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2016-10-14 08:26
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如何获取image.RGBA或任何其他类型的特定像素的几何?

I wish there was something like image.Point struct but instead it was pixel based, if that makes sense.

Say I have loaded and decoded an image.RGBA with size(bounds) of 300x300. How can I get the exact coordinate of the middle of the image in image.Point or fixed.Point26_6?

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我希望有类似image.Point的结构,但是如果可行的话,它是基于像素的。

说我已经加载并解码了尺寸(边界)为300x300的 image.RGBA 。 如何在 image.Point fixed.Point26_6 中获取图像中间的确切坐标?

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  • duanhe6464 2016-10-14 09:28
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    image.RGBA is a concrete implementation of the general image.Image interface.

    It has an Image.Bounds() method:

    // Bounds returns the domain for which At can return non-zero color.
    // The bounds do not necessarily contain the point (0, 0).
    Bounds() Rectangle
    

    Important to note that the top-left corner of the image might not be at the zero point (0, 0) (although generally it is).

    So the geometry of the image is handed to you as a value of image.Rectangle:

    type Rectangle struct {
        Min, Max Point
    }
    

    To handle the general case (where top-left might not be (0, 0)), you have to account both the Min and Max points to to calculate the center point:

    cx := (r.Min.X + r.Max.X)/2
    cy := (r.Min.Y + r.Max.Y)/2
    

    Another solution is to use Rectangle.Dx() and Rectangle.Dy():

    cx := r.Min.X + r.Dx()/2
    cy := r.Min.Y + r.Dy()/2
    

    And there is an image.Point struct type. To get the center point as a value of image.Point:

    cp := image.Point{(r.Min.X + r.Max.X) / 2, (r.Min.Y + r.Max.Y) / 2}
    

    Or:

    cp := image.Point{r.Min.X + r.Dx()/2, r.Min.Y + r.Dy()/2}
    

    See this example:

    r := image.Rect(0, 0, 300, 100)
    fmt.Println(r)
    cp := image.Point{(r.Min.X + r.Max.X) / 2, (r.Min.Y + r.Max.Y) / 2}
    fmt.Println(cp)
    
    cp = image.Point{r.Min.X + r.Dx()/2, r.Min.Y + r.Dy()/2}
    fmt.Println(cp)
    

    Output (try it on the Go Playground):

    (0,0)-(300,100)
    (150,50)
    (150,50)
    
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