图像插值

发布于 2024-10-05 09:14:33 字数 881 浏览 6 评论 0原文

您好,在我的项目中,我想在这些图像中提供一些静止图像,我应该将洗牌图像作为输入,但输出将像连续图像一样,它可以通过使用插值方法比较最近的邻居,我如何实现该功能,请告诉我.我有一些代码告诉我它是如何工作的。

    public InterpolationMethod method = InterpolationMethod.Bilinear;
    public int newWidth = 0;
    public int newHeight = 0;
    try {
      this.newWidth = Math.Max(1, Math.Min(5000, int.Parse(widthBox.Text)));
      this.newHeight = Math.Max(1, Math.Min(5000, int.Parse(heightBox.Text)));
      this.method = (methodCombo.SelectedIndex == 0) 
        ? InterpolationMethod.NearestNeighbor 
        : (methodCombo.SelectedIndex == 1) 
            ? InterpolationMethod.Bilinear 
            : InterpolationMethod.Bicubic;
      this.DialogResult = DialogResult.OK;
      this.Close();
    }
    catch (Exception)
    {
      MessageBox.Show(this, "Incorrect values are entered", "Error", MessageBoxButtons.OK, MessageBoxIcon.Error);
    }
}

hi in my project i want to give some number of still images in those images i should take shuffling images as input but the out put will cme like sequential images it can compare the nearest neighbor by using interpolation methods how can i implement that plz tell me .i have some code tell me how it wrks.

    public InterpolationMethod method = InterpolationMethod.Bilinear;
    public int newWidth = 0;
    public int newHeight = 0;
    try {
      this.newWidth = Math.Max(1, Math.Min(5000, int.Parse(widthBox.Text)));
      this.newHeight = Math.Max(1, Math.Min(5000, int.Parse(heightBox.Text)));
      this.method = (methodCombo.SelectedIndex == 0) 
        ? InterpolationMethod.NearestNeighbor 
        : (methodCombo.SelectedIndex == 1) 
            ? InterpolationMethod.Bilinear 
            : InterpolationMethod.Bicubic;
      this.DialogResult = DialogResult.OK;
      this.Close();
    }
    catch (Exception)
    {
      MessageBox.Show(this, "Incorrect values are entered", "Error", MessageBoxButtons.OK, MessageBoxIcon.Error);
    }
}

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傲世九天 2024-10-12 09:14:33

从最近邻插值开始——这是最简单的。基本上,该方法的工作原理如下:

  • 为新图像分配内存(显然,除非已经为您完成了)
  • 迭代新图像的所有像素。您可以按照您想要的任何方式执行此操作,但最常见的方式称为“扫描线遍历”或“光栅扫描”——首先跨列读取,当到达最后一列时,向下移动一行,然后返回到第一列。
    • 在每个像素位置(row, col),计算出旧图像中相应的位置(row_o, col_o)。例如,如果旧图像大小为一半,则 row_o = row/2col_o = col/2。请注意,由于所有像素位置都是整数,因此会涉及一些舍入。对于最近邻插值来说,这是一种不可避免的罪恶,你无法避免它。
    • 抓取旧图像中(row_o, col_o)位置的像素值
    • 将此值分配给位置(row, col)的新图像

一旦你从这样,其他方法应该更有意义。

扫描线遍历:

for (i=0; i < image.height; ++i)
for (j=0; j < image.width;  ++j)
{
    // Do stuff here
}

它看起来像什么:

alt text

Start with nearest neighbor interpolation -- it's the simplest. Basically, the approach works like this:

  • Allocate memory for your new image (unless it's already been done for you, obviously)
  • Iterate over all the pixels of your new image. You can do this in any way you want, but the most common way is called a "scanline traversal" or "raster scan" -- you read across columns first, and when you get to the last column, you move down a row and go back to the first column.
    • At each pixel position (row, col), work out the corresponding position (row_o, col_o) in the old image. For example, if the old image is half the size, then row_o = row/2 and col_o = col/2. Note that since all the pixel positions are integers, there will be some rounding involved. With nearest neighbor interpolation, it's a necessary evil and you can't avoid it.
    • Grab the pixel value at position (row_o, col_o) in the old image
    • Assign this value to the new image at position (row, col)

Once you get that out of the way, the other approaches should make more sense.

Scanline traversal:

for (i=0; i < image.height; ++i)
for (j=0; j < image.width;  ++j)
{
    // Do stuff here
}

What it looks like:

alt text

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