duanlongnao0028 2017-06-11 21:20
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如何在Go中打开图像以获取黑白像素的二进制数据?

I've been trying for sometime to open an image in binary mode with Go. In Python I'd use the Pillow and image.open() (rb mode). Example.

img = Image.open("PNG.png")
pix = img.getdata()  #where 0 is black and 1 is white pixel

That would open the image with very clean binary of white and black dots like the image below. Example In go I've tried os.Open(file.jpg) to open the file.. I've tried decoding it with image.Decode(), I've loaded the file into bytes.Buffer, I've tried fmt.Sprintf("%b", data), all of the solutions give a byte array. Converting that byte array to binary looks nothing like the image above. I've also tried encoding/binary and its the same story with just getting bytes and the binary generated isn't what i want...

Most recently I've tried this

package main

import (
    "fmt"
    "image"
    "image/jpeg"
    "io"
    "log"
    "os"
)

// Pixel struct example
type Pixel struct {
    R int
    G int
    B int
    A int
}

func main() {
    // You can register another format here
    image.RegisterFormat("jpg", "jpg", jpeg.Decode, jpeg.DecodeConfig)

    file, err := os.Open("/Users/marcsantiago/Desktop/2033bb1b194adace86f99c7bb7d72e81.jpg")

    if err != nil {
        log.Fatalln("Error: File could not be opened")

    }

    defer file.Close()

    pixels, err := getPixels(file)

    if err != nil {
        log.Fatalln("Error: Image could not be decoded")
    }
    black := Pixel{0, 0, 0, 255}

    for i := range pixels {
        if pixels[i] == black {
            fmt.Print("0")
        } else {
            fmt.Print("1")
        }

    }
}

func getPixels(file io.Reader) ([]Pixel, error) {
    img, _, err := image.Decode(file)

    if err != nil {
        return nil, err
    }

    bounds := img.Bounds()
    width, height := bounds.Max.X, bounds.Max.Y

    var pixels []Pixel
    for y := 0; y < height; y++ {
        for x := 0; x < width; x++ {
            pixels = append(pixels, rgbaToPixel(img.At(x, y).RGBA()))
        }
    }
    return pixels, nil
}

// img.At(x, y).RGBA() returns four uint32 values; we want a Pixel
func rgbaToPixel(r uint32, g uint32, b uint32, a uint32) Pixel {
    return Pixel{int(r / 257), int(g / 257), int(b / 257), int(a / 257)}
}

So that I can compare the binary against what I expect I converted the rgba to 1 and 0s where 0 == black... it still doesn't match up not even close. Example enter image description here

Help please. I'm out of ideas. PS. This site http://www.dcode.fr/binary-image, also opens the image and generates the data I'm expecting.

UPDATE: This is the image i'm working with.. Image to decode

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1条回答 默认 最新

  • doujiao7520 2017-06-11 23:24
    关注

    For example,

    package main
    
    import (
        "bytes"
        "fmt"
        "image"
        "os"
    
        _ "image/jpeg"
    )
    
    func main() {
        fName := "ggk3Z.jpg"
        f, err := os.Open(fName)
        if err != nil {
            fmt.Fprintln(os.Stderr, err)
            os.Exit(1)
        }
        defer f.Close()
        img, _, err := image.Decode(f)
        if err != nil {
            fmt.Fprintln(os.Stderr, err)
            os.Exit(1)
        }
    
        // http://www.dcode.fr/binary-image
        var txt bytes.Buffer
        bounds := img.Bounds()
        for y := bounds.Min.Y; y < bounds.Max.Y; y++ {
            for x := bounds.Min.X; x < bounds.Max.X; x++ {
                r, g, b, _ := img.At(x, y).RGBA()
                bin := "0"
                if float64((r+g+b))/3 > 0.5 {
                    bin = "1"
                }
                txt.WriteString(bin)
            }
            txt.WriteString("
    ")
        }
        fmt.Fprint(os.Stdout, txt.String())
    }
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

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