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Detecting lines and shapes in OpenCV using Python

I've been playing around with OpenCV (cv2) and detecting lines and shapes. Say that my daughter drew a drawing, like so:

enter image description here

I am trying to write a Python script that would analyze the drawing and convert it into hard lines/shapes, something like:

enter image description here

That being said, I have installed opencv and tried to play around with it, but have had no luck aside from being able to draw a single vertical line through the image. Below is my code so far, any pointers or suggestions as to how I should go about doing this with opencv would be greatly appreciated.

import cv2
import numpy as np

class File(object):
    def __init__(self, filename):
        self.filename = filename

    def open(self, filename=None, mode='r'):
        if filename is None:
            filename = self.filename

        return cv2.imread(filename), open(filename, mode)

    def save(self, image=None, filename_override=None):
        filename = "output/" + self.filename.split('/')[-1]

        if filename_override:
            filename = "output/" + filename_override

        return cv2.imwrite(filename, image)

class Image(object):
    def __init__(self, image):
        self.image = image

    def grayscale(self):
        return cv2.cvtColor(self.image, cv2.COLOR_BGR2GRAY)

    def edges(self):
        return cv2.Canny(self.image, 0, 255)

    def lines(self):
        lines = cv2.HoughLinesP(self.image, 1, np.pi/2, 6, None, 50, 10)
        for line in lines[0]:
            pt1 = (line[0],line[1])
            pt2 = (line[2],line[3])
            cv2.line(self.image, pt1, pt2, (0,0,255), 2)

if __name__ == '__main__':
    File = File('images/a.png')
    Image = Image(File.open()[0])
    Image.image = Image.grayscale()
    Image.lines()
    File.save(Image.image)

Unfortunately, for a simple square drawing, all I get back is:

enter image description here

where the vertical line in the box is the output from the code.

like image 637
jsanc623 Avatar asked Aug 12 '15 20:08

jsanc623


2 Answers

Here is my attempt. It's in C++, but can be easily ported to python since most are OpenCV functions.

A brief outline of the method, comments in the code should help, too.

  1. Load the image
  2. Convert to grayscale
  3. Binaryze the image (threshold)
  4. Thinning, to have thin contours and help findContours
  5. Get contours
  6. For each contour, get convex hull (to handle open contours), and classify according to circularity. Handle each shape differently.

    • Circle : find the minimum encolsing circle, or the best fitting ellipse
    • Recrangle : find the boundinx box, or the minimum oriented bounding box.
    • Triangle : search for the intersection of the minimum enclosing circle with the original shape, as they would intersect in the three vertices of the triangle.

NOTES:

  • I needed to modify the original image to 3 channel RGB from a png with transparency.
  • The thinning code is from here. There is also the Python version.
  • Circularity is defined as: A measures how close to a circle the shape is. E.g. a regular hexagon has higher circularity than say a square. Is defined as (\frac{4*\pi*Area}{perimeter * perimeter}). This means that a circle has a circularity of 1, circularity of a square is 0.785, and so on.
  • Because of the contours, there may be multiple detection for each shape. These can be filtered out according to, for example, intersection over union condition. I did't inserted this part in the code for now, since it requires additional logic that isn't strictly related to the main task of finding the shapes.

UPDATE - Just noticed that in OpenCV 3.0.0 there is the function minEnclosingTriangle. This might be helpful to use instead of my procedure to find the triangle vertices. However, since inserting this function in the code would be trivial, I'll leave my procedure in the code in case one doesn't have OpenCV 3.0.0.

The code:

#include <opencv2\opencv.hpp>
#include <vector>
#include <iostream>

using namespace std;
using namespace cv;

/////////////////////////////////////////////////////////////////////////////////////////////
// Thinning algorithm from here:
// https://github.com/bsdnoobz/zhang-suen-thinning
/////////////////////////////////////////////////////////////////////////////////////////////

void thinningIteration(cv::Mat& img, int iter)
{
    CV_Assert(img.channels() == 1);
    CV_Assert(img.depth() != sizeof(uchar));
    CV_Assert(img.rows > 3 && img.cols > 3);

    cv::Mat marker = cv::Mat::zeros(img.size(), CV_8UC1);

    int nRows = img.rows;
    int nCols = img.cols;

    if (img.isContinuous()) {
        nCols *= nRows;
        nRows = 1;
    }

    int x, y;
    uchar *pAbove;
    uchar *pCurr;
    uchar *pBelow;
    uchar *nw, *no, *ne;    // north (pAbove)
    uchar *we, *me, *ea;
    uchar *sw, *so, *se;    // south (pBelow)

    uchar *pDst;

    // initialize row pointers
    pAbove = NULL;
    pCurr = img.ptr<uchar>(0);
    pBelow = img.ptr<uchar>(1);

    for (y = 1; y < img.rows - 1; ++y) {
        // shift the rows up by one
        pAbove = pCurr;
        pCurr = pBelow;
        pBelow = img.ptr<uchar>(y + 1);

        pDst = marker.ptr<uchar>(y);

        // initialize col pointers
        no = &(pAbove[0]);
        ne = &(pAbove[1]);
        me = &(pCurr[0]);
        ea = &(pCurr[1]);
        so = &(pBelow[0]);
        se = &(pBelow[1]);

        for (x = 1; x < img.cols - 1; ++x) {
            // shift col pointers left by one (scan left to right)
            nw = no;
            no = ne;
            ne = &(pAbove[x + 1]);
            we = me;
            me = ea;
            ea = &(pCurr[x + 1]);
            sw = so;
            so = se;
            se = &(pBelow[x + 1]);

            int A = (*no == 0 && *ne == 1) + (*ne == 0 && *ea == 1) +
                (*ea == 0 && *se == 1) + (*se == 0 && *so == 1) +
                (*so == 0 && *sw == 1) + (*sw == 0 && *we == 1) +
                (*we == 0 && *nw == 1) + (*nw == 0 && *no == 1);
            int B = *no + *ne + *ea + *se + *so + *sw + *we + *nw;
            int m1 = iter == 0 ? (*no * *ea * *so) : (*no * *ea * *we);
            int m2 = iter == 0 ? (*ea * *so * *we) : (*no * *so * *we);

            if (A == 1 && (B >= 2 && B <= 6) && m1 == 0 && m2 == 0)
                pDst[x] = 1;
        }
    }

    img &= ~marker;
}

void thinning(const cv::Mat& src, cv::Mat& dst)
{
    dst = src.clone();
    dst /= 255;         // convert to binary image

    cv::Mat prev = cv::Mat::zeros(dst.size(), CV_8UC1);
    cv::Mat diff;

    do {
        thinningIteration(dst, 0);
        thinningIteration(dst, 1);
        cv::absdiff(dst, prev, diff);
        dst.copyTo(prev);
    } while (cv::countNonZero(diff) > 0);

    dst *= 255;
}


int main()
{
    RNG rng(123);

    // Read image
    Mat3b src = imread("path_to_image");

    // Convert to grayscale
    Mat1b gray;
    cvtColor(src, gray, COLOR_BGR2GRAY);

    // Binarize
    Mat1b bin;
    threshold(gray, bin, 127, 255, THRESH_BINARY_INV);

    // Perform thinning
    thinning(bin, bin);

    // Create result image
    Mat3b res = src.clone();

    // Find contours
    vector<vector<Point>> contours;
    findContours(bin.clone(), contours, CV_RETR_LIST, CV_CHAIN_APPROX_NONE);

    // For each contour
    for (vector<Point>& contour : contours)
    {
        // Compute convex hull
        vector<Point> hull;
        convexHull(contour, hull);

        // Compute circularity, used for shape classification
        double area = contourArea(hull);
        double perimeter = arcLength(hull, true);
        double circularity = (4 * CV_PI * area) / (perimeter * perimeter);

        // Shape classification

        if (circularity > 0.9)
        {
            // CIRCLE

            //{
            //  // Fit an ellipse ...
            //  RotatedRect rect = fitEllipse(contour);
            //  Scalar color = Scalar(rng.uniform(0, 255), rng.uniform(0, 255), rng.uniform(0, 255));
            //  ellipse(res, rect, color, 5);
            //}
            {
                // ... or find min enclosing circle
                Point2f center;
                float radius;
                minEnclosingCircle(contour, center, radius);
                Scalar color = Scalar(rng.uniform(0, 255), rng.uniform(0, 255), rng.uniform(0, 255));
                circle(res, center, radius, color, 5);
            }
        }
        else if (circularity > 0.75)
        {
            // RECTANGLE

            //{
            //  // Minimum oriented bounding box ...
            //  RotatedRect rect = minAreaRect(contour);
            //  Point2f pts[4];
            //  rect.points(pts);

            //  Scalar color = Scalar(rng.uniform(0, 255), rng.uniform(0, 255), rng.uniform(0, 255));
            //  for (int i = 0; i < 4; ++i)
            //  {
            //      line(res, pts[i], pts[(i + 1) % 4], color, 5);
            //  }
            //}
            {
                // ... or bounding box
                Rect box = boundingRect(contour);
                Scalar color = Scalar(rng.uniform(0, 255), rng.uniform(0, 255), rng.uniform(0, 255));
                rectangle(res, box, color, 5);
            }
        }
        else if (circularity > 0.7)
        {
            // TRIANGLE

            // Select the portion of the image containing only the wanted contour
            Rect roi = boundingRect(contour);
            Mat1b maskRoi(bin.rows, bin.cols, uchar(0));
            rectangle(maskRoi, roi, Scalar(255), CV_FILLED);
            Mat1b triangle(roi.height, roi.height, uchar(0));
            bin.copyTo(triangle, maskRoi);

            // Find min encolsing circle on the contour
            Point2f center;
            float radius;
            minEnclosingCircle(contour, center, radius);

            // decrease the size of the enclosing circle until it intersects the contour
            // in at least 3 different points (i.e. the 3 vertices)
            vector<vector<Point>> vertices;
            do
            {
                vertices.clear();
                radius--;

                Mat1b maskCirc(bin.rows, bin.cols, uchar(0));
                circle(maskCirc, center, radius, Scalar(255), 5);

                maskCirc &= triangle;
                findContours(maskCirc.clone(), vertices, CV_RETR_LIST, CV_CHAIN_APPROX_NONE);

            } while (vertices.size() < 3);

            // Just get the first point in each vertex blob.
            // You could get the centroid for a little better accuracy

            Scalar color = Scalar(rng.uniform(0, 255), rng.uniform(0, 255), rng.uniform(0, 255));
            line(res, vertices[0][0], vertices[1][0], color, 5);
            line(res, vertices[1][0], vertices[2][0], color, 5);
            line(res, vertices[2][0], vertices[0][0], color, 5);

        }
        else
        {
            cout << "Some other shape..." << endl;
        }

    }

    return 0;
}

The results (minEnclosingCircle and boundingRect): enter image description here

The results (fitEllipse and minAreaRect): enter image description here

like image 107
Miki Avatar answered Nov 05 '22 02:11

Miki


You might check out a couple of resources.

First, you might consider asking questions at answers.opencv.org. There is probably a higher concentration of opencv specialists there.

Second, the book Practical OpenCV by Samarth Brahmbhatt is available as a free pdf and is easily found on google. It is contains many examples related to what you are looking for.

For example, you can separate different (non-overlapping) contours, as is shown in example 6.1 on page 68. He has a simple program for finding circles and lines in example 6.4 on page 78. You can also find a RANSAC-based ellipse-finder (much more complicated, but would be very useful here) in example 6.5 on page 82.

The book is in C++, but I imagine it will be very relevant, only you will need an API reference to translate it to python.

Personally, for your proejct, I would analyze one contour at a time, starting with his ellipse finder, and where a suitable ellipse can not be found, you could use a Hough transform of adjustable threshold, and truncate the resulting lines at their intersections, and bam! You have polygons.

like image 38
rexroni Avatar answered Nov 05 '22 01:11

rexroni