How to change font size using the Python ImageDraw Library

Question

I am trying to change the font size using python's ImageDraw library.

You can do something like this:

fontPath = "/usr/share/fonts/dejavu-lgc/DejaVuLGCSansCondensed-Bold.ttf"
sans16  =  ImageFont.truetype ( fontPath, 16 )

im  =  Image.new ( "RGB", (200,50), "#ddd" )
draw  =  ImageDraw.Draw ( im )
draw.text ( (10,10), "Run awayyyy!", font=sans16, fill="red" )

The problem is that I don't want to specify a font. I want to use the default font and just change the size of the font. This seems to me that it should be simple, but I can't find documentation on how to do this.

Answer 1

Per PIL's docs, ImageDraw's default font is a bitmap font, and therefore it cannot be scaled. For scaling, you need to select a true-type font. I hope it's not difficult to find a nice truetype font that "looks kinda like" the default font in your desired font-size!

Answer 2

Just do this

from PIL import Image
from PIL import ImageDraw
from PIL import ImageFont

def image_char(char,image_size, font_size):
        img = Image.new("RGB", (image_size, image_size), (255,255,255))
        print img.getpixel((0,0))
        draw = ImageDraw.Draw(img)
        font_path = "/Users/admin/Library/Fonts/InputSans-Regular.ttf"
        font = ImageFont.truetype(font_path, font_size)

        draw.text((5, 5), char, (0,0,0),font=font)
        img.show()

main():
       image_char("A",36,16)

if __name__ == '__main__':
       sys.exit(main())