title: PIL Image mode I is grayscale?
body: <p>I'm trying to specify the colours of my image in Integer format instead of (R,G,B) format. I assumed that I had to create an image in mode &quot;I&quot; since according to the <a href="https://pillow.readthedocs.io/en/4.1.x/handbook/concepts.html#modes" rel="noreferrer">documentation</a>:</p>
  <blockquote>
  <p>The mode of an image defines the type and depth of a pixel in the
  image. The current release supports the following standard modes:</p>
  <ul>
  <li>1 (1-bit pixels, black and white, stored with one pixel per byte)</li>
  <li>L (8-bit pixels, black and white)</li>
  <li>P (8-bit pixels, mapped to any other mode using a colour palette)</li>
  <li>RGB (3x8-bit pixels, true colour)</li>
  <li>RGBA (4x8-bit pixels, true colour with transparency mask)</li>
  <li>CMYK (4x8-bit pixels, colour separation)</li>
  <li>YCbCr (3x8-bit pixels, colour video format)</li>
  <li><strong>I (32-bit signed integer pixels)</strong></li>
  <li>F (32-bit floating point pixels)</li>
  </ul>
  </blockquote>
  <p>However this seems to be a grayscale image. Is this expected? Is there a way of specifying a coloured image based on a 32-bit integer? In my MWE I even let PIL decide how to convert &quot;red&quot; to the &quot;I&quot; format.</p>
  <hr />
  <h3>MWE</h3>
  <pre><code>from PIL import Image
  
  ImgRGB=Image.new('RGB', (200,200),&quot;red&quot;) # create a new blank image
  ImgI=Image.new('I', (200,200),&quot;red&quot;) # create a new blank image
  ImgRGB.show()
  ImgI.show()
  </code></pre>