Compter le nombre de groupes de 1 dans une carte booléenne de numpy.array

16

Je m'occupe actuellement d'un traitement d'image en Python via PIL (Python Image Library). Mon objectif principal est de compter le nombre de cellules colorées dans une image d'immunohistochimie. Je sais qu'il existe des programmes, des bibliothèques, des fonctions et des didacticiels pertinents à ce sujet, et j'ai vérifié presque tous. Mon objectif principal est d'écrire le code manuellement à partir de zéro, autant que possible. Par conséquent, j'essaie d'éviter d'utiliser de nombreuses bibliothèques et fonctions extérieures. J'ai écrit la plupart du programme. Voici donc ce qui se passe pas à pas:

Le programme prend le fichier image:exemple

Et le traite pour les globules rouges (en gros, il désactive les valeurs RVB en dessous d'un certain seuil pour le rouge): entrez la description de l'image ici

Et crée la carte booléenne de celui-ci (je vais en coller une partie car il est grand) qui met simplement 1 à chaque fois qu'il rencontre un pixel rouge dans la deuxième image traitée ci-dessus. 

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

J'ai intentionnellement généré ce trame comme une chose sur les bordures avec 2 pour m'aider à compter le nombre de groupes de 1 dans cette carte booléenne.

Ma question est la suivante: comment se fait-il que je puisse compter efficacement le nombre de cellules (groupes de 1) dans ce type de carte booléenne? J'ai trouvé http://en.wikipedia.org/wiki/Connected-component_labeling qui semble extrêmement apparenté et similaire, mais pour autant que je vois, c'est au niveau des pixels. Le mien est au niveau booléen. Juste 1 et 0.

Merci beaucoup.

python Ibrahim C. Kurt
la source
L'étiquetage des composants connectés est exactement ce dont vous avez besoin. Je ne sais pas pourquoi vous pensez que c'est différent, car l'article de Wikipédia contient également des exemples commençant par des tableaux de 1 et de 0.
Je sais que cela ressemble (ou peut-être la même), je ne peux tout simplement pas saisir entièrement la page wikipedia car l'anglais n'est pas ma langue maternelle. La partie "algorithme séquentiel" de la page semble traiter des 1 et des 0 mais je n'ai toujours pas vu la logique derrière. Forex, pourquoi commence-t-il par vérifier le nord, le nord-est, le nord-ouest et l'ouest?
Ce problème est résolu.
Et si les cellules d'intérêt se chevauchent? Ne devriez-vous pas rechercher spécifiquement des caractéristiques circulaires afin de pouvoir différencier deux cellules qui semblent connectées, et pas seulement trouver des taches continues?
endolith
les choses deviennent beaucoup plus compliquées lorsque les images d'immunohistochimie que vous avez sont de mauvaise qualité en termes de nombre de cellules qui se chevauchent, de résolution, de déviation standard des valeurs des pixels et ainsi de suite ... J'ai essayé d'écrire un petit programme qui fonctionne bien à de telles fins, mais il ressemble à l'image d'entrée et ses conditions sont très importantes pour un résultat parfait ..
Ibrahim C. Kurt

Réponses:

6

Quelque chose d'une approche par force brute, mais réalisée en inversant le problème d'indexation sur des collections de pixels pour trouver des régions, au lieu de tramer sur le tableau.

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

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

def points_adjoin(p1, p2):
    # 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)

def adjoins(pts, pt):
    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:
        # find all adjoining regions
        adjregs = [r for r in regions if adjoins(r,dp)]
        if adjregs:
            adjregs[0].add(dp)
            if len(adjregs) > 1:
                # joining more than one reg, merge
                regions[:] = [r for r in regions if r not in adjregs]
                regions.append(reduce(set.union, 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)

Tirages:

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

la source
wow ... je ne sais pas quoi dire. Très cool. Et fonctionne totalement. Merci Paul.
Il y a tellement de travail là-dedans, quand il y a déjà une fonction pour le Scipyfaire, ce qui est probablement aussi plus rapide ^^ 'Mais probablement un bon exercice de toute façon et cela montre comment le faire en général. Je voterai alors.
Zelphir Kaltstahl
13

Vous pouvez utiliser ndimage.label, ce qui est une bonne façon de le faire. Il renvoie un nouveau tableau, chaque fonctionnalité ayant une valeur unique et le nombre de fonctionnalités. Vous pouvez également spécifier un élément de connexion.

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

#flatten to make greyscale, using your second red-black image as input.
im = scipy.misc.imread('blobs.jpg',flatten=1)
#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

entrez la description de l'image ici

j08lue
la source
Merci fraxel. Cela fonctionne totalement comme une solution rapide et sale, mais je devrais peut-être améliorer la qualité de l'image, car comme vous pouvez le voir, il y a beaucoup de cellules fusionnées. La réponse devrait être de 30 cellules. Merci encore une fois. (modifier: j'ai essayé d'améliorer la qualité de la résolution de l'image, puis je l'ai effacée avec votre code, mais il fusionne toujours beaucoup de cellules. Il doit s'agir de la façon dont flatten = 1 ou imread fonctionne?)
1
@ Ibrahim C. Kurt - J'ai mis à jour pour corriger cela (je viens juste de le remarquer!). Le problème était que l'image téléchargée était en jpg, donc il y avait beaucoup d'artefacts. Une petite quantité de lissage et de seuillage résout cela .. Devrait fonctionner parfaitement pour l'image png (je pense ...)
sans besoin de lissage et de seuillage.
Cela fonctionne totalement. Tu as raison. Sans avoir besoin de modifications supplémentaires, il suffit de le changer en png à partir de jpg. Merci beaucoup. J'ai maintenant plus de 2 réponses parfaites. Je ne sais pas quoi faire ni dire: D
@Ibrahim C. Kurt - vous chanceux;)
6

Voici un algorithme qui est O (nombre total de pixels + nombre de pixels des cellules). Nous scannons simplement l'image pour les pixels des cellules, et lorsque nous en trouvons un, nous remplissons la cellule pour l'effacer.

Implémentation en Common Lisp, mais vous pourrez la traduire trivialement en Python.

(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
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--> 30

la source
Hey Pascal, merci beaucoup pour la réponse. Je vois que votre programme trouve très clairement la bonne réponse. Le problème est que je n'ai aucune idée de ce langage Common Lisp, mais je vais essayer de comprendre cela et d'écrire un scripty similaire en Python.
3

Plus un commentaire étendu qu'une réponse:

Comme @interjay l'a laissé entendre, dans une image binaire, c'est-à-dire dans laquelle seulement 2 couleurs sont présentes, les pixels prennent la valeur 1 ou 0. Cela peut ou peut ne pas être vrai dans le format de représentation d'image que vous utilisez, mais c'est vrai dans la représentation «conceptuelle» de votre image; ne laissez pas les détails de mise en œuvre vous dérouter sur ce problème. L'un de ces détails d'implémentation est votre utilisation de 2 autour de la bordure de l'image - un moyen parfaitement sensé d'identifier la zone morte autour de l'image, mais sans affecter qualitativement la binaire de l'image.

Quant à l'examen des pixels N, NE, NW et W: il s'agit de la connectivité des pixels dans la formation du composant. Chaque pixel (sauf les cas spéciaux de bordure) a 8 voisins (N, S, E, W, NE, NW, SE, SW) mais lesquels sont candidats à l'inclusion dans le même composant? Parfois, les composants qui ne se rencontrent qu'aux coins (NE, NW, SE, SW) ne sont pas considérés comme connectés, parfois ils le sont.

Vous devez décider de ce qui convient à votre candidature. Je vous suggère de travailler, à la main, quelques opérations de l'algorithme séquentiel, en vérifiant différents voisins pour chaque pixel, pour avoir une idée de ce qui se passe.


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