the standard observer functions.
Having thought about it a little more I felt the obvious solution is to generate a list of xy points around the edge of spectral locus, corresponding to pure monochromatic colors. It seems to me that this can be done by directly inputting the visible frequencies (~380-780nm) into the CIE XYZ standard observer color matching functions. Treating these points like a convex polygon you could determine if a point is within the spectral locus using one algorithm or another. In my case, since what I really wanted to do is simply generate the chromaticity diagram, I simply input these points into a graphics library's polygon drawing routine and then for each pixel of the polygon I can transform it into sRGB.
I believe this solution is similar to the one used by the library that Kel linked in a comment. I'm not entirely sure, as I am not familiar with Python.
function RGBfromXYZ(X, Y, Z) {
const R = 3.2404542 * X - 1.5371385 * Y - 0.4985314 * Z
const G = -0.969266 * X + 1.8760108 * Y + 0.0415560 * Z
const B = 0.0556434 * X - 0.2040259 * Y + 1.0572252 * Z
return [R, G, B]
}
function XYZfromYxy(Y, x, y) {
const X = Y / y * x
const Z = Y / y * (1 - x - y)
return [X, Y, Z]
}
function srgb_from_linear(x) {
if (x <= 0.0031308) {
return x * 12.92
} else {
return 1.055 * Math.pow(x, 1/2.4) - 0.055
}
}
// Analytic Approximations to the CIE XYZ Color Matching Functions
// from Sloan http://jcgt.org/published/0002/02/01/paper.pdf
function xFit_1931(x) {
const t1 = (x - 442) * (x < 442 ? 0.0624 : 0.0374)
const t2 = (x -599.8) * (x < 599.8 ? 0.0264 : 0.0323)
const t3 = (x - 501.1) * (x < 501.1 ? 0.0490 : 0.0382)
return 0.362 * Math.exp(-0.5 * t1 * t1) + 1.056 * Math.exp(-0.5 * t2 * t2) - 0.065 * Math.exp(-0.5 * t3 * t3)
}
function yFit_1931(x) {
const t1 = (x - 568.8) * (x < 568.8 ? 0.0213 : 0.0247)
const t2 = (x - 530.9) * (x < 530.9 ? 0.0613 : 0.0322)
return 0.821 * Math.exp(-0.5 * t1 * t1) + 0.286 * Math.exp(-0.5 * t2 * t2)
}
function zFit_1931(x) {
const t1 = (x - 437) * (x < 437 ? 0.0845 : 0.0278)
const t2 = (x - 459) * (x < 459 ? 0.0385 : 0.0725)
return 1.217 * Math.exp(-0.5 * t1 * t1) + 0.681 * Math.exp(-0.5 * t2 * t2)
}
const canvas = document.createElement("canvas")
document.body.append(canvas)
canvas.width = canvas.height = 512
const ctx = canvas.getContext("2d")
const locus_points = []
for (let i = 440; i < 650; ++i) {
const [X, Y, Z] = [xFit_1931(i), yFit_1931(i), zFit_1931(i)]
const x = (X / (X + Y + Z)) * canvas.width
const y = (Y / (X + Y + Z)) * canvas.height
locus_points.push([x, y])
}
ctx.beginPath()
ctx.moveTo(...locus_points[0])
locus_points.slice(1).forEach(point => ctx.lineTo(...point))
ctx.closePath()
ctx.fill()
const imageData = ctx.getImageData(0, 0, canvas.width, canvas.height)
for (let y = 0; y < canvas.height; ++y) {
for (let x = 0; x < canvas.width; ++x) {
const alpha = imageData.data[(y * canvas.width + x) * 4 + 3]
if (alpha > 0) {
const [X, Y, Z] = XYZfromYxy(1, x / canvas.width, y / canvas.height)
const [R, G, B] = RGBfromXYZ(X, Y, Z)
const r = Math.round(srgb_from_linear(R / Math.sqrt(R**2 + G**2 + B**2)) * 255)
const g = Math.round(srgb_from_linear(G / Math.sqrt(R**2 + G**2 + B**2)) * 255)
const b = Math.round(srgb_from_linear(B / Math.sqrt(R**2 + G**2 + B**2)) * 255)
imageData.data[(y * canvas.width + x) * 4 + 0] = r
imageData.data[(y * canvas.width + x) * 4 + 1] = g
imageData.data[(y * canvas.width + x) * 4 + 2] = b
}
}
}
ctx.putImageData(imageData, 0, 0)
I want to extract the red ball from one picture and get the detected ellipse matrix in picture.
Here is my example:
I threshold the picture, find the contour of red ball by using findContour() function and use fitEllipse() to fit an ellipse.
But what I want is to get coefficient of this ellipse. Because the fitEllipse() return a rotation rectangle (RotatedRect), so I need to re-write this function.
One Ellipse can be expressed as Ax^2 + By^2 + Cxy + Dx + Ey + F = 0; So I want to get u=(A,B,C,D,E,F) or u=(A,B,C,D,E) if F is 1 (to construct an ellipse matrix).
I read the source code of fitEllipse(), there are totally three SVD process, I think I can get the above coefficients from the results of those three SVD process. But I am quite confused what does each result (variable cv::Mat x) of each SVD process represent and why there are three SVD here?
Here is this function:
cv::RotatedRect cv::fitEllipse( InputArray _points )
{
Mat points = _points.getMat();
int i, n = points.checkVector(2);
int depth = points.depth();
CV_Assert( n >= 0 && (depth == CV_32F || depth == CV_32S));
RotatedRect box;
if( n < 5 )
CV_Error( CV_StsBadSize, "There should be at least 5 points to fit the ellipse" );
// New fitellipse algorithm, contributed by Dr. Daniel Weiss
Point2f c(0,0);
double gfp[5], rp[5], t;
const double min_eps = 1e-8;
bool is_float = depth == CV_32F;
const Point* ptsi = points.ptr<Point>();
const Point2f* ptsf = points.ptr<Point2f>();
AutoBuffer<double> _Ad(n*5), _bd(n);
double *Ad = _Ad, *bd = _bd;
// first fit for parameters A - E
Mat A( n, 5, CV_64F, Ad );
Mat b( n, 1, CV_64F, bd );
Mat x( 5, 1, CV_64F, gfp );
for( i = 0; i < n; i++ )
{
Point2f p = is_float ? ptsf[i] : Point2f((float)ptsi[i].x, (float)ptsi[i].y);
c += p;
}
c.x /= n;
c.y /= n;
for( i = 0; i < n; i++ )
{
Point2f p = is_float ? ptsf[i] : Point2f((float)ptsi[i].x, (float)ptsi[i].y);
p -= c;
bd[i] = 10000.0; // 1.0?
Ad[i*5] = -(double)p.x * p.x; // A - C signs inverted as proposed by APP
Ad[i*5 + 1] = -(double)p.y * p.y;
Ad[i*5 + 2] = -(double)p.x * p.y;
Ad[i*5 + 3] = p.x;
Ad[i*5 + 4] = p.y;
}
solve(A, b, x, DECOMP_SVD);
// now use general-form parameters A - E to find the ellipse center:
// differentiate general form wrt x/y to get two equations for cx and cy
A = Mat( 2, 2, CV_64F, Ad );
b = Mat( 2, 1, CV_64F, bd );
x = Mat( 2, 1, CV_64F, rp );
Ad[0] = 2 * gfp[0];
Ad[1] = Ad[2] = gfp[2];
Ad[3] = 2 * gfp[1];
bd[0] = gfp[3];
bd[1] = gfp[4];
solve( A, b, x, DECOMP_SVD );
// re-fit for parameters A - C with those center coordinates
A = Mat( n, 3, CV_64F, Ad );
b = Mat( n, 1, CV_64F, bd );
x = Mat( 3, 1, CV_64F, gfp );
for( i = 0; i < n; i++ )
{
Point2f p = is_float ? ptsf[i] : Point2f((float)ptsi[i].x, (float)ptsi[i].y);
p -= c;
bd[i] = 1.0;
Ad[i * 3] = (p.x - rp[0]) * (p.x - rp[0]);
Ad[i * 3 + 1] = (p.y - rp[1]) * (p.y - rp[1]);
Ad[i * 3 + 2] = (p.x - rp[0]) * (p.y - rp[1]);
}
solve(A, b, x, DECOMP_SVD);
// store angle and radii
rp[4] = -0.5 * atan2(gfp[2], gfp[1] - gfp[0]); // convert from APP angle usage
if( fabs(gfp[2]) > min_eps )
t = gfp[2]/sin(-2.0 * rp[4]);
else // ellipse is rotated by an integer multiple of pi/2
t = gfp[1] - gfp[0];
rp[2] = fabs(gfp[0] + gfp[1] - t);
if( rp[2] > min_eps )
rp[2] = std::sqrt(2.0 / rp[2]);
rp[3] = fabs(gfp[0] + gfp[1] + t);
if( rp[3] > min_eps )
rp[3] = std::sqrt(2.0 / rp[3]);
box.center.x = (float)rp[0] + c.x;
box.center.y = (float)rp[1] + c.y;
box.size.width = (float)(rp[2]*2);
box.size.height = (float)(rp[3]*2);
if( box.size.width > box.size.height )
{
float tmp;
CV_SWAP( box.size.width, box.size.height, tmp );
box.angle = (float)(90 + rp[4]*180/CV_PI);
}
if( box.angle < -180 )
box.angle += 360;
if( box.angle > 360 )
box.angle -= 360;
return box;
}
The source code link: https://github.com/Itseez/opencv/blob/master/modules/imgproc/src/shapedescr.cpp
The function fitEllipse returns a RotatedRect that contains all the parameters of the ellipse.
An ellipse is defined by 5 parameters:
xc : x coordinate of the center
yc : y coordinate of the center
a : major semi-axis
b : minor semi-axis
theta : rotation angle
You can obtain these parameters like:
RotatedRect e = fitEllipse(points);
float xc = e.center.x;
float yc = e.center.y;
float a = e.size.width / 2; // width >= height
float b = e.size.height / 2;
float theta = e.angle; // in degrees
You can draw an ellipse with the function ellipse using the RotatedRect:
ellipse(image, e, Scalar(0,255,0));
or, equivalently using the ellipse parameters:
ellipse(res, Point(xc, yc), Size(a, b), theta, 0.0, 360.0, Scalar(0,255,0));
If you need the values of the coefficients of the implicit equation, you can do like (from Wikipedia):
So, you can get the parameters you need from the RotatedRect, and you don't need to change the function fitEllipse.
The solve function is used to solve linear systems or least-squares problems. Using the SVD decomposition method the system can be over-defined and/or the matrix src1 can be singular.
For more details on the algorithm, you can see the paper of Fitzgibbon that proposed this fit ellipse method.
Here is some code that worked for me which I based on the other responses on this thread.
def getConicCoeffFromEllipse(e):
# ellipse(Point(xc, yc),Size(a, b), theta)
xc = e[0][0]
yc = e[0][1]
a = e[1][0]/2
b = e[1][1]/2
theta = math.radians(e[2])
# See https://en.wikipedia.org/wiki/Ellipse
# Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 is the equation
A = a*a*math.pow(math.sin(theta),2) + b*b*math.pow(math.cos(theta),2)
B = 2*(b*b - a*a)*math.sin(theta)*math.cos(theta)
C = a*a*math.pow(math.cos(theta),2) + b*b*math.pow(math.sin(theta),2)
D = -2*A*xc - B*yc
E = -B*xc - 2*C*yc
F = A*xc*xc + B*xc*yc + C*yc*yc - a*a*b*b
coef = np.array([A,B,C,D,E,F]) / F
return coef
def getConicMatrixFromCoeff(c):
C = np.array([[c[0], c[1]/2, c[3]/2], # [ a, b/2, d/2 ]
[c[1]/2, c[2], c[4]/2], # [b/2, c, e/2 ]
[c[3]/2, c[4]/2, c[5]]]) # [d/2], e/2, f ]
return C
Im trying to apply a Sharpen Kernel to a raster picture, Here is my kernel:
{ 0.0f,-1.0f,0.0f,
-1.0f,5.0f,-1.0f,
0.0f,-1.0f,0.0f }
And here is my Code:
struct Pixel{
GLubyte R, G, B;
float x, y;
};
. . .
for (unsigned i = 1; i < iWidth - 1; i++){
for (unsigned j = 1; j < iHeight - 1; j++){
float r = 0, g = 0, b = 0;
r += -(float)pixels[i + 1][j].R;
g += -(float)pixels[i + 1][j].G;
b += -(float)pixels[i + 1][j].B;
r += -(float)pixels[i - 1][j].R;
g += -(float)pixels[i - 1][j].G;
b += -(float)pixels[i - 1][j].B;
r += -(float)pixels[i][j + 1].R;
g += -(float)pixels[i][j + 1].G;
b += -(float)pixels[i][j + 1].B;
r += -(float)pixels[i][j - 1].R;
g += -(float)pixels[i][j - 1].G;
b += -(float)pixels[i][j - 1].B;
pixels[i][j].R = (GLubyte)((pixels[i][j].R * 5) + r);
pixels[i][j].G = (GLubyte)((pixels[i][j].G * 5) + g);
pixels[i][j].B = (GLubyte)((pixels[i][j].B * 5) + b);
}
}
But the colors get mixed up when I apply this kernel, Here is an example:
What am I doing wrong?
NOTE : I know that OpenGL can do this fast and easy, but I just wanted to experiment on this kind of masks.
EDIT : The first code had a bug:
pixels[i][j].R = (GLubyte)((pixels[i][j].R * 5) + r);
pixels[i][j].G = (GLubyte)((pixels[i][j].R/*G*/ * 5) + g);
pixels[i][j].B = (GLubyte)((pixels[i][j].R/*B*/ * 5) + b);
I fixed it but I still got that problem.
Iv changed the last three lines to this:
r = (float)((pixels[i][j].R * 5) + r);
g = (float)((pixels[i][j].G * 5) + g);
b = (float)((pixels[i][j].B * 5) + b);
if (r < 0) r = 0;
if (g < 0) g = 0;
if (b < 0) b = 0;
if (r > 255) r = 255;
if (g > 255) g = 255;
if (b > 255) b = 255;
pixels[i][j].R = r;
pixels[i][j].G = g;
pixels[i][j].B = b;
And now the output looks like this:
You have a copy-paste bug here:
pixels[i][j].R = (GLubyte)((pixels[i][j].R * 5) + r);
pixels[i][j].G = (GLubyte)((pixels[i][j].R * 5) + g);
pixels[i][j].B = (GLubyte)((pixels[i][j].R * 5) + b);
^
This should be:
pixels[i][j].R = (GLubyte)((pixels[i][j].R * 5) + r);
pixels[i][j].G = (GLubyte)((pixels[i][j].G * 5) + g);
pixels[i][j].B = (GLubyte)((pixels[i][j].B * 5) + b);
Also it looks like you may have iWidth/iHeight transposed, but it's hard to say without seeing the rest of the code. Typically though the outer loop iterates over rows, so the upper bound would be the number of rows, i.e. the image height.
Most importantly though you have a fundamental problem in that you're trying to perform a neighbourhood operation in-place. Each output pixel depends on its neighbours, but you're modifying these neighbours as you iterate through the image. You need to do this kind of operation out-of-place, i.e. have a separate output image:
out_pixels[i][j].R = r;
out_pixels[i][j].G = g;
out_pixels[i][j].B = b;
so that the input image does not get modified. (Note also that you'll want to copy the edge pixels over from the input image to the output image.)
( face )
( body )
Hi, i am new to image processing and openCV C/C++. I am wondering that is it possible to extract skin tone from the first image (face). And then applied to the second image (body).
In other words, user upload his face image and the program extract the skin tone from that image and apply it to the body.
Thanks,
Aisha
This is a hard problem to solve, especially given the variation of colours depending on lighting and reflection. I have worked previously on finding skin in images, and generally the Cr (chroma red) component of the YCbCr colour space stands out prominently on skin. You might be able to exploit this information to find skin regions.
Here are a couple of papers that attempt to use colour for locating human skin:
1. Interaction between hands and wearable cameras
2. Markerless inspection of augmented reality objects
For finding skin you can use one of this formulas:
1) With normilized RGB space:
for(int i = 0; i < m_image->height; ++i)
{
for(int j = 0; j < m_image->width; ++j)
{
if (m_image->nChannels == 3)
{
int valueR = (reinterpret_cast<uchar*>(m_image->imageData + i * m_image->widthStep))[j * 3 + 2];
int valueG = (reinterpret_cast<uchar*>(m_image->imageData + i * m_image->widthStep))[j * 3 + 1];
int valueB = (reinterpret_cast<uchar*>(m_image->imageData + i * m_image->widthStep))[j * 3];
float normR = static_cast<float>(valueR) / static_cast<float>(valueR + valueG + valueB);
float normG = static_cast<float>(valueG) / static_cast<float>(valueR + valueG + valueB);
float normB = static_cast<float>(valueB) / static_cast<float>(valueR + valueG + valueB);
if ((normB / normG < 1.249) &&
(( normR + normG + normB ) / ( 3 * normR ) > 0.696 ) &&
( 1/3.0 - normB/( normR + normG + normB ) > 0.014 ) &&
(normG/(3* (normR + normG + normB)) < 0.108 ))
{
//pixel is skin
}
}
}
2) in RGB space:
for(size_t i = 0; i < m_image->height; ++i)
{
for(size_t j = 0; j < m_image->width; ++j)
{
if (m_image->nChannels == 3)
{
int R = (reinterpret_cast<uchar*>(m_image->imageData + i * m_image->widthStep))[j * 3 + 2];
int G = (reinterpret_cast<uchar*>(m_image->imageData + i * m_image->widthStep))[j * 3 + 1];
int B = (reinterpret_cast<uchar*>(m_image->imageData + i * m_image->widthStep))[j * 3];
if (( R > 95) && ( G > 40 ) && ( B > 20 ) &&
(std::max(R, std::max( G, B) ) - std::min(R, std::min(G, B) ) > 15) &&
(std::abs(R - G) > 15) && (R > G) && (R > B))
{
//skin pixel
}
}
}
3) in YCrCb space:
for(size_t i = 0; i < m_image->height; ++i)
{
for(size_t j = 0; j < m_image->width; ++j)
{
if (m_image->nChannels == 3)
{
int Cr = (reinterpret_cast<uchar*>(image->imageData + i * image->widthStep))[j * 3 + 2];
int Cb = (reinterpret_cast<uchar*>(image->imageData + i * image->widthStep))[j * 3 + 1];
int Y = (reinterpret_cast<uchar*>(image->imageData + i * image->widthStep))[j * 3];
if (( Y > 80 ) && ( Cb > 85 ) && ( Cb < 135 ) &&
(Cr > 135) && (Cr < 180))
{
//skin pixel
}
}
}
}