Python 如何计算细胞核的数量？

我正在使用Python 3.5和OpenCV 3分析生物学中的细胞图片。我的照片是这样的：

我想计算出细胞核面积与整个细胞面积的比率

在我的幻灯片中，细胞核是深紫色，细胞的其他区域是浅蓝色。还有一些棕褐色的红血球，我想完全忽略它们。为了清晰起见，这里有一个带标签的图像：

如何使用图像分割来识别和测量感兴趣的区域

我尝试了跟踪，但它返回一个完全黑色的图像

# light purple color segmentation (to get cells)
cell_hsvmin = (110,40,145)
cell_hsvmax = (150,190,255)

hsv = cv2.cvtColor(frame, cv2.COLOR_BGR2HSV)
color_thresh = cv2.inRange(hsv, cell_hsvmin, cell_hsvmax)
# masked = cv2.bitwise_and(frame,frame, mask=color_thresh)
# cv2.imshow('masked0', masked)
ksize = 5
open_thresh = cv2.morphologyEx(color_thresh, cv2.MORPH_OPEN, np.ones((ksize,ksize),'uint8'), iterations=1)
masked = cv2.bitwise_and(frame,frame, mask=open_thresh)
cv2.imshow('masked', masked)

# dark purple color segmentation (to get nucleus)
nucleus_hsvmin = (125,65,160)
nucleus_hsvmax = (150,190,255)

nucleus_color_thresh = cv2.inRange(hsv, nucleus_hsvmin, nucleus_hsvmax)
ksize = 3
nucleus_open_thresh = cv2.morphologyEx(nucleus_color_thresh, cv2.MORPH_OPEN, np.ones((ksize,ksize),'uint8'), iterations=1)
nucleus_masked = cv2.bitwise_and(masked,masked, mask=nucleus_open_thresh)
cv2.imshow('nucleus_masked', nucleus_masked)

"""
HULL APPROXIMATES THE CELLS TO A CIRCLE TO FILL IN GAPS CREATED BY THRESHOLDING AND CLOSING.
FOR NON-CIRCULAR CELLS LIKE IN YOUR SECOND IMAGE, THIS MIGHT CAUSE BAD AREA CALCULATIONS
"""
# doHULL = False
doHULL = True

cells = []
cells_ratio = []
minArea = frame.shape[0]*frame.shape[1]* 0.01
_, contours, _ = cv2.findContours(open_thresh,cv2.RETR_EXTERNAL,cv2.CHAIN_APPROX_SIMPLE)
for cnt in contours:
    area = cv2.contourArea(cnt)
    if area >= minArea:
        cells.append(cnt)
        nucleus_mask = np.zeros(frame.shape[:2], 'uint8')
        if doHULL:
            hull = cv2.convexHull(cnt)
            cv2.drawContours(nucleus_mask, [hull], -1, 255, -1);
        else:
            cv2.drawContours(nucleus_mask, [cnt], -1, 255, -1);
        nucleus_mask = cv2.bitwise_and(nucleus_open_thresh, nucleus_mask)
        nucleus_area = np.count_nonzero(nucleus_mask)

        ratio = nucleus_area / area
        cells_ratio.append(ratio)

        # nucleus_img = cv2.bitwise_and(frame, frame, mask=nucleus_mask)
        # cv2.imshow('nucleus_img', nucleus_img)
        # cv2.waitKey(0)

doDRAWCELLS = False
# doDRAWCELLS = True
if doDRAWCELLS:
    for cell_cnt in cells:
        cells_mask = np.zeros(frame.shape[:2], 'uint8')
        if doHULL:
            hull = cv2.convexHull(cell_cnt)
            cv2.drawContours(cells_mask, [hull], -1, 255, -1);
        else:
            cv2.drawContours(cells_mask, [cell_cnt], -1, 255, -1);
        cells_img = cv2.bitwise_and(frame, frame, mask=cells_mask)
        cv2.imshow('cells_img', cells_img)
        cv2.waitKey(0)

这仅适用于未连接的单元。您可以将其用作使用分水岭算法的基础。 此外，颜色分割参数已根据您发布的2幅图像进行了调整。其他幻灯片可能会偏离颜色范围，因此您可能需要调整它们。如果调整它们不能让你得到一个很好的折衷，你可能不得不研究大津二值化或自适应阈值分割的颜色

另一个选择是查看cv2.MORPH_GRADIENT，它的工作原理类似于边缘检测器。或

gray = cv2.cvtColor(frame, cv2.COLOR_BGR2GRAY)
kernel = np.array([[1,1,1],[1,-8,1],[1,1,1]],dtype='float32')
laplace = cv2.filter2D(cv2.GaussianBlur(gray,(blur_ksize,blur_ksize),0), -1, kernel)
cv2.imshow('laplace', laplace)

---

并使用边缘分割单元格？

首先，我们将使用以下一些初步代码：

import numpy as np
import cv2
from matplotlib import pyplot as plt
from skimage.morphology import extrema
from skimage.morphology import watershed as skwater

def ShowImage(title,img,ctype):
  if ctype=='bgr':
    b,g,r = cv2.split(img)       # get b,g,r
    rgb_img = cv2.merge([r,g,b])     # switch it to rgb
    plt.imshow(rgb_img)
  elif ctype=='hsv':
    rgb = cv2.cvtColor(img,cv2.COLOR_HSV2RGB)
    plt.imshow(rgb)
  elif ctype=='gray':
    plt.imshow(img,cmap='gray')
  elif ctype=='rgb':
    plt.imshow(img)
  else:
    raise Exception("Unknown colour type")
  plt.title(title)
  plt.show()

以下是您的原始图像供参考：

#Read in image
img         = cv2.imread('cells.jpg')
ShowImage('Original',img,'bgr')

您链接到的文章建议使用颜色分割。该方法假设图像像素的强度可以绘制成双峰直方图，并为该直方图找到最佳分隔符。我采用下面的方法

#Convert to a single, grayscale channel
gray        = cv2.cvtColor(img,cv2.COLOR_BGR2GRAY)
#Threshold the image to binary using Otsu's method
ret, thresh = cv2.threshold(gray,0,255,cv2.THRESH_BINARY_INV+cv2.THRESH_OTSU)
ShowImage('Grayscale',gray,'gray')
ShowImage('Applying Otsu',thresh,'gray')

图像的二进制形式不是很好！查看灰度图像，您可以看到原因：大津变换产生三类像素：暗背景像素、圆环细胞和细胞内部以及细胞核。下面的柱状图说明了这一点：

#Make a histogram of the intensities in the grayscale image
plt.hist(gray.ravel(),256)
plt.show()

因此，您已经打破了正在使用的算法的假设，因此您得到的结果并不意外。由于丢掉了颜色信息，我们已经失去了区分甜甜圈和细胞内部的能力

处理这种情况的一种方法是基于颜色阈值进行分割。为此，请选择要使用的颜色空间。对不同的空间有很好的图像描述

让我们选择HSV。这有一个优点，即单个通道

H

，用于描述图像的颜色。一旦我们将图像转换到这个空间，我们就可以找到我们感兴趣的颜色的界限。例如，要找到细胞核，我们可以做如下操作：

cell_hsvmin  = (110,40,145)  #Lower end of the HSV range defining the nuclei
cell_hsvmax  = (150,190,255) #Upper end of the HSV range defining the nuclei
#Transform image to HSV color space
hsv          = cv2.cvtColor(img,cv2.COLOR_BGR2HSV) 
#Threshold based on HSV values
color_thresh = cv2.inRange(hsv, cell_hsvmin, cell_hsvmax) 
ShowImage('Color Threshold',color_thresh,'gray')

masked = cv2.bitwise_and(img,img, mask=color_thresh)
ShowImage('Color Threshold Maksed',masked,'bgr')

这看起来好多了！尽管如此，请注意，细胞内部的某些部分被标记为Nucleuii，尽管它们不应该被标记为Nucleuii。也有人可能会说，它不是很自动的：你仍然需要仔细地手工挑选你的颜色。在HSV空间中操作消除了许多猜测，但也许我们可以利用四种不同颜色的事实来避免对范围的需要！为此，我们将HSV像素通过一个

请注意，这在无需手动指定的情况下就可以很好地分离颜色！（除了指定簇的数目。）

现在，我们需要找出哪些标签对应于单元格的哪些部分

为此，我们找到了两个像素的颜色：一个是核像素，另一个是细胞像素。然后，我们找出哪个簇中心最接近这些像素

#(Distance,Label) pairs
nucleus_colour = np.array([139, 106, 192])
cell_colour    = np.array([130, 41,  207])
nuclei_label  = (np.inf,-1)
cell_label    = (np.inf,-1)
for l,c in enumerate(center):
  print(l,c)
  dist_nuc = np.sum(np.square(c-nucleus_colour)) #Euclidean distance between colours
  if dist_nuc<nuclei_label[0]:
        nuclei_label=(dist_nuc,l)
  dist_cell = np.sum(np.square(c-cell_colour)) #Euclidean distance between colours
  if dist_cell<cell_label[0]:
        cell_label=(dist_cell,l)
nuclei_label = nuclei_label[1]
cell_label   = cell_label[1]
print("Nuclei label={0}, cell label={1}".format(nuclei_label,cell_label))

我们现在可以消除单像素噪声：

#Remove noise by eliminating single-pixel patches
kernel  = np.ones((3,3),np.uint8)
opening = cv2.morphologyEx(thresh,cv2.MORPH_OPEN, kernel, iterations = 2)
ShowImage('Opening',opening,'gray')

我们现在需要确定流域的峰值，并给它们单独的标签。这样做的目的是生成一组像素，使得每个细胞核+细胞内都有一个像素，并且没有两个细胞核的标识符像素相接触

为了实现这一点，我们可以进行距离变换，然后过滤掉距离细胞核+细胞中心两倍远的距离

然而，我们必须小心，因为具有高阈值的长而窄的细胞可能会完全消失。在下图中，我们分离了右下角接触的两个细胞，但完全消除了右上角的长而窄的细胞

#Identify areas which are surely foreground
fraction_foreground = 0.75
dist         = cv2.distanceTransform(opening,cv2.DIST_L2,5)
ret, sure_fg = cv2.threshold(dist,fraction_foreground*dist.max(),255,0)
ShowImage('Distance',dist_transform,'gray')
ShowImage('Surely Foreground',sure_fg,'gray')

降低阈值会使长而窄的单元格返回，但会使右下角的单元格保持连接

我们可以通过使用自适应方法来解决这个问题，该方法可以识别每个局部区域的峰值。这样就不需要为阈值设置单个全局常量。为此，我们使用

h_axima

函数，该函数返回大于指定截止值的所有局部最大值。这与距离函数形成对比，距离函数返回大于给定值的所有像素

#Identify areas which are surely foreground
h_fraction = 0.1
dist     = cv2.distanceTransform(opening,cv2.DIST_L2,5)
maxima   = extrema.h_maxima(dist, h_fraction*dist.max())
print("Peaks found: {0}".format(np.sum(maxima)))
#Dilate the maxima so we can see them
maxima   = cv2.dilate(maxima, kernel, iterations=2)
ShowImage('Distance',dist_transform,'gray')
ShowImage('Surely Foreground',maxima,'gray')

现在，我们通过减去最大值来识别未知区域，即将由分水岭算法标记的区域：

# Finding unknown region
unknown = cv2.subtract(opening,maxima)
ShowImage('Unknown',unknown,'gray')

接下来，我们给每个maxima唯一标签，然后在最终执行分水岭变换之前标记未知区域：

# Marker labelling
ret, markers = cv2.connectedComponents(maxima)
ShowImage('Connected Components',markers,'rgb')

# Add one to all labels so that sure background is not 0, but 1
markers = markers+1

# Now, mark the region of unknown with zero
markers[unknown==np.max(unknown)] = 0

ShowImage('markers',markers,'rgb')

dist    = cv2.distanceTransform(opening,cv2.DIST_L2,5)
markers = skwater(-dist,markers,watershed_line=True)

ShowImage('Watershed',markers,'rgb')
imgout = img.copy()
imgout[markers == 0] = [0,0,255] #Label the watershed_line

ShowImage('img',imgout,'bgr')

这为我们提供了一组表示单元的标记区域。接下来，我们迭代这些区域，将它们用作标记数据的遮罩，并计算分数：

for l in np.unique(markers):
    if l==0:      #Watershed line
        continue
    if l==1:      #Background
        continue
    #For displaying individual cells
    #temp=khsv.copy()
    #temp[markers!=l]=0
    #ShowImage('out',temp,'hsv')
    temp = label.copy()
    temp[markers!=l]=-1
    nucleus_area = np.sum(temp==nuclei_label)
    cell_area    = np.sum(temp==cell_label)
    print("Nucleus fraction for cell {0} is {1}".format(l,nucleus_area/(cell_area+nucleus_area)))

这使得：

Nucleus fraction for cell 2 is 0.9002795899347623
Nucleus fraction for cell 3 is 0.7953321364452424
Nucleus fraction for cell 4 is 0.7525925925925926
Nucleus fraction for cell 5 is 0.8151515151515152
Nucleus fraction for cell 6 is 0.6808656818962556
Nucleus fraction for cell 7 is 0.8276481149012568
Nucleus fraction for cell 8 is 0.878500237304224
Nucleus fraction for cell 9 is 0.8342518016108521
Nucleus fraction for cell 10 is 0.9742324561403509
Nucleus fraction for cell 11 is 0.8728733459357277
Nucleus fraction for cell 12 is 0.7968570333461096
Nucleus fraction for cell 13 is 0.8226831716293075
Nucleus fraction for cell 14 is 0.7491039426523297
Nucleus fraction for cell 15 is 0.839096357768557
Nucleus fraction for cell 16 is 0.7589670014347202
Nucleus fraction for cell 17 is 0.8559168925022583
Nucleus fraction for cell 18 is 0.7534142640364189
Nucleus fraction for cell 19 is 0.8036734693877551
Nucleus fraction for cell 20 is 0.7566037735849057

---

（请注意，如果您将此用于学术目的，学术诚信需要正确的归属。有关详细信息，请与我联系。）

能否发布您尝试使用的代码？该指南的链接很好，但我们仍在猜测您可能做了什么导致了您所看到的问题。

for l in np.unique(markers):
    if l==0:      #Watershed line
        continue
    if l==1:      #Background
        continue
    #For displaying individual cells
    #temp=khsv.copy()
    #temp[markers!=l]=0
    #ShowImage('out',temp,'hsv')
    temp = label.copy()
    temp[markers!=l]=-1
    nucleus_area = np.sum(temp==nuclei_label)
    cell_area    = np.sum(temp==cell_label)
    print("Nucleus fraction for cell {0} is {1}".format(l,nucleus_area/(cell_area+nucleus_area)))

Nucleus fraction for cell 2 is 0.9002795899347623
Nucleus fraction for cell 3 is 0.7953321364452424
Nucleus fraction for cell 4 is 0.7525925925925926
Nucleus fraction for cell 5 is 0.8151515151515152
Nucleus fraction for cell 6 is 0.6808656818962556
Nucleus fraction for cell 7 is 0.8276481149012568
Nucleus fraction for cell 8 is 0.878500237304224
Nucleus fraction for cell 9 is 0.8342518016108521
Nucleus fraction for cell 10 is 0.9742324561403509
Nucleus fraction for cell 11 is 0.8728733459357277
Nucleus fraction for cell 12 is 0.7968570333461096
Nucleus fraction for cell 13 is 0.8226831716293075
Nucleus fraction for cell 14 is 0.7491039426523297
Nucleus fraction for cell 15 is 0.839096357768557
Nucleus fraction for cell 16 is 0.7589670014347202
Nucleus fraction for cell 17 is 0.8559168925022583
Nucleus fraction for cell 18 is 0.7534142640364189
Nucleus fraction for cell 19 is 0.8036734693877551
Nucleus fraction for cell 20 is 0.7566037735849057