Contando il numero di gruppi di 1 in una mappa booleana di numpy.array

16

In questo momento mi occupo dell'elaborazione delle immagini in Python tramite PIL (Python Image Library). Il mio obiettivo principale è contare il numero di cellule colorate in un'immagine immunoistochimica. So che ci sono programmi, librerie, funzioni ed esercitazioni pertinenti su di esso, e ho controllato quasi tutti. Il mio obiettivo principale è scrivere il codice manualmente da zero, per quanto possibile. Quindi sto cercando di evitare di usare molte librerie e funzioni esterne. Ho scritto la maggior parte del programma. Quindi ecco cosa sta succedendo passo dopo passo:

Il programma accetta il file immagine:esempio

E lo elabora per i globuli rossi (sostanzialmente, disattiva i valori RGB al di sotto di una certa soglia per il rosso): inserisci qui la descrizione dell'immagine

E crea la sua mappa booleana, (ne incollerà una parte poiché è grande) che sostanzialmente mette 1 ovunque si trovi con un pixel rosso nella seconda immagine elaborata sopra. 

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

Ho intenzionalmente generato quel frame come qualcosa sui bordi con 2s per aiutarmi a contare il numero di gruppi di 1 in quella mappa booleana.

La mia domanda per voi ragazzi è: come mai posso contare in modo efficiente il numero di celle (gruppi di 1) in quel tipo di mappa booleana? Ho trovato http://en.wikipedia.org/wiki/Connected-component_labeling che sembrano estremamente correlati e simili, ma per quanto vedo, è a livello di pixel. Il mio è a livello booleano. Solo 1s e 0s.

Molte grazie.

python 
— Ibrahim C. Kurt
fonte
L'etichettatura dei componenti collegati è esattamente ciò di cui hai bisogno. Non so perché pensi che sia diverso, poiché l'articolo di Wikipedia ha anche esempi che iniziano con matrici di 1 e 0.
So che sembra simile (o forse lo stesso), non riesco a cogliere appieno l'intera pagina di Wikipedia poiché l'inglese non è la mia lingua madre. La parte "Algoritmo sequenziale" della pagina sembra avere a che fare con 1 e 0 ma non ho ancora visto la logica dietro di essa. Forex, perché inizia controllando nord, nord-est, nord-ovest e ovest?
Questo problema è stato risolto
Cosa succede se le celle di interesse si sovrappongono? Non dovresti cercare specificamente le caratteristiche circolari in modo da poter differenziare due celle che sembrano connesse, non solo trovare BLOB continui?
— endolith
le cose diventano molto più complicate quando le immagini immunoistochimiche che hai sono di cattiva qualità in termini di numero di celle sovrapposte, risoluzione, deviazione standard dei valori dei pixel così e così forte ... Ho cercato di scrivere un piccolo programma che funzioni bene per tali scopi ma sembra l'immagine di input e le sue condizioni sono molto importanti per un risultato perfetto ..
— Ibrahim C. Kurt

Risposte:

6

Qualcosa di un approccio a forza bruta, ma fatto invertendo il problema per indicizzare raccolte di pixel per trovare regioni, invece di rasterizzare sull'array.

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)

stampe:

4
........................................
........00000000........................
........00000000........................
........00000000........................
.........000000.........1.........33....
..........00000........11........33333..
...........000...........2......333333..
................................333333..
................................3333333.
...............................33333333.
...............................33333333.
................................333333..
.................................3......
........................................
wow ... non so cosa dire. Molto bello. E funziona totalmente. Grazie Paolo
Tanto lavoro in questo, quando c'è già una funzione nel Scipyfare questo, che è probabilmente anche più veloce ^^ 'Ma probabilmente comunque un buon esercizio e mostra come farlo in generale. Voterò allora.
— Zelphir Kaltstahl,
13

Puoi usare ndimage.label, che è un bel modo per farlo. Restituisce un nuovo array, con ogni funzione con un valore univoco e il numero di funzionalità. È inoltre possibile specificare un elemento di connessione.

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

inserisci qui la descrizione dell'immagine

— j08lue
fonte
Grazie fraxel. Funziona totalmente come una soluzione rapida e sporca, ma forse dovrei migliorare la qualità dell'immagine, perché come puoi vedere ci sono molte celle unite. La risposta dovrebbe essere di 30 celle. Grazie mille ancora una volta. (modifica: ho provato a migliorare la qualità della risoluzione dell'immagine e poi l'ho cancellata con il tuo codice ma unisce ancora molte celle. Deve essere nel modo in cui flatten = 1 o imread funziona?)
1
@Ibrahim C. Kurt - L'ho aggiornato per correggerlo (l'ho appena notato!). Il problema era che l'immagine caricata era jpg, quindi c'erano molti artefatti. Una piccola quantità di livellamento e soglia risolve il problema. Dovrebbe funzionare perfettamente per l'immagine png (penso ...)
senza necessità di livellamento e soglia.
Funziona totalmente. Hai ragione. Senza la necessità di ulteriori modifiche, è sufficiente cambiarlo in png da jpg. Grazie mille. Ora ho più di 2 risposte perfette. Non so cosa fare e dire: D
@Ibrahim C. Kurt - lucky you;)
6

Ecco un algoritmo che è O (numero totale di pixel + numero di pixel delle celle). Scansioniamo semplicemente l'immagine per i pixel delle celle e quando ne troviamo una riempiamo la cella per inondarla per cancellarla.

Implementazione in Common Lisp, ma sarai in grado di tradurlo banalmente in 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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                #171*000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000111111111000000000000000000001110000
                #171*000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000011111110000000000000000000011111100
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                #171*000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000111111100
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                #171*000000000000000000000000001111111100011111110000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000)))
--> 30
Ehi Pascal, grazie mille per la risposta. Vedo che il tuo programma trova la risposta corretta in modo molto chiaro. Il problema è che non ho idea di quel linguaggio Common Lisp, ma proverò a capirlo e scriverò un simile script in Python.
3

Più di un commento esteso che di una risposta:

Come ha suggerito @interjay, in un'immagine binaria, ovvero in una immagine in cui sono presenti solo 2 colori, i pixel assumono il valore 1 o 0. Ciò può o non può essere vero nel formato di rappresentazione dell'immagine che si sta utilizzando ma è vero nella rappresentazione "concettuale" della tua immagine; non lasciare che i dettagli di implementazione ti confondano su questo problema. Uno di questi dettagli di implementazione è l'uso di 2s attorno al bordo dell'immagine, un modo perfettamente sensato di identificare la zona morta intorno all'immagine, ma non influisce qualitativamente sulla binarietà dell'immagine.

Per quanto riguarda l'esame dei pixel N, NE, NW e W: ciò ha a che fare con la connettività dei pixel nella formazione del componente. Ogni pixel (esclude i casi speciali del bordo) ha 8 vicini (N, S, E, W, NE, NW, SE, SW) ma quali sono candidati per l'inclusione nello stesso componente? A volte i componenti che si incontrano solo agli angoli (NE, NW, SE, SW) non sono considerati collegati, a volte lo sono.

Devi decidere cosa è appropriato per la tua applicazione. Ti suggerisco di elaborare, a mano, alcune operazioni dell'algoritmo sequenziale, controllando diversi vicini per ciascun pixel, per avere un'idea di ciò che sta succedendo.

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