# Counting the number of groups of 1s in a boolean map of numpy.array

I am right now dealing with some image processing in Python via PIL (Python Image Library). My main aim is counting the number of colored cells in an immunohistochemistry image. I know that there are relevant programs, libraries, functions and tutorials about it, and I checked almost all of them. My major aim is writing the code manually from scratch, as much as possible. Hence I am trying to avoid using lots of exterior libraries and functions. I have written the most of the program. So here is what's going on step by step:

Program takes in the image file:

And processes it for the red cells (basically, it turns off the RGB values below a certain threshold for red):

And creates the boolean map of it, (gonna paste a part of it since it is big) which basically just puts 1 wherever it encounters with a red pixel in the processed second image above.

22222222222222222222222222222222222222222
20000000111111110000000000000000000000002
20000000111111110000000000000000000000002
20000000111111110000000000000000000000002
20000000011111100000000000000000001100002
20000000001111100000000000000000011111002
20000000000110000000000000000000011111002
20000000000000000000000000000000111111002
20000000000000000000000000000000111111102
20000000000000000000000000000001111111102
20000000000000000000000000000001111111102
20000000000000000000000000000000111111002
20000000000000000000000000000000010000002
20000000000000000000000000000000000000002
22222222222222222222222222222222222222222


I intentionally generated that frame like thing on the borders with 2s to help me with counting the number of groups of 1s in that boolean map.

My question to you guys is, how come I can efficiently count the number of cells (groups of 1s) in that kind of boolean map? I have found http://en.wikipedia.org/wiki/Connected-component_labeling which look extremely related and similar, but as far as I see, it is at the pixel level. Mine is at the boolean level. Just 1s and 0s.

Thanks a lot.

## migrated from stackoverflow.comMay 31 '12 at 11:50

This question came from our site for professional and enthusiast programmers.

• Connected component labeling is exactly what you need. I don't know why you think it's different, since the Wikipedia article also has examples starting with arrays of 1s and 0s. – interjay May 31 '12 at 9:04
• I know it looks similar (or maybe same), I just can't fully grasp the whole wikipedia page since English is not my native language. The "Sequential algorithm" part of the page looks like dealing with 1s and 0s but still I didn't see the logic behind it. Forex, why does it start by checking north, northeast, northwest, and west? – Ibrahim C. Kurt May 31 '12 at 9:16
• This problem is solved. – Ibrahim C. Kurt May 31 '12 at 11:05
• What if cells of interest are overlapping? Shouldn't you be looking specifically for circular features so you can differentiate two cells that look connected, not just finding continuous blobs? – endolith May 31 '12 at 13:36
• things get much more complicated when the immunohistochemistry pictures that you have are bad quality in terms of number of cells overlapping, resolution, standart deviation of the pixel values so and so fort... I've been trying to write a little program that works fine for such purposes but it looks like the input image and it's conditions are very important for a perfect result.. – Ibrahim C. Kurt May 31 '12 at 13:54

Something of a brute force approach, but done by inverting the problem to index over collections of pixels to find regions, instead of rasterizing over the array.

data = """\
000000011111111000000000000000000000000
000000011111111000000000000000000000000
000000011111111000000000000000000000000
000000001111110000000001000000000110000
000000000111110000000011000000001111100
000000000011100000000000100000011111100
000000000000000000000000000000011111100
000000000000000000000000000000011111110
000000000000000000000000000000111111110
000000000000000000000000000000111111110
000000000000000000000000000000011111100
000000000000000000000000000000001000000
000000000000000000000000000000000000000"""

from collections import namedtuple
Point = namedtuple('Point', 'x y')

# to accept diagonal adjacency, use this form
#return -1 <= p1.x-p2.x <= 1 and -1 <= p1.y-p2.y <= 1
return (-1 <= p1.x-p2.x <= 1 and p1.y == p2.y or
p1.x == p2.x and -1 <= p1.y-p2.y <= 1)

return any(points_adjoin(p,pt) for p in pts)

def locate_regions(datastring):
data = map(list, datastring.splitlines())
regions = []
datapts = [Point(x,y)
for y,row in enumerate(data)
for x,value in enumerate(row) if value=='1']
for dp in datapts:
# joining more than one reg, merge
regions[:] = [r for r in regions if r not in adjregs]
else:
# not adjoining any, start a new region
regions.append(set([dp]))
return regions

def region_index(regs, p):
return next((i for i,reg in enumerate(regs) if p in reg), -1)

def print_regions(regs):
maxx = max(p.x for r in regs for p in r)
maxy = max(p.y for r in regs for p in r)
allregionpts = reduce(set.union, regs)
for y in range(-1,maxy+2):
line = []
for x in range(-1,maxx+2):
p = Point(x, y)
if p in allregionpts:
line.append(str(region_index(regs, p)))
else:
line.append('.')
print ''.join(line)
print

# test against data set
regs = locate_regions(data)
print len(regs)
print_regions(regs)


Prints:

4
........................................
........00000000........................
........00000000........................
........00000000........................
.........000000.........1.........33....
..........00000........11........33333..
...........000...........2......333333..
................................333333..
................................3333333.
...............................33333333.
...............................33333333.
................................333333..
.................................3......
........................................

• wow... I don't know what to say. Very cool. And totally works. Thanks Paul. – Ibrahim C. Kurt May 31 '12 at 10:44
• So much work in this, when there is already a function in Scipy doing this, which is probably faster too ^^' But probably a good exercise anyway and it shows how to do this in general. I shall vote up then. – Zelphir Kaltstahl Apr 18 '16 at 20:35

You can use ndimage.label, which is a nice way to do this. It returns a new array, with every feature having a unique value, and the number of features. You can also specify a connection element.

import scipy
from scipy import ndimage
import matplotlib.pyplot as plt

#flatten to make greyscale, using your second red-black image as input.
#smooth and threshold as image has compression artifacts (jpg)
im = ndimage.gaussian_filter(im, 2)
im[im<10]=0
blobs, number_of_blobs = ndimage.label(im)
print 'Number of blobls:', number_of_blobs

plt.imshow(blobs)
plt.show()

#Output is:
Number of blobls: 30


• Thanks fraxel. That totally works as a quick and dirty solution, but maybe I should improve the quality of the image, cause as you can see there are lots of merged cells. The answer should be 30 cells. Thanks very much once again. (edit: I tried improving the resolution quality of the image and then blotted it with your code but it still merges lots of cells. It must be about the way flatten=1 or imread works?) – Ibrahim C. Kurt May 31 '12 at 10:33
• @Ibrahim C. Kurt - I've updated to correct that (I only just noticed!). The problem was the uploaded image was jpg, so had loads of artifacts. Small amount of smoothing and thresholding resolves that.. Should work perfectly for png image (i think...) – fraxel May 31 '12 at 10:38
• without the need for smoothing and thresholding. – fraxel May 31 '12 at 10:46
• It totally works. You are right. Without the need for any further modification, just changing it to png from jpg is sufficient. Thanks very much. I have more than 2 perfect answers now. I don't know what to do and say :D – Ibrahim C. Kurt May 31 '12 at 10:49
• @Ibrahim C. Kurt - lucky you ;) – fraxel May 31 '12 at 10:59

Here is an algorithm that's O(total number of pixel + number of cell pixels). We simply scan the picture for cell pixels, and when we find one, we flood-fill the cell to erase it.

Implementation in Common Lisp, but you'll be able to translate it to Python trivially.

(defun flood-fill (picture i j target-color replacement-color)
;; http://en.wikipedia.org/wiki/Flood_fill
(when (= (aref picture i j) target-color)
(setf (aref picture i j) replacement-color)
(when (plusp i)
(flood-fill picture (1- i) j target-color replacement-color))
(when (< (1+ i) (array-dimension picture 0))
(flood-fill picture (1+ i) j target-color replacement-color))
(when (plusp j)
(flood-fill picture i (1- j) target-color replacement-color))
(when (< (1+ j) (array-dimension picture 1))
(flood-fill picture i (1+ j) target-color replacement-color)))
picture)

(defun count-cells (picture)
(loop
:with cell-count = 0
:for i :from 0 :below (array-dimension picture 0)
:do (loop
:for j :from 0 :below (array-dimension picture 1)
:unless (zerop (aref picture i j))
:do (progn (incf cell-count)
(flood-fill picture i j 1 0)))
:finally (return cell-count)))

(count-cells
(make-array '(128 171) :element-type 'bit
:initial-contents
#(#171*000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000111111111000000000000000000000000000
#171*000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000111111111000000000000000000000000000
#171*000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000111111111000000000000000000001110000
#171*000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000011111110000000000000000000011111100
#171*000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000111000000000000000000000011111100
#171*000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000111111100
#171*000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000111111110
#171*000000000001110000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001111111110
#171*000000000011111000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001111111110
#171*000000000011111100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001111111100
#171*000000000011111110000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001
#171*000000000001111110000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001
#171*000000000001111110000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000011
#171*000000000000011110000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000011
#171*000000000000011000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001
#171*000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000011111000000000000000000011
#171*000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000111111100000000000000000000
#171*000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000111111100000000000000000000
#171*000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000111111110000000000000000000
#171*000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001111000000000000000000000111111110000000000000000000
#171*000000000000000000111111000000000000000000000000000000000000000000000000000000111110000000000000000000000000000000000001111111100001111000000000011111110000000000000000000
#171*000000000000000000111111100000000000000000000000000000000000000000000000000011111110000000000000000000000000000000000111111111100011111100000000011111100000000000000000000
#171*000000000000000000111111100000000000000000000000000000000000000000000000000011111110000000000000000000000000000000000111111111110111111110000000000100000000000000000000000
#171*000000000000000000111111100000000000000000000000000000000000000000000000000011111100000000000000000000000000000000000111111111110111111110000000000000000000000000000000000
#171*000000000000000000111111100000000000000000000000000000000000000000000000000011111100000000000000000000000000000000000011111111100111111110000000000000000000000000000000000
#171*000000000000000000111111100000000000000000000000000000000000000000000000000011111100000000000000000000000000000000000001111111100111111110000000000000000000000000000000000
#171*000000000000000000011111100000000000000000000000000000000000000000000000000011111001110000000000000000000000000000000000011111000011111110000000000000000000000000000000000
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#171*000000000000000000000000000111111100111111111100000000000000000011111100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
#171*000000000000000000000000000111111100011111111100000000000000000001110000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
#171*000000000000000000000000001111111100011111110000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000)))
--> 30

• Hey Pascal, thanks a lot for the reply. I see your program finds the correct answer very cleanly. The problem is that I have no idea about that Common Lisp language, but I'll try figuring that out and write a similar scripty in Python. – Ibrahim C. Kurt May 31 '12 at 10:37

More of an extended comment than an answer:

As @interjay has hinted, in a binary image, ie one in which only 2 colours are present, the pixels take the value either 1 or 0. This may or may not be true in the image representation format you are using but it is true in the 'conceptual' representation of your image; don't let implementation details confuse you on this issue. One of those implementation details is your use of 2s around the border of the image -- a perfectly sensible way of identifying the dead zone around the image, but not qualitatively affecting the binary-ness of the image.

As to the examination of the N, NE, NW and W pixels: this is to do with the connectivity of pixels in the formation of the component. Each pixel (bar the border special cases) has 8 neighbours (N,S,E,W,NE,NW,SE,SW) but which ones are candidates for inclusion in the same component ? Sometimes components which meet only at corners (NE,NW,SE,SW) are not regarded as connected, sometimes they are.

You have to decide what is appropriate for your application. I suggest you work out, by hand, a few operations of the sequential algorithm, checking different neighbours for each pixel, to get a feel for what is going on.