如何计算环形内部的内部面积(像素数)? [英] How to calculate the internal area(count of pixels) inside a ring-like shape?

问题描述

• 我有以下图像，我想对环内的像素进行计数以获得面积.

• 我进行了一些形态学操作作为一种后处理，以尽可能使图像具有清晰的平滑边缘.
• 我尝试用不同的方式来做到这一点，就像您在代码中所看到的那样，但是没有一种是最佳的.
• 可以请教我如何计算圆的像素内部面积吗?注意:里面的一些像素不是全黑的，它们的强度很低，这就是为什么我要进行Otsu阈值处理.
• 预先感谢

• I did some morphological operations as a kind of post-processing to make the image as much as possible with clear smooth edges.
• I tried to do that in different ways as you can see in the code down, but none of them was optimal.
• Can you please advice me how to count the pixels inner area of the circle? Note: some pixels inside are not totally black, they are with low intensity, that's why I was trying to do Otsu thresholding.
• Thanks in advance

import numpy as np
import matplotlib.pyplot as plt
from skimage.io import imread, imsave
# import scipy.ndimage as ndi 
from skimage import morphology, filters, feature

seg = np.squeeze(imread('prediction.png')[...,:1])
# meijering alpha=None,
# rem2 = morphology.remove_small_objects(seg, 4)
resf = filters.meijering(seg, sigmas=range(1, 3, 1),  black_ridges=False)

sobel = filters.sobel(resf)
# diam = morphology.diameter_closing(sobel, 64, connectivity=2)
gaussian = filters.gaussian(sobel, sigma= 1)
val = filters.threshold_otsu(gaussian)
resth = gaussian < val 

# Morphology
SE = morphology.diamond(2)
# SE = np.ones((3,3))
# SE = morphology.disk(2)
# SE = square(7)
# SE = rectangle(3,3)
# SE = octagon(3, 3)

erosion  = morphology.binary_erosion( resth, SE).astype(np.uint8)
dilation = morphology.binary_dilation(resth, SE).astype(np.uint8)
opening  = morphology.binary_opening( resth, SE).astype(np.uint8)
closing  = morphology.binary_closing( resth, SE).astype(np.uint8)
#thinner = morphology.thin(erosion, max_iter=4)

rem  = morphology.remove_small_holes(resth, 2)

# entropy  = filters.rank.entropy(resth, SE) 
# print(seg.shape)

plt.figure(num='PProc')
# 1
plt.subplot('335')
plt.imshow(rem,cmap='gray')
plt.title('rem')
plt.axis('off')
# 2
plt.subplot('336')
plt.imshow(dilation,cmap='gray')
plt.title('dilation')
plt.axis('off')
# 3
plt.subplot('337')
plt.imshow(opening,cmap='gray')
plt.title('opening')
plt.axis('off')
# 4
plt.subplot('338')
plt.imshow(closing,cmap='gray')
plt.title('closing')
plt.axis('off')
# 5
plt.subplot('332')
plt.imshow(seg,cmap='gray')
plt.title('segmented')
plt.axis('off')
# 6
plt.subplot('333')
plt.imshow(resf,cmap='gray')
plt.title('meijering')
plt.axis('off')
# 7
# 8
plt.subplot('334')
plt.imshow(resth,cmap='gray')
plt.title('threshold_otsu')
plt.axis('off')
# 9
plt.subplot('339')
plt.imshow(erosion,cmap='gray')
plt.title('erosion')
plt.axis('off')
#
plt.show()

推荐答案

我确定我遗漏了一些东西，但是为什么不能仅仅使用 regionprops 阈值，标记图像并计算区域?>?

I am sure I am missing something, but why can't you just threshold, label the image, and compute your areas with regionprops?

#!/usr/bin/env python
"""
Determine areas in image of ring.

SO: https://stackoverflow.com/q/61681565/2912349
"""
import numpy as np
import matplotlib.pyplot as plt

from skimage.io import imread
from skimage.filters import threshold_otsu
from skimage.measure import label, regionprops
from skimage.color import label2rgb

if __name__ == '__main__':

    raw = imread('prediction.png', as_gray=True)
    threshold = threshold_otsu(raw)
    thresholded = raw > threshold
    # Label by default assumes that zeros correspond to "background".
    # However, we actually want the background pixels in the center of the ring,
    # so we have to "disable" that feature.
    labeled = label(thresholded, background=2)
    overlay = label2rgb(labeled)

    fig, axes = plt.subplots(1, 3)
    axes[0].imshow(raw, cmap='gray')
    axes[1].imshow(thresholded, cmap='gray')
    axes[2].imshow(overlay)

    convex_areas = []
    areas = []
    for properties in regionprops(labeled):
        areas.append(properties.area)
        convex_areas.append(properties.convex_area)

    # take the area with the smallest convex_area
    idx = np.argmin(convex_areas)
    area_of_interest = areas[idx]
    print(f"My area of interest has {area_of_interest} pixels.")
    # My area of interest has 714 pixels.
    plt.show()

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