Drawing Simple Rectangles in PIL

Quick Color Checks

When I found myself doing analysis on color using numpy, I kept getting turned around and having trouble conceptualizing the color representation for the R, G, B values I was seeing.

PIL makes this really easy to do. All you have to do is call Image.new() and specify the color parameter accordingly.

from PIL import Image

rect = Image.new(mode='RGB', size=(200, 200),
                 color=(0, 74, 127))


Layering in More Colors

If we wanted to look at more colors than one at a time, we could probably leverage some matplotlib “span” method, or use the built-in tools that PIL provides.

Specifically, we’ll leverage the ImageDraw.Draw object, which takes an existing Image object.

from PIL import ImageDraw
drawer = ImageDraw.Draw(rect)

Which gives us access to a ton of simple drawing utillities.

print([x for x in dir(drawer) if x[0] != '_'])
['arc', 'bitmap', 'chord', 'draw', 'ellipse', 'fill', 'font', 'fontmode', 'getfont', 'im', 'ink', 'line', 'mode', 'multiline_text', 'multiline_textsize', 'palette', 'pieslice', 'point', 'polygon', 'rectangle', 'shape', 'text', 'textsize']

Note how we leverage the rectangle() method below.

def draw_rectangle(color_list):
    Make a long rectangle, composed of the colors
    detailed in color_list, a list of (R, G, B) tuples
    n = len(color_list)

    im = Image.new('RGBA', (100*n, 100))
    draw = ImageDraw.Draw(im)

    for idx, color in enumerate(color_list):
        # ensure that numbers are all ints
        color = tuple([int(x) for x in color])
        # draw the colors by array-indexing
        draw.rectangle([(100*idx, 0), (100*(idx+1), 100*(idx+1))],

    return im
neapolitan = ([161, 91, 65],
              [252, 192, 181],
              [251, 233, 209])