计数齿轮（Python，OpenCV）

Question

For a prototype I need to build a 3d model of a gear. This have a "many" number of teeth. So I am trying to count them using OpenCV and Python. I found this (only?) post which explain how to do it in C++.

I am following the steps and, for now this is the code I made.

import numpy as np
import cv2

img = cv2.imread('C:\\Users\\Link\\Desktop\\gear.png')

gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)

ret, thresh = cv2.threshold(gray, 127, 255, cv2.THRESH_BINARY_INV | cv2.THRESH_OTSU)

kernel = np.ones((3, 3), np.uint8)

img_erosion = cv2.erode(thresh, kernel, iterations=1)

edges = cv2.Canny(img_erosion, 50, 150)

img_dilate = cv2.dilate(edges, kernel, iterations=1)

cv2.imshow('i', thresh)
cv2.waitKey(0)
cv2.imshow('i', img_erosion)
cv2.waitKey(0)
cv2.imshow('i', edges)
cv2.waitKey(0)
cv2.imshow('i', img_dilate)
cv2.waitKey(0)

What stopped me from go ahead is this: the image at some point became really a mess.

This is the original on which I am working:

And this is the output of image_dilate

As you can see, the teeth at the bottom is not displayed properly, maybe because of the shaddow in the original image. How can I get rid of this ?

Answer 1

Because your source image is cleaner than the link your post, so you can do approx on the max-area-contour , then get half number of points , the result is 84 .

---

Sample code:

#!/usr/bin/python3
# 2018.01.22 11:53:24 CST
import cv2
import myutils

## Read
img = cv2.imread("img13_2.jpg")
gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)

## threshold and find contours
ret, threshed = cv2.threshold(gray, 50, 255, cv2.THRESH_BINARY_INV)
cnts= cv2.findContours(threshed, cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_NONE)[-2]

## Find the max-area-contour
cnt = max(contours, key=cv2.contourArea)

## Approx the contour
arclen = cv2.arcLength(cnt, True)
approx = cv2.approxPolyDP(cnt, 0.002*arclen, True)

## Draw and output the result
for pt in approx:
    cv2.circle(img, (pt[0][0],pt[0][1]), 3, (0,255,0), -1, cv2.LINE_AA)

msg = "Total: {}".format(len(approx)//2)
cv2.putText(img, msg, (20,40),cv2.FONT_HERSHEY_PLAIN, 2, (0,0,255), 2, cv2.LINE_AA)

## Display
cv2.imshow("res", img);cv2.waitKey()

---

Result:

Answer 2

Solved it..

This is the code. The count is wrong by one because one teeth, on the right is lower than the others and because it found two points by itself. Don't know why this happens.

Also, it has been made with another image. It's not the source I posted above as long as it is in low definition.

import numpy as np
import cv2

img = cv2.imread('C:\\Users\\Link\\Desktop\\gear.png')
img2 = cv2.imread('C:\\Users\\Link\\Desktop\\gear.png')

gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)

ret, thresh = cv2.threshold(gray, 127, 255, cv2.THRESH_BINARY_INV | cv2.THRESH_OTSU)

kernel = np.ones((3, 3), np.uint8)

img_dilate = cv2.dilate(thresh, kernel, iterations=1)

im2, contours, hierarchy = cv2.findContours(img_dilate, cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)
cnts = cv2.drawContours(img, contours, -1, (0, 255, 0), -1)

edges = cv2.Canny(cnts, 350, 350)

cnt = contours[0]

hull = cv2.convexHull(cnt, returnPoints=False)
defects = cv2.convexityDefects(cnt, hull)

for i in range(defects.shape[0]):
    s, e, f, d = defects[i, 0]
    start = tuple(cnt[s][0])
    end = tuple(cnt[e][0])
    far = tuple(cnt[f][0])
    cv2.line(edges, start, end, [0, 255, 255], 1)
    circles = cv2.circle(img2, end, 5, [0, 255, 0], -1)


# print(len(defects)) - number of points

cv2.imshow('thresh', thresh)
cv2.waitKey(0)
cv2.imshow('dilate', img_dilate)
cv2.waitKey(0)
cv2.imshow('edges', edges)
cv2.waitKey(0)
cv2.imshow('cnts', cnts)
cv2.waitKey(0)
cv2.imshow('points', circles)
cv2.waitKey(0)