How to determine length of spline (e.g. snake) Hello, I'm trying to automatically determine length of a spline - well, in this case it's the length of a snake. I thought I could modify existing code and then divide the contour into segments but I'm not having much luck.

The black rectangle in the photo is 12" x 2". The snake is more than 5' long.

Can someone point me in the right direction?

Command line used:

python object_size.py --image images/img_1718_useforcalib_w2.5.jpg --width 2.5

Code:

# USAGE
# python object_size.py --image images/example_01.png --width 0.955
# python object_size.py --image images/example_02.png --width 0.955
# python object_size.py --image images/example_03.png --width 3.5

# import the necessary packages
from scipy.spatial import distance as dist
from imutils import perspective
from imutils import contours
import numpy as np
import argparse
import imutils
import cv2

def midpoint(ptA, ptB):
return ((ptA + ptB) * 0.5, (ptA + ptB) * 0.5)

# construct the argument parse and parse the arguments
ap = argparse.ArgumentParser()
help="path to the input image")
ap.add_argument("-w", "--width", type=float, required=True,
help="width of the left-most object in the image (in inches)")
args = vars(ap.parse_args())

# load the image, convert it to grayscale, and blur it slightly
gray = cv2.cvtColor(image, cv2.COLOR_BGR2GRAY)
gray = cv2.GaussianBlur(gray, (7, 7), 0)

# perform edge detection, then perform a dilation + erosion to
# close gaps in between object edges
edged = cv2.Canny(gray, 50, 100)
edged = cv2.dilate(edged, None, iterations=1)
edged = cv2.erode(edged, None, iterations=1)

# find contours in the edge map
cnts = cv2.findContours(edged.copy(), cv2.RETR_EXTERNAL,
cv2.CHAIN_APPROX_SIMPLE)
cnts = cnts if imutils.is_cv2() else cnts

# sort the contours from left-to-right and initialize the
# 'pixels per metric' calibration variable
(cnts, _) = contours.sort_contours(cnts)
pixelsPerMetric = None

# loop over the contours individually
for c in cnts:
# if the contour is not sufficiently large, ignore it
if cv2.contourArea(c) < 100:
continue

# compute the rotated bounding box of the contour
orig = image.copy()
box = cv2.minAreaRect(c)
box = cv2.cv.BoxPoints(box) if imutils.is_cv2() else cv2.boxPoints(box)
box = np.array(box, dtype="int")

# order the points in the contour such that they appear
# in top-left, top-right, bottom-right, and bottom-left
# order, then draw the outline of the rotated bounding
# box
box = perspective.order_points(box)
cv2.drawContours(orig, [box.astype("int")], -1, (0, 255, 0), 2)

# loop over the original points and draw them
for (x, y) in box:
cv2.circle(orig, (int(x), int(y)), 5, (0, 0, 255), -1)

# unpack the ordered bounding box, len compute the midpoint
# between the top-left and top-right coordinates, followed by
# the midpoint between bottom-left and bottom-right coordinates
(tl, tr, br, bl) = box
(tltrX, tltrY) = midpoint(tl, tr)
(blbrX, blbrY) = midpoint(bl, br)

# compute the midpoint between the top-left and top-right points,
# followed by the midpoint between the top-righ and bottom-right
(tlblX, tlblY) = midpoint(tl, bl ...
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I don't know python but I tried this :

1. invert image
2. threshold image
3. apply thinning algoirthm
4. look for connected components
5. snake is index surface is 513=length and and black rectangle is index 2 length is 79 (there is an error here because skeleton is not rectangle height)

There is no calibration so width of pixel far from center is not equal to pixel at center more

This solution doesn't use Contours, correct? What tool are you using - is that a OpenCV GUI that I'm not aware of?

1
1

@LBerger - I've tried 3 forms of thresholding but don't get the same results as you. Can you tell me what thresholding you're using - perhaps the exact commands? I'm a noob here.

manual threshold = 180. If you want to use OTSU you need to crop image

It's been a while and I still haven't figured this out. Any way you could share your spreadsheet with me - maybe it would help me get Python code done? Thanks!

I can't tell you how many hours I tried to find a function within OpenCV to no avail. Found several in other packages. Thank you! @DougL. python3 rectangle_size.py --image images/snake.jpg --width 2.5. Uncertainly, measurement isn't accurately. But i set another is 1.019685. more I got the C++ code for LBerger's algorithm! Hopefully it's not hard to convert it to Python. Does Python even support the opencv contrib modules?

Here's the input image: Here's the code:

#include <opencv2/opencv.hpp>
#include <opencv2/ximgproc.hpp>
using namespace cv;
using namespace cv::ximgproc;
#pragma comment(lib, "opencv_world331.lib")

#include <iostream>
#include <map>
using namespace std;

int main(void)
{
Mat frame = imread("snake.png");

if (frame.empty())
{
cout << "Error loading image file" << endl;
return -1;
}

cvtColor(frame, frame, CV_RGB2GRAY);

bitwise_not(frame, frame);

threshold(frame, frame, 190, 255, CV_THRESH_BINARY);

thinning(frame, frame);

Mat labels;
connectedComponents(frame, labels, 8);

map<size_t, size_t> counts;

for (int j = 0; j < labels.rows; j++)
{
for (int i = 0; i < labels.cols; i++)
{
counts[labels.at<int>(j, i)]++;

if (labels.at<int>(j, i) == 2)
frame.at<unsigned char>(j, i) = 255;
else if (labels.at<int>(j, i) == 1)
frame.at<unsigned char>(j, i) = 127;
else if (labels.at<int>(j, i) == 0)
frame.at<unsigned char>(j, i) = 0;
}
}

cout << counts.size() << endl;

cout << counts << endl; // The snake is this long in pixels.

imshow("Frame", frame);

waitKey();

return 0;
}
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Asked: 2017-09-14 14:47:15 -0500

Seen: 1,043 times

Last updated: Dec 12 '17