Storing a pixel lookuptable in a dict, although unordered, is probably faster for lookup and manipulation. How true is this? Well, throw some sample images at it and timeit() compared to some other structures.
px_list = [
1, 2, 3, 4, 5, 6, 7, 8,
9, 10,11,12,13,14,15,16,
17,18,19,20,21,22,23,24,
25,26,27,28,29,30,31,32]
def idx_to_co(idx, width):
r = int(idx / width)
c = idx % width
return r, c
def px_list_to_dict(px_list, width):
px_dict = {}
for idx, px in enumerate(px_list):
px_dict[idx_to_co(idx, width)] = px
return px_dict
image_width = 8
px_dict = px_list_to_dict(px_list, image_width)
# unordered, but much faster lookup than list
for i in px_dict:
print(i)
import bpy
D = bpy.data
img = D.images['moop2.png']
img_width = img.size[0]
img_height = img.size[1]
# slowest part is the transfer from pixel array to list
pxs = list(img.pixels)
num_values = len(pxs)
px_list = [pxs[i:i+4] for i in range(num_values)[::4]]
def idx_to_co(idx, width):
r = int(idx / width)
c = idx % width
return r, c
def px_list_to_dict(px_list, width):
px_dict = {}
for idx, px in enumerate(px_list):
px_dict[idx_to_co(idx, width)] = px
return px_dict
px_dict = px_list_to_dict(px_list, img_width)
print(px_dict[(30,30)])
How about using tuple for lookup? ..still have to time this, but it looks interesting, it has to be faster in some way because it is an immutable structure.
