COUNTLESS8 + Avoiding trouble with 0

zero_corrected_countless3d.py

from itertools import combinations
from functools import reduce
import numpy as np

def countless3d(data):
  """Now write countless8 in such a way that it could be used
  to process an image."""
  sections = []

  # shift zeros up one so they don't interfere with bitwise operators
  # we'll shift down at the end
  data = data + 1 # don't use += as it'll modify the original image
  
  # This loop splits the 2D array apart into four arrays that are
  # all the result of striding by 2 and offset by (0,0,0), (0,0,1), (0,1,0), etc
  factor = (2,2,2)
  for offset in np.ndindex(factor):
    part = data[tuple(np.s_[o::f] for o, f in zip(offset, factor))]
    sections.append(part)

  p2 = lambda q,r: q * (q == r) # PICK(A,B)
  p3 = lambda q,r,s: q * ( (q == r) & (r == s) ) # PICK(A,B,C)
  p4 = lambda p,q,r,s: p * ( (p == q) & (q == r) & (r == s) ) # PICK(A,B,C,D)

  lor = lambda x,y: x + (x == 0) * y # logical or

   # important to use generator expressions here to reduce peak memory usage
  results4 = ( p4(x,y,z,w) for x,y,z,w in combinations(sections, 4) )
  results4 = reduce(lor, results4)

  results3 = ( p3(x,y,z) for x,y,z in combinations(sections, 3) )
  results3 = reduce(lor, results3)

  # We can always omit the last section because we'll "|| section[-1]" at the end
  results2 = ( p2(x,y) for x,y in combinations(sections[:-1], 2) )
  results2 = reduce(lor, results2)

  return reduce(lor, (results4, results3, results2, sections[-1])) - 1 # combine and shift down