PIL simple gradient and TypeError: an integer is required

Question

I have seen numpy-->PIL int type issue, but it doesn't answer my question which is simpler, as it doesn't use numpy. Consider this example:

import Image
import math

img = Image.new('L', (100, 50), 'white')
a = 0.1 # factor
for x in xrange(img.size[0]):
  for y in xrange(img.size[1]):
    # val: 0 to 255; 255/2 = 127.5;
    val = int( 127.5*math.sin(a*y) + 127.5 )
    print x, y, val, type(x), type(y), type(val)
    img.putpixel((x, y), (val, val, val))
img.save('singrad.png', 'png')

This fails with:

$ python test.py 
0 0 127 <type 'int'> <type 'int'> <type 'int'>
Traceback (most recent call last):
  File "test.py", line 11, in <module>
    img.putpixel((x, y), (val, val, val))
  File "/usr/lib/python2.7/dist-packages/PIL/Image.py", line 1267, in putpixel
    return self.im.putpixel(xy, value)
TypeError: an integer is required

I don't see what "integer is required", - given that all of the arguments to putpixel are reported as <type 'int'>?

How do I get this to work?

Answer 1

You are creating an Image with mode 'L' (single 8-bit pixel value), hence the value you put needs to be an int < 255. You are putting a tuple which requires the 'RGB' mode. I would think changing your code to img.putpixl((x,y), val) solves the issue.

Answer 2

Ah, got it: image type 'L' is grayscale/monochrome, so it needs a single integer value per pixel, not an RGB tuple:

img.putpixel((x, y), val)

If the image was type 'RGB' (i.e. Image.new('RGB',...), then the command as written works:

img.putpixel((x, y), (val, val, val))