Count pixels of different colors in C#

[count pixels]

The following CountPixels method counts pixels in an image that match a target color.

// Return the number of matching pixels.
private int CountPixels(Bitmap bm, Color target_color)
    // Loop through the pixels.
    int matches = 0;
    for (int y = 0; y < bm.Height; y++)
        for (int x = 0; x < bm.Width; x++)
            if (bm.GetPixel(x, y) == target_color) matches++;
    return matches;

This code is reasonably straightforward. It loops through the pixels calling the GetPixels method to get each pixel’s color. It then compares the pixel’s returned value to the target color.

The only strangeness here is that the Color class’s Equals method, which is used by == to determine equality, treats named colors differently from those obtained in other ways, including colors obtained with the GetPixel method. That means this method won’t work if you pass it a named color such as Color.Black. If you used such a color, the method would find no matching pixels.

The following code shows how this program uses the CountPixels method to count pixels that are black or white in the image.

// Count the black and white pixels.
private void btnCount_Click(object sender, EventArgs e)
    Bitmap bm = new Bitmap(picImage.Image);
    int black_pixels =
        CountPixels(bm, Color.FromArgb(255, 0, 0, 0));
    int white_pixels =
        CountPixels(bm, Color.FromArgb(255, 255, 255, 255));
    lblBlack.Text = black_pixels + " black pixels";
    lblWhite.Text = white_pixels + " white pixels";
    lblTotal.Text = white_pixels + black_pixels + " total pixels"; 

The code is also reasonably straightforward. The only thing to note is that it uses the colors returned by Color.FromArgb(255, 0, 0, 0) and Color.FromArgb(255, 255, 255, 255) instead of the named colors Color.Black and Color.White.

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17 Responses to Count pixels of different colors in C#

  1. Rod Stephens says:

    Lots of other posts show how to do this. The steps are:

    – Add a OpenFileDialog to the form.
    – Add something to let the user tell the program to open a file. For example, add a button or a File > Open menu item.
    – When the user wants to open a file, call the OpenFileDialog’s ShowDialog method. If ShowDialog returns OK, then open the file given by the dialog’s FileName property.

    For example, see today’s post Pixellate parts of an image in C#. It lets you open (save save) an image file.

  2. Rod Stephens says:

    Sorry but if you’re asking a question I don’t understand what it is.

  3. Rod Stephens says:

    Do you mean you want to calculate the area of the non-transparent pixels? You can use this same technique. Just look for pixels where the alpha component is 0. Those are the transparent ones. In other words:

     if (GetPixel(x, y).A == 0) num_transparent++; 

    This won’t work with pixels that are translucent (i.e. only partly transparent). Although you could do something like the following, which looks for pixels that are mostly transparent.

     if (GetPixel(x, y).A < 128) num_transparent++; 
  4. Abdul Amin Khan says:


  5. hieu nguyen says:

    How to count similar pixel blacks
    For example : Count these color : RGB(6,4,7) RGB(3,9,11) RGB(4,6,2) …v.v
    Because these color is very similar with RGB(0,0,0) . How to count for all the pixel like that.

    • RodStephens says:

      You need to decide how to tell if a pixel is close enough. For example, you could calculate the squares of the differences between the red, green, and blue color components and count the pixel if the result is within a certain distance.

          int dr = target_r - pixel.R;
          int dg = target_g - pixel.G;
          int db = target_b - pixel.B;
          if (dr * dr + dg * dg + db * db < difff_squared) matches++;
      • Carlos Rum says:

        Hello RodStephens, i’ve a question. How you have gotten this definition?

        • RodStephens says:

          What definition?

          This example was for a very specific problem. It’s not very generalizable.

          • Carlos Rum says:
            if (dr * dr + dg * dg + db * db < diff_squared) matches++;

            Ok, is a Euclidian method ?.


          • RodStephens says:

            Yes. It’s the Euclidean distance formula in three dimensions. The dimensions are basically the red, green, and blue values.

            The code compares the square of the distance to diff_squared so it doesn’t need to take the square root and that saves a little time. A program like this one performs a LOT of calculations for a large image, so the savings is worthwhile.

            Sorry I didn’t make this clearer in the original post.

  6. Carlos Rum says:

    @RodStephens .. thanks a lot about your time and help

  7. Regina says:

    Can you use scanline to do this? How?

    • RodStephens says:

      I’m not sure what you mean. This program loops through the pixels in the image by scanline, more or less.

      This example is really only for small images. If you have a larger image, it will be slow. To make it faster, see this post:

      Use the Bitmap32 class to manipulate image pixels very quickly in C#

      • Regina says:

        And where should i put the Bitmap32 class in this code? Sorry im new to this

        • RodStephens says:

          Download the example program to see all of the details.

          If you’re asking where the class’s code goes, the Bitmap32 class is a separate class. You can put it in a separate file or inside the main program but outside of any other class.

          If you’re asking where you create an instance of the class, you normally do that where you want to manipulate the image’s pixels.

          • Regina says:

            Thank you so much for your help it really save me!! Last question, can you tell me the argb value for Red? I’ve tried with (120,250,0,0) but it didn’t return anything

          • RodStephens says:

            Opaque red would be (255, 255, 0, 0). Note that the color must match exactly for this program to count a pixel. For example, if a pixel in the image has color (255, 254, 0, 0), you would not notice that it wasn’t perfectly red but the program would.

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