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pythonopencvhough-transform
dd210
dd210

Reputation: 565

Probabilistic Hough transform in OpenCV 2.4.9 (Python)

My question is about Hough transform in OpenCV 2.4.9 (Python).

Here is an extract from tutorial:

cv2.HoughLinesP(image, rho, theta, threshold[, lines[, minLineLength[, maxLineGap]]]) → lines

I do not really understand what "lines[," means. I use the function in the following manner:

lines = cv2.HoughLinesP(edges, 1, np.pi/180, 25, 2,25,115)

But what does parameter "2" here really mean? Seems nothing change when I assign different values for that parameter.

Tnanks..

Upvotes: 4

Views: 2604

Answers (1)

Froyo
Froyo

Reputation: 18477

You have to use it like this

lines = cv2.HoughLinesP(edge_image, rho=1.0, theta=math.pi/180.0,
                                    threshold=thresholdVal,
                                    minLineLength=minlinelengthVal,
                                    maxLineGap=maxlinegapVal)

If you read up about Hough Transforms and probabilistic hough transforms, you'd realize that an accumulator is used to accumulate all the edge points. rho is the distance resolution of the accumulator in pixels and theta is the angle resolution of the accumulator in radians.

And as far as cv2.HoughLinesP(image, rho, theta, threshold[, lines[, minLineLength[, maxLineGap]]]) → lines docs is concerned, it is sort of function overloading but since python provides optional arguments, this is used. lines[ just means that you can pass a numpy array where the lines will be stored. So now, if you want pass other parameters and skip lines, you'd have to pass them by the parameter name.

Upvotes: 6

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