如何计算细胞的核量？
How to calculate nucleus amount of cell?
我正在使用 Python 3.5 和 OpenCV 3 来分析生物学中的细胞图片。我的照片是这样的：
我希望能够计算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。还有棕褐色的红细胞，我想完全忽略它们。为清楚起见，这里有一张带标签的图片：
如何使用图像分割来识别和测量我感兴趣的区域？
我试过跟随 this guide，但它 returns 是一个全黑的图像。
# 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格。您可以以此为基础来使用分水岭算法。
此外，颜色分割参数已根据您发布的 2 张图像进行了调整。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好的妥协，你可能需要研究 otsu 二值化或自适应阈值来分割颜色。
另一种选择是查看 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，我们将在下面使用一些初步代码：
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()
作为参考，这是您的原始图片：
#Read in image
img = cv2.imread('cells.jpg')
ShowImage('Original',img,'bgr')
您链接到的文章建议使用 Otsu's method 进行颜色分割。该方法假设图像像素的强度可以绘制成双峰直方图，并找到该直方图的最佳分隔符。我使用下面的方法。
#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太好！查看灰度图像，您会明白原因：Otsu 变换产生 三个 类 像素：深色背景像素、甜甜圈细胞和细胞内部以及细胞核。下面的直方图证明了这一点：
#Make a histogram of the intensities in the grayscale image
plt.hist(gray.ravel(),256)
plt.show()
因此，您已经打破了所用算法的假设，因此得到不好的结果并不意外。通过丢弃颜色信息，我们失去了区分甜甜圈和细胞内部的能力。
处理此问题的一种方法是根据颜色阈值执行分割。为此，您需要选择一种颜色 space 来工作。This guide 对不同的 space 有出色的图片描述。
让我们选择 HSV。这具有单个通道 H 描述图像颜色的优点。一旦我们将图像转换成这个 space，我们就可以找到我们感兴趣的颜色的边界。例如，要找到细胞的细胞核，我们可以这样做：
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！但是，请注意细胞内部的某些部分被标记为核，即使它们不应该被标记。也有人可能会争辩说它不是很自动：您仍然需要仔细挑选您的颜色。在 HSV space 中操作消除了很多猜测，但也许我们可以利用有四种不同颜色这一事实来避免对范围的需求！为此，我们通过 k-means clustering algorithm.
传递我们的 HSV 像素
#Convert pixel space to an array of triplets. These are vectors in 3-space.
Z = hsv.reshape((-1,3))
#Convert to floating point
Z = np.float32(Z)
#Define the K-means criteria, these are not too important
criteria = (cv2.TERM_CRITERIA_EPS + cv2.TERM_CRITERIA_MAX_ITER, 10, 1.0)
#Define the number of clusters to find
K = 4
#Perform the k-means transformation. What we get back are:
#*Centers: The coordinates at the center of each 3-space cluster
#*Labels: Numeric labels for each cluster
#*Ret: A return code indicating whether the algorithm converged, &c.
ret,label,center = cv2.kmeans(Z,K,None,criteria,10,cv2.KMEANS_RANDOM_CENTERS)
#Produce an image using only the center colours of the clusters
center = np.uint8(center)
khsv = center[label.flatten()]
khsv = khsv.reshape((img.shape))
ShowImage('K-means',khsv,'hsv')
#Reshape labels for masking
label = label.reshape(img.shape[0:2])
ShowImage('K-means Labels',label,'gray')
请注意，这在分离颜色方面做得非常出色，无需手动指定！ （除了指定集群的数量。）
现在，我们需要弄清楚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：一个明显是细胞核像素，一个明显是细胞像素。然后我们找出哪个聚类中心最接近这些像素中的每一个。
#(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))
现在，让我们构建二元分类器，我们需要为分水岭算法识别整个单元格：
#Multiply by 1 to keep image in an integer format suitable for OpenCV
thresh = cv2.bitwise_or(1*(label==nuclei_label),1*(label==cell_label))
thresh = np.uint8(thresh)
ShowImage('Binary',thresh,'gray')
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')
我们现在需要识别分水岭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+细胞中都有一个像素，并且没有两个细胞核的标识符像素接触。
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。
但是，我们必须小心，因为具有高阈值的细长单元格可能会完全消失。在下图中，我们分离了右下角接触的两个单元格，但完全消除了右上角的细长单元格。
#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')
降低阈值会使又长又窄的单元格恢复原样，但会保留右下角的单元格连接。
我们可以使用一种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。这消除了为我们的阈值设置单个全局常量的需要。为此，我们使用 h_axima 函数，其中 returns 所有大于指定截止值的局部最大值。这与距离函数形成对比，距离函数返回所有大于给定值的像素。
#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，这些区域将被分水岭算法标记：
# Finding unknown region
unknown = cv2.subtract(opening,maxima)
ShowImage('Unknown',unknown,'gray')
接下来，我们给每个最大值赋予唯一标签，然后在最后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代表细胞的标记区域。接下来，我们遍历这些区域，将它们用作标记数据的掩码，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)))
这给出：
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，如果您将其用于学术目的，学术诚信需要适当的归属。请联系我了解详情。）
我正在使用 Python 3.5 和 OpenCV 3 来分析生物学中的细胞图片。我的照片是这样的：
我希望能够计算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。还有棕褐色的红细胞，我想完全忽略它们。为清楚起见，这里有一张带标签的图片：
如何使用图像分割来识别和测量我感兴趣的区域？
我试过跟随 this guide，但它 returns 是一个全黑的图像。
# 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格。您可以以此为基础来使用分水岭算法。 此外，颜色分割参数已根据您发布的 2 张图像进行了调整。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好的妥协，你可能需要研究 otsu 二值化或自适应阈值来分割颜色。
另一种选择是查看 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，我们将在下面使用一些初步代码：
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()
作为参考，这是您的原始图片：
#Read in image
img = cv2.imread('cells.jpg')
ShowImage('Original',img,'bgr')
您链接到的文章建议使用 Otsu's method 进行颜色分割。该方法假设图像像素的强度可以绘制成双峰直方图，并找到该直方图的最佳分隔符。我使用下面的方法。
#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太好！查看灰度图像，您会明白原因：Otsu 变换产生 三个 类 像素：深色背景像素、甜甜圈细胞和细胞内部以及细胞核。下面的直方图证明了这一点：
#Make a histogram of the intensities in the grayscale image
plt.hist(gray.ravel(),256)
plt.show()
因此，您已经打破了所用算法的假设，因此得到不好的结果并不意外。通过丢弃颜色信息，我们失去了区分甜甜圈和细胞内部的能力。
处理此问题的一种方法是根据颜色阈值执行分割。为此，您需要选择一种颜色 space 来工作。This guide 对不同的 space 有出色的图片描述。
让我们选择 HSV。这具有单个通道 H 描述图像颜色的优点。一旦我们将图像转换成这个 space，我们就可以找到我们感兴趣的颜色的边界。例如，要找到细胞的细胞核，我们可以这样做：
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！但是，请注意细胞内部的某些部分被标记为核，即使它们不应该被标记。也有人可能会争辩说它不是很自动：您仍然需要仔细挑选您的颜色。在 HSV space 中操作消除了很多猜测，但也许我们可以利用有四种不同颜色这一事实来避免对范围的需求！为此，我们通过 k-means clustering algorithm.
传递我们的 HSV 像素#Convert pixel space to an array of triplets. These are vectors in 3-space.
Z = hsv.reshape((-1,3))
#Convert to floating point
Z = np.float32(Z)
#Define the K-means criteria, these are not too important
criteria = (cv2.TERM_CRITERIA_EPS + cv2.TERM_CRITERIA_MAX_ITER, 10, 1.0)
#Define the number of clusters to find
K = 4
#Perform the k-means transformation. What we get back are:
#*Centers: The coordinates at the center of each 3-space cluster
#*Labels: Numeric labels for each cluster
#*Ret: A return code indicating whether the algorithm converged, &c.
ret,label,center = cv2.kmeans(Z,K,None,criteria,10,cv2.KMEANS_RANDOM_CENTERS)
#Produce an image using only the center colours of the clusters
center = np.uint8(center)
khsv = center[label.flatten()]
khsv = khsv.reshape((img.shape))
ShowImage('K-means',khsv,'hsv')
#Reshape labels for masking
label = label.reshape(img.shape[0:2])
ShowImage('K-means Labels',label,'gray')
请注意，这在分离颜色方面做得非常出色，无需手动指定！ （除了指定集群的数量。）
现在，我们需要弄清楚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：一个明显是细胞核像素，一个明显是细胞像素。然后我们找出哪个聚类中心最接近这些像素中的每一个。
#(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))
现在，让我们构建二元分类器，我们需要为分水岭算法识别整个单元格：
#Multiply by 1 to keep image in an integer format suitable for OpenCV
thresh = cv2.bitwise_or(1*(label==nuclei_label),1*(label==cell_label))
thresh = np.uint8(thresh)
ShowImage('Binary',thresh,'gray')
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')
我们现在需要识别分水岭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+细胞中都有一个像素，并且没有两个细胞核的标识符像素接触。
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。
但是，我们必须小心，因为具有高阈值的细长单元格可能会完全消失。在下图中，我们分离了右下角接触的两个单元格，但完全消除了右上角的细长单元格。
#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')
降低阈值会使又长又窄的单元格恢复原样，但会保留右下角的单元格连接。
我们可以使用一种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。这消除了为我们的阈值设置单个全局常量的需要。为此，我们使用 h_axima 函数，其中 returns 所有大于指定截止值的局部最大值。这与距离函数形成对比，距离函数返回所有大于给定值的像素。
#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，这些区域将被分水岭算法标记：
# Finding unknown region
unknown = cv2.subtract(opening,maxima)
ShowImage('Unknown',unknown,'gray')
接下来，我们给每个最大值赋予唯一标签，然后在最后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代表细胞的标记区域。接下来，我们遍历这些区域，将它们用作标记数据的掩码，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)))
这给出：
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，如果您将其用于学术目的，学术诚信需要适当的归属。请联系我了解详情。）