Using the Color package to create new Color from RGB value?

Question

I'm trying to create a new Color object using RGB values I have in variables:

package main

import (
    "fmt"
    "image"
    _ "image/gif"
    _ "image/jpeg"
    _ "image/png"
    "os"
)

func main() {
    reader, err := os.Open("test-image.jpg")
    if err != nil {
        fmt.Fprintf(os.Stderr, "%v\n", err)
    }

    image, _, err := image.Decode(reader)
    if err != nil {
        fmt.Fprintf(os.Stderr, "%s", err)
    }

    bounds := image.Bounds()

    for i := 0; i <= bounds.Max.X; i++ {
        for j := 0; j <= bounds.Max.Y; j++ {
            pixel := image.At(i, j)
            if i == 0 && j == 0 {
                red, green, blue, _ := pixel.RGBA()
                averaged := (red + green + blue) / 3

                            // This FromRGBA function DOES NOT EXIST!
                grayColor := Color.FromRGBA(averaged, averaged, averaged, 1)

                // Then I could do something like:
                grayColor.RGBA() // This would work since it's a type Color.

            }
        }
    }
}

I can't seem to find any package Function that generates a new Color object given rgba values.

Any recommendations?

mikespook · Answer 1 · 2014-03-02T02:28:17.823

2

The image.Color actually is an interface. You can use any structure which satisfies it. Even your own structures.

For example, you could use image.Gary:

grayColor := image.Gray{averaged}

or your own grayColor:

type MyGray struct {
    y uint32
}

func (gray *MyGray) FromRGBA(r, g, b, a uint32) {
    gray.y = (r + g + b) / 3
}

func (gray *MyGray) RGBA() (r, g, b, a uint32) { // to satisfy image.Color
    return gray.y, gray.y, gray.y, 1
}

grayColor := &MyGray{}
grayColor.FromRGBA(pixel.RGBA())
grayColor.RGBA()
// blablabla

edited Mar 02 '14 at 02:28

answered Feb 28 '14 at 07:17

mikespook

768
4
13

Gray-Gary is confusing. It's not clear if these are typos or a joke about a new color name, but it's confusing in any case. – Yves Junqueira Feb 28 '14 at 08:26
Typo and copy-paste fault. Sorry for that! -_-! – mikespook Mar 02 '14 at 02:30
NP but there's still an instance there `image.Gary`. – robstarbuck Jun 09 '17 at 22:19

James Henstridge · Answer 2 · 2014-02-28T02:48:53.713

1

The types in the image/color package have exported fields, so you can instantiate them directly. For your example, you could create the colour value with:

grayColor := color.Gray16{Y: uint16(averaged)}

(The red, green and blue values are all in the 0..0xffff range, so the 16-bit Gray colour implementation seems appropriate).

edited Feb 28 '14 at 02:48

answered Feb 28 '14 at 01:50

James Henstridge

42,244
6
132
114

Using the Color package to create new Color from RGB value?

2 Answers2