Contando o número de grupos de 1s em um mapa booleano de numpy.array

16

Agora, estou lidando com algum processamento de imagem em Python via PIL (Python Image Library). Meu principal objetivo é contar o número de células coloridas em uma imagem imuno-histoquímica. Sei que existem programas, bibliotecas, funções e tutoriais relevantes sobre o assunto, e verifiquei quase todos eles. Meu principal objetivo é escrever o código manualmente do zero, o máximo possível. Por isso, estou tentando evitar o uso de muitas bibliotecas e funções externas. Eu escrevi a maior parte do programa. Então, aqui está o que está acontecendo passo a passo:

O programa recebe o arquivo de imagem:exemplo

E processa para os glóbulos vermelhos (basicamente, desativa os valores RGB abaixo de um certo limite para o vermelho): insira a descrição da imagem aqui

E cria o mapa booleano dele ((vai colar uma parte dele porque é grande), que basicamente coloca 1 onde quer que encontre um pixel vermelho na segunda imagem processada acima. 

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

Eu intencionalmente gerei esse quadro como algo nas bordas com 2s para me ajudar a contar o número de grupos de 1s naquele mapa booleano.

Minha pergunta para vocês é: como é que eu posso contar com eficiência o número de células (grupos de 1s) nesse tipo de mapa booleano? Eu descobri http://en.wikipedia.org/wiki/Connected-component_labeling que parecem extremamente relacionados e semelhantes, mas, tanto quanto vejo, é no nível de pixel. O meu está no nível booleano. Apenas 1s e 0s.

Muito obrigado.

python Ibrahim C. Kurt
fonte
A etiquetagem de componentes conectados é exatamente o que você precisa. Não sei por que você acha que é diferente, pois o artigo da Wikipedia também tem exemplos começando com matrizes de 1s e 0s.
Sei que parece semelhante (ou talvez seja o mesmo), mas não consigo entender completamente a página da Wikipedia, pois o inglês não é minha língua nativa. A parte "Algoritmo sequencial" da página parece lidar com 1s e 0s, mas ainda não vi a lógica por trás disso. Forex, por que começa verificando norte, nordeste, noroeste e oeste?
Este problema está resolvido.
E se as células de interesse estiverem sobrepostas? Você não deveria procurar especificamente por recursos circulares para diferenciar duas células que parecem conectadas, e não apenas encontrar blobs contínuos?
Endolith
as coisas ficam muito mais complicadas quando as imagens imuno-histoquímicas que você tem são de má qualidade em termos de número de células sobrepostas, resolução, desvio padrão dos valores de pixel, tanto e mais forte ... Eu tenho tentado escrever um pequeno programa que funcione bem para tais fins, mas parece que a imagem de entrada e é condições são muito importantes para um resultado perfeito ..
Ibrahim C. Kurt

Respostas:

6

Uma abordagem de força bruta, mas feita ao inverter o problema para indexar sobre coleções de pixels para encontrar regiões, em vez de rasterizar sobre a matriz.

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)

Impressões:

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

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uau ... eu não sei o que dizer. Muito legal. E totalmente funciona. Obrigado Paul.
Tanto trabalho nisso, quando já existe uma função para Scipyfazer isso, o que provavelmente é mais rápido também ^^ 'Mas provavelmente é um bom exercício de qualquer maneira e mostra como fazer isso em geral. Vou votar então.
Zelphir Kaltstahl
13

Você pode usar ndimage.label, que é uma boa maneira de fazer isso. Ele retorna uma nova matriz, com todos os recursos com um valor único e o número de recursos. Você também pode especificar um elemento de conexão.

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

insira a descrição da imagem aqui

j08lue
fonte
Obrigado Fraxel. Isso funciona totalmente como uma solução rápida e suja, mas talvez eu deva melhorar a qualidade da imagem, pois como você pode ver, existem muitas células mescladas. A resposta deve ser 30 células. Muito obrigado mais uma vez. (edit: Eu tentei melhorar a qualidade da resolução da imagem e depois a borrou com seu código, mas ele ainda mescla muitas células. Deve ser da mesma forma que achatar = 1 ou ler) funciona?)
11
@Ibrahim C. Kurt - eu atualizei para corrigir isso (eu só notei!). O problema era que a imagem carregada era jpg, e havia muitos artefatos. Pequena quantidade de suavização e limiarizar resolve isso .. deve funcionar perfeitamente para png imagem (eu acho ...)
sem a necessidade de suavização e limiarização.
Isso funciona totalmente. Você está certo. Sem a necessidade de qualquer modificação adicional, basta alterá-lo para png de jpg é suficiente. Muito obrigado. Eu tenho mais de 2 respostas perfeitas agora. Não sei o que fazer e dizer: D
@Ibrahim C.Kurt - lucky you;)
6

Aqui está um algoritmo que é O (número total de pixels + número de pixels da célula). Simplesmente digitalizamos a imagem em busca de pixels da célula e, quando a encontramos, preenchemos a célula para apagá-la.

Implementação no Common Lisp, mas você poderá traduzi-lo para o Python trivialmente.

(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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                #171*000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000111111111000000000000000000001110000
                #171*000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000011111110000000000000000000011111100
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                #171*000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000111111100
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                #171*000000000000000000000000001111111100011111110000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000)))
--> 30

fonte
Ei Pascal, muito obrigado pela resposta. Vejo que seu programa encontra a resposta correta de maneira muito limpa. O problema é que não tenho idéia sobre essa linguagem Common Lisp, mas tentarei descobrir isso e escrever um script semelhante em Python.
3

Mais um comentário extenso do que uma resposta:

Como @interjay sugeriu, em uma imagem binária, ou seja, na qual apenas duas cores estão presentes, os pixels assumem o valor 1 ou 0. Isso pode ou não ser verdade no formato de representação de imagem que você está usando, mas é verdade na representação "conceitual" da sua imagem; não deixe que os detalhes da implementação o confundam sobre esse problema. Um desses detalhes de implementação é o uso de 2s na borda da imagem - uma maneira perfeitamente sensata de identificar a zona morta ao redor da imagem, mas não afetando qualitativamente o binário da imagem.

Quanto ao exame dos pixels N, NE, NW e W: trata-se da conectividade dos pixels na formação do componente. Cada pixel (com exceção dos casos especiais da borda) tem 8 vizinhos (N, S, E, W, NE, NW, SE, SW), mas quais são candidatos à inclusão no mesmo componente? Às vezes, os componentes que se encontram apenas nos cantos (NE, NW, SE, SW) não são considerados conectados, às vezes são.

Você precisa decidir o que é apropriado para sua aplicação. Sugiro que você trabalhe, manualmente, algumas operações do algoritmo seqüencial, verificando vizinhos diferentes para cada pixel, para ter uma ideia do que está acontecendo.


fonte