11
\$\begingroup\$

I wanted to write a template matching algorithm from scratch.

The key area's I'm interesting in are readability, maintainability, DRY, and if the program performance/efficiency could be improved in any way.

Image scanning class:

import sys
from itertools import izip


class ScanImage:
  def __init__(self, image, template, process='sad'):
    self.image = image 
    self.temp = template
    self.pro = process #sad or ssd 
    self.x_axis_leng = len(image[0]) - len(template[0]) # how far to scan on x axis of image
    self.y_axis_hgt = len(image) - len(template) # how far to scan on y axis
    self.best_location = list() # will hold the coordinates of the final location 
    self.score = sys.maxint # similarity strength score (the less the better)
    self.temp_matrix_holder = list() # holds a matrix of the same size as the template
    self.calculate = ImageCalculation() # class that is used to calculate strength score using ssd or sad 

  def _empty_list(self):
    del self.temp_matrix_holder[:]

  def run(self):
    return self._scan_image(self.image, self.temp, self.pro)

  def _scan_image(self, image, template, process):
    """main function: will iterate over the image one pixel at a time"""


    for i in xrange(len(image)):
      for j in xrange(len(image[i])):
        if (j <= self.x_axis_leng) and \
          (i <= self.y_axis_hgt):

          self.temp_matrix_holder.append(\
            image[i][j:j+len(template[0])])

          self.temp_matrix_holder.append(\
            image[i + 1][j:j+len(template[0])])



          new_score = self.calculate.run(\
            self.temp_matrix_holder, self.temp, process)


          if new_score <= self.score:
            self.score = new_score
            self.best_location = [j, i]

        self._empty_list()


    return self.score, self.best_location

Score calculation class:

class ImageCalculation:
  def __init__(self):
    self.score = 0.0

  def _reset_score(self):
    self.score = 0.0

  def _minus(self, atuple):
    return abs(atuple[0]-atuple[1])

  def _sad(self, atuple):
    """sum of absolute difference""" 

    # this is not nessesary but keeps the code a little neater 
    # and also shows the explicit difference between sad and ssd
    return self._minus(atuple)

  def _ssd(self, atuple):
    """sum of squared difference"""

    diff = self._minus(atuple)
    return diff * diff 

  def _block_template_comparison(self, image_block, template, process):

    if (len(image_block) != len(template)) or \
        (len(image_block[0]) != len(template[0])):

      return 'Error'

    for i in xrange(len(image_block)):
      for j in xrange(len(image_block[i])):
        one_pixel_score = 0.0
        for k in izip(image_block[i][j], template[i][j]):

          if process == 'sad':
            one_pixel_score += self._sad(k)

          elif process == 'ssd':
            one_pixel_score += self._ssd(k)


        self.score += (one_pixel_score/3.0) # average out all 3 (RGB) channels for each pixle    

    return self.score

  def run(self, image_block, template, process='sad'):

    if process == 'ssd':
      process = 'ssd'
    elif process == 'sad':
      process = 'sad'
    else:
      raise ValueError, 'Proccess must be either sad or ssd'

    if not self.score == 0.0:
      self._reset_score()

    return self._block_template_comparison(image_block, template, process)

Outputs:

if __name__ == '__main__':

  p_i = [
      [[0], [1], [2], [10]],
      [[3], [4], [5], [11]],
      [[6], [7], [8], [12]],
      ]


  t_i = [ 
        [[4], [5]],
        [[7], [12]],
        ]


  s = ScanImage(p_i, t_i)

  print s.run()


  # ssd outputs : 5.333 (sength score), [1,1] (x, y coords)
  # sad outputs : 1.3333, [1, 1] 

Template matching is taking a small 'image' and scanning it over a larger image one pixel at a time until a match is found.

During each iteration of the scan the template and a block of the image, which is the same size as the template, are compared. They are given a score. The closer the score (in this specific scoring metric) to 0 the better. A perfect match gets a score of 0.

Special note:

This program is designed to work with pixels with 3 channels each (RGB), this is just a dummy example but it also works for fully built out 'images'.

\$\endgroup\$
3
  • \$\begingroup\$ Can you edit your question and describe in a few words what template matching is and how it works? \$\endgroup\$ Commented Jun 8, 2018 at 12:57
  • \$\begingroup\$ Is Python2 required, or are you open to using Python3? \$\endgroup\$
    – scnerd
    Commented Jun 8, 2018 at 14:27
  • \$\begingroup\$ @scnerd python 2.7.9 \$\endgroup\$
    – arm93
    Commented Jun 8, 2018 at 14:34

1 Answer 1

5
\$\begingroup\$

I am not very experienced in Python, so this review might miss some Python-specific issues. But I am very experienced with this type of algorithm, and so will primarily discuss the algorithmic and computational aspects of the code.

Image storage

If I understand correctly, the image here is stored as a list of lists. This is not an efficient storage model, because each image line needs its own allocation, and is potentially stored in disparate memory locations. It is always better to store an image as a single memory block, so that all pixels are consecutive in memory. This leads to fewer cache misses. There is also one fewer level of indirection (looking up the pointer to the image line (list) in the outer list, then looking up the pixel in the inner list).

And looking through the code again, it might even be the case that the image is stored as a list of lists of lists, where each pixel is a list containing the RGB triplet. This is even worse, as each pixel is potentially in a different location in memory.

Using a NumPy array to store the image should lead to a significant improvement in speed just because of the data locality and fewer levels of indirection.

Also, extracting the image block can be done in a single statement. The resulting NumPy array references the data in the original array, meaning that no data copy is necessary.

Additional advantages to using NumPy are the possibility to compute things such as done in ImageCalculation._block_template_comparison with a single statement rather than three nested loops. This is both faster (the looping happens in compiled code, which is much faster than loops in Python), and easier to maintain (less code).

Generalization

These two lines imply that the template always has exactly two image lines:

self.temp_matrix_holder.append(\
  image[i][j:j+len(template[0])])
self.temp_matrix_holder.append(\
  image[i + 1][j:j+len(template[0])])

It would be a good idea to add another loop here to copy an image block of arbitrary size, to be able to match templates of arbitrary size.

Loop limits

for i in xrange(len(image)):
  for j in xrange(len(image[i])):
    if (j <= self.x_axis_leng) and \
      (i <= self.y_axis_hgt):

You are looping over all pixels, but not doing anything for the pixels at the right and bottom edges. Why not directly loop only over those pixels that you need to address?

for i in xrange(self.y_axis_hgt + 1):
  for j in xrange(self.x_axis_leng + 1):

Square difference

The function ImageCalculation._ssd computes the square of the absolute difference, as self._minus calls abs. That call to abs is redundant (unless you have complex-valued images).

Short-cut

If you compute a score of 0, you can stop the algorithm, as it won't find a better match.

Method selection

In ImageCalculation._block_template_comparison, you test for which of the two process values is given within the double loop. This is not at all efficient, you should do this test outside. In fact, for every block being compared, the process value is tested for validity as well, which needs to be done only once.

Instead, the ImageCalculation class could take the process value in its constructor, and store a pointer to the method it should use in computation. This way, no further string comparisons are needed during computation.

Readability and maintainability

You specifically ask for these. I think the code is readable, you use clear variable names and add comments where necessary.

However (and this is probably more about me than your code), I think that the classes here make the code more convoluted than necessary. I've never been a fan of the classes that just have a constructor and a run method. I don't see the advantage over a normal function, and only see unnecessary complication, both in usage:

s = ScanImage(p_i, t_i)
print s.run()

vs

print scan_image(p_i, t_i)

and in the code itself. Your first block of code would be identical, but with lots of lines deleted:

def scan_image(image, template, process='sad'):
    x_axis_leng = len(image[0]) - len(template[0]) # how far to scan on x axis of image
    y_axis_hgt = len(image) - len(template) # how far to scan on y axis
    best_location = list() # will hold the coordinates of the final location 
    score = sys.maxint # similarity strength score (the less the better)
    calculate = ImageCalculation() # class that is used to calculate strength score using ssd or sad 

    for i in xrange(len(image)):
      for j in xrange(len(image[i])):
        if (j <= x_axis_leng) and \
          (i <= y_axis_hgt):

          temp_matrix_holder = list()
          temp_matrix_holder.append(\
            image[i][j:j+len(template[0])])
          temp_matrix_holder.append(\
            image[i + 1][j:j+len(template[0])])

          new_score = calculate.run(\
            temp_matrix_holder, template, process)

          if new_score <= score:
            score = new_score
            best_location = [j, i]

    return score, best_location

Because there are fewer lines of code, it's more readable and more maintainable. It's also easier to follow the data throughout the code, since there are no variables being shared across different methods.

\$\endgroup\$
1
  • \$\begingroup\$ Thank you very much. This is the exact sort of thing I was looking for. Your SSD point out was such a face palm moment :) \$\endgroup\$
    – arm93
    Commented Jun 11, 2018 at 15:38

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.