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Question

我一直在尝试根据序列对轮廓进行排序（这里的序列无关紧要）。我有一个非常小的问题，那就是我应该在下面的代码段中传递的正确的numpy数组，为此我要同时获得正确的行/列（非零）像素。

row_pixels=cv2.countNonZero(blur[cy][:])

col_pixels=cv2.countNonZero(blur[:,cx])

我所做的结果如下：对于所有5个轮廓，我得到的非零像素数量几乎相同，我意识到这是因为我正在传递“整个”图像（如您在上方看到的那样，是模糊的图像）作为用于计算像素的numpy数组，这是错误的，我意识到这一点。

当前输入：下图（无标记）

预期输出：对于所有5个轮廓，行/列（非零）像素。

我目前正在做的是：

import cv2
import numpy as np
from imutils import perspective
from imutils import contours 
import imutils 
from scipy.spatial import distance as dist
import argparse
import pandas as pd
import time

parser = argparse.ArgumentParser(description='Object Detection and Tracking using YOLO in OPENCV')
parser.add_argument('--image', help='Path to image file.')

args = parser.parse_args()
font=cv2.FONT_HERSHEY_SIMPLEX


start=time.time()
im_in = cv2.imread(args.image, 0)
_, thres2=cv2.threshold(im_in, 140, 255,cv2.THRESH_BINARY_INV)
dilate = cv2.dilate(thres2,None)
erode = cv2.erode(dilate,None)

im_3=erode.copy()

blur=cv2.medianBlur(im_3,5)

a=[]
r=[] 
row_col_pixel_values=[]
cl=[]
data=[]
global mainar
#find contours 
_,contour2,_=cv2.findContours(blur,cv2.RETR_CCOMP,cv2.CHAIN_APPROX_NONE)
# print(contour2)
for c in contour2:
    area=cv2.contourArea(c)
    if area>10000 and area <30000:
        a.append(area)

        cv2.drawContours(blur, [c], 0, (128, 128, 128), 1)


        M=cv2.moments(c)
        cx=int((M["m10"]/M["m00"]))
        cy=int((M["m01"]/M["m00"]))
        center =(cx,cy)
        data.append((cx,cy))
        cv2.circle(blur,(cx,cy), 5,(128,128,128),-1)
        print("",cx,cy)
        print(len(blur[cy][:])) 
        # one=blur[c]
        row_pixels=cv2.countNonZero(blur[cy][:])

        col_pixels=cv2.countNonZero(blur[:,cx])
        comb=(row_pixels,col_pixels)
        cl.append(comb)

nparea=np.array(a)
npcentercoord=np.array(data)

row_col_pixel_values=np.array(cl)
print("Area of 5 contours :",nparea)
print("Center coordinates of 5 contours:",npcentercoord)

print("Row and Column pixel values of 5 contours:",row_col_pixel_values)

mainar=np.column_stack((nparea,npcentercoord,row_col_pixel_values))
# print(mainar)

mainar[:,[1]] = (mainar[:,[1]]).astype(int)

MinX = int(min([_[1] for _ in mainar]))
MinlowerX = (MinX - 10) 
MinupperX = (MinX + 10)
MinY = int(min([_[2] for _ in mainar]))
MinlowerY = (MinY - 10) 
MinupperY = (MinY + 10)
MaxX = int(max([_[1] for _ in mainar]))
MaxlowerX = (MaxX - 10) 
MaxupperX = (MaxX + 10)
MaxY = int(max([_[2] for _ in mainar]))
MaxlowerY = (MaxY - 10)
MaxupperY = (MaxY + 10)

print("", MinX,MinY,MaxX,MaxY)


def PixeltoNumeric(channel,rowMM,colMM):

    if channel=="4S":
        for i in range(0, len(mainar[:,1])):
            cx=mainar[i,1]
            cy=mainar[i,2]
            if (cx in range(MinlowerX,MinupperX+1)) and (cy in range(MinlowerY,MinupperY+1)):
                rowp=mainar[i,3]
                colp=mainar[i,4]
                print("The center coordinates(x,y) and (Row/Col) pixels of 4Schannel: ")
                print("1 pixel has {:.5f} many mm in row:".format(rowMM/rowp))
                print("1 pixel has {:.5f} many mm in col:".format(colMM/colp))
                print(cx,cy,rowp,colp)

    if channel == '1':
            for i in range(0, len(mainar[:,1])):
                cx=mainar[i,1]
                cy=mainar[i,2]
                if (cx in range(MaxlowerX,MaxupperX+1)) and (cy in range(MaxlowerY,MaxupperY+1)):
                    rowp=mainar[i,3]
                    colp=mainar[i,4]
                    print("The center coordinates(x,y) and (Row/Col) pixels of 1Channel: ")
                    print("1 pixel has {:.5f} many mm in row:".format(rowMM/rowp))
                    print("1 pixel has {:.5f} many mm in col:".format(colMM/colp))
                    print(cx,cy,rowp,colp)

    if channel == '2':
        for i in range(0, len(mainar[:,1])):
            cx=mainar[i,1]
            cy=mainar[i,2]
            if (cx in range(MinlowerX,MinupperX+1)) and (cy in range(MaxlowerY,MaxupperY+1)):
                rowp=mainar[i,3]
                colp=mainar[i,4]
                print("The center coordinates(x,y) and (Row/Col) pixels of 2Channel: ")
                print("1 pixel has {:.5f} many mm in row:".format(rowMM/rowp))
                print("1 pixel has {:.5f} many mm in col:".format(colMM/colp))
                print(cx,cy,rowp,colp)

    if channel == '3':
        for i in range(0, len(mainar[:,1])):
            cx=mainar[i,1]
            cy=mainar[i,2]
            if (cx in range(((MinlowerX+MaxlowerX)//2),((MinupperX+MaxupperX+1)//2)) and (cy in range(((MinlowerY+MaxlowerY)//2),((MinupperY+MaxupperY+1)//2)))):
                rowp=mainar[i,3]
                colp=mainar[i,4]
                print("The center coordinates(x,y) and (Row/Col) pixels of 3Channel: ")
                print("1 pixel has {:.5f} many mm in row:".format(rowMM/rowp))
                print("1 pixel has {:.5f} many mm in col:".format(colMM/colp))
                print(cx,cy,rowp,colp)

    if channel == '4N':
        for i in range(0, len(mainar[:,1])):
            cx=mainar[i,1]
            cy=mainar[i,2]
            if (cx in range(MaxlowerX,MaxupperX+1)) and (cy in range(MinlowerY,MinupperY+1)):
                rowp=mainar[i,3]
                colp=mainar[i,4]
                print("The center coordinates(x,y) and (Row/Col) pixels of 4NChannel: ")
                print("1 pixel has {:.5f} many mm in row:".format(rowMM/rowp))
                print("1 pixel has {:.5f} many mm in col:".format(colMM/colp))
                print(cx,cy,rowp,colp)

    return (cv2.imshow("4",blur))


cv2.waitKey(0)
cv2.destroyAllWindows()

Answer 1

我不能完全确定我是否正确理解您，但是我的理解是您想找到所有轮廓内像素的所有坐标（x，y）。如果这是您的问题，则可以使用以下代码来实现：

import cv2
import matplotlib.pyplot as plt
import numpy as np

im_in = cv2.imread(r'image.png', 0)
_, thres2 = cv2.threshold(im_in, 140, 255, cv2.THRESH_BINARY_INV)
dilate = cv2.dilate(thres2, None)
erode = cv2.erode(dilate, None)
im_3 = erode.copy()
blur = cv2.medianBlur(im_3, 5)

# I am using OpenCV 4 therefore it returns only 4 parameters
contour2, _ = cv2.findContours(blur, cv2.RETR_CCOMP, cv2.CHAIN_APPROX_NONE)
extracted = np.zeros(blur.shape, np.uint8)

for c in contour2:
    area = cv2.contourArea(c)
    # I have modified these values to make it work for attached picture
    if 10000 < area < 300000: 
        cv2.drawContours(extracted, [c], 0, (255), cv2.FILLED)

contour_x, contour_y = np.nonzero(extracted)

plt.imshow(extracted, 'gray')
plt.show()

这是提取的image。

更新1

经过您的解释，我了解到您想计算每个单独轮廓的宽度和高度。根据您提供的示例代码，假设您要使用与轮廓中心交叉的线来测量宽度和高度。您可以通过在清晰的图像上绘制和测量轮廓来实现。请参见下面的代码：

# I am using OpenCV 4 therefore it returns only 4 parameters
contour2, _ = cv2.findContours(blur, cv2.RETR_CCOMP, cv2.CHAIN_APPROX_NONE)
extracted = np.zeros(blur.shape, np.uint8)
contoursSize = []
for c in contour2:
    area = cv2.contourArea(c)
    # I have modified these values to make it work for attached picture
    if 10000 < area < 300000:
        M = cv2.moments(c)
        cx = int((M["m10"] / M["m00"]))
        cy = int((M["m01"] / M["m00"]))
        extracted.fill(0) 
        cv2.drawContours(extracted, [c], 0, 255, cv2.FILLED)
        width = cv2.countNonZero(extracted[cy][:])
        height = cv2.countNonZero(extracted[:, cx])
        contoursSize.append((width, height))

收集所有不同轮廓的非零像素

1 个答案:

更新1