Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
161 views
in Technique[技术] by (71.8m points)

objective c - Shift hue of an RGB Color

I'm trying to write a function to shift the hue of an RGB color. Specifically I'm using it in an iOS app, but the math is universal.

The graph below shows how the R, G, and B values change with respect to the hue.

Graph of RGB values across hues

Looking at that it seems like it should be a relatively simple to write a function to shift the hue without doing any nasty conversions to a different color format which would introduce more error (which could be an issue if continue applying small shifts to a color), and I suspect would be more computationally expensive.

Here is what I have so far which sort of works. It works perfectly if you're shifting from pure yellow or cyan or magenta but otherwise it gets a little squiffy in some places.

Color4f ShiftHue(Color4f c, float d) {
    if (d==0) {
        return c;
    }
    while (d<0) {
        d+=1;
    }

    d *= 3;

    float original[] = {c.red, c.green, c.blue};
    float returned[] = {c.red, c.green, c.blue};

    // big shifts
    for (int i=0; i<3; i++) {
        returned[i] = original[(i+((int) d))%3];
    }
    d -= (float) ((int) d);
    original[0] = returned[0];
    original[1] = returned[1];
    original[2] = returned[2];

    float lower = MIN(MIN(c.red, c.green), c.blue);
    float upper = MAX(MAX(c.red, c.green), c.blue);

    float spread = upper - lower;
    float shift  = spread * d * 2;

    // little shift
    for (int i = 0; i < 3; ++i) {
        // if middle value
        if (original[(i+2)%3]==upper && original[(i+1)%3]==lower) {
            returned[i] -= shift;
            if (returned[i]<lower) {
                returned[(i+1)%3] += lower - returned[i];
                returned[i]=lower;
            } else
                if (returned[i]>upper) {
                    returned[(i+2)%3] -= returned[i] - upper;
                    returned[i]=upper;
                }
            break;
        }
    }

    return Color4fMake(returned[0], returned[1], returned[2], c.alpha);
}

I know you can do this with UIColors and shift the hue with something like this:

CGFloat hue;
CGFloat sat;
CGFloat bri;
[[UIColor colorWithRed:parent.color.red green:parent.color.green blue:parent.color.blue alpha:1] getHue:&hue saturation:&sat brightness:&bri alpha:nil];
hue -= .03;
if (hue<0) {
    hue+=1;
}
UIColor *tempColor = [UIColor colorWithHue:hue saturation:sat brightness:bri alpha:1];
const float* components= CGColorGetComponents(tempColor.CGColor);
color = Color4fMake(components[0], components[1], components[2], 1);

but I'm not crazy about that as It only works in iOS 5, and between allocating a number of color objects and converting from RGB to HSB and then back it seems pretty overkill.

I might end up using a lookup table or pre-calculate the colors in my application, but I'm really curious if there's a way to make my code work. Thanks!

question from:https://stackoverflow.com/questions/8507885/shift-hue-of-an-rgb-color

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

Edit per comment changed "are all" to "can be linearly approximated by".
Edit 2 adding offsets.


Essentially, the steps you want are

RBG->HSV->Update hue->RGB

Since these can be approximated by linear matrix transforms (i.e. they are associative), you can perform it in a single step without any nasty conversion or loss of precision. You just multiple the transform matrices with each other, and use that to transform your colors.

There's a quick step by step here http://beesbuzz.biz/code/hsv_color_transforms.php

Here's the C++ code (With the saturation and value transforms removed):

Color TransformH(
    const Color &in,  // color to transform
    float H
)
{
  float U = cos(H*M_PI/180);
  float W = sin(H*M_PI/180);

  Color ret;
  ret.r = (.299+.701*U+.168*W)*in.r
    + (.587-.587*U+.330*W)*in.g
    + (.114-.114*U-.497*W)*in.b;
  ret.g = (.299-.299*U-.328*W)*in.r
    + (.587+.413*U+.035*W)*in.g
    + (.114-.114*U+.292*W)*in.b;
  ret.b = (.299-.3*U+1.25*W)*in.r
    + (.587-.588*U-1.05*W)*in.g
    + (.114+.886*U-.203*W)*in.b;
  return ret;
}

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...