How to get an expanded or contracted contour in OpenCV?

Question

Is it possible to get the expanded or contracted version of a contour?

For example in the below image, I have used cv::findContour() and cv::drawContour on a binary image to get the contours:

I would like to draw another contour which has a customed pixel distance from the original contour, like these:

Except for eroding, which I think it might not be a good idea as it seems hard to control the pixel distance using eroding, I have no idea on how to solve this problem. May I know what should be the correct direction?

Donetsk

You can use cv::dilate() and cv::erode() then detect the contours again. – Lovellalovelock

How large are the initial contours? How exact has the resulting contour to reflect the shape of original contour? These two issues will determine, how complex a solution will be. I would agree, that cv::dilate or cv::erode will do the job to a certain degree (or detail). The super-fine solution will incorporate something like finding the center of mass, projecting the x, y coordinates of the original contour into the right direction, and determining new x, y coordinates for the resuling contour, thus a lot of interpolating and extrapolating, I assume. – Incontrovertible

Answer 1

Using cv::erode with a small kernel and multiple iterations may be enough for your needs, even if it's not exact.

C++ code:

cv::Mat img = ...;
int iterations = 10;
cv::erode(img, img,
   cv::getStructuringElement(cv::MORPH_RECT, cv::Size(3,3)),
   cv::Point(-1,-1),
   iterations);

Demo:

# img is the image containing the original black contour
for form in [cv.MORPH_RECT, cv.MORPH_CROSS]:
    eroded = cv.erode(img, cv.getStructuringElement(form, (3,3)), iterations=10)
    contours, hierarchy = cv.findContours(~eroded, cv.RETR_LIST, cv.CHAIN_APPROX_SIMPLE)
    vis = cv.cvtColor(img, cv.COLOR_GRAY2BGR)
    cv.drawContours(vis, contours, 0, (0,0,255))
    cv.drawContours(vis, contours, 1, (255,0,0))
    show_image(vis)

10 iterations with cv.MORPH_RECT with a 3x3 kernel:

10 iterations with cv.MORPH_CROSS with a 3x3 kernel:

You can change the offset by adjusting the number of iterations.

A much more accurate approach would be to use cv::distanceTransform to find all pixels that lie approximately 10px away from the contour:

dist = cv.distanceTransform(img, cv.DIST_L2, cv.DIST_MASK_PRECISE)
ring = cv.inRange(dist, 9.5, 10.5) # take all pixels at distance between 9.5px and 10.5px
show_image(ring)
contours, hierarchy = cv.findContours(ring, cv.RETR_LIST, cv.CHAIN_APPROX_SIMPLE)

vis = cv.cvtColor(img, cv.COLOR_GRAY2BGR)
cv.drawContours(vis, contours, 0, (0,0,255))
cv.drawContours(vis, contours, 2, (255,0,0))
show_image(vis)

You'll get two contours on each side of the original contour. Use findContours with RETR_EXTERNAL to recover only the outer contour. To also recover the inner contour, use RETR_LIST

Gregggreggory

Answer 2

I think the solution can be easier, without dilataion and new contours.

1. For each contour search mass center: cv::moments(contours[i]) -> cv::Point2f mc(mu.m10 / mu.m00), mu.m01 / mu.m00));

2. For each point point of contour: make shift for mass center -> multiply by coefficient K -> shift backward: pt_new = (k * (pt - mc) + mc);

But coefficient k must be individual for each point. I will calculate it a little later...

Spurtle

But this will shift every point away from the mass center, whereas I would expect it moves away perpendicular from the outer line. For convex bodies this will likely not matter much, the trouble will probably arise when there are insets. For example a hole in the contour that is getting "fixed". – Mcnair