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Count Pixel

asked 2013-04-04 21:50:05 -0600

Vuong Bui gravatar image

I load image in openCV , then I draw line Point(x1,y1), Point(x2,y2). How to count pixel on line that. Thank you

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answered 2013-04-05 14:21:31 -0600

AlexanderShishkov gravatar image

updated 2013-04-05 14:22:36 -0600

You should use LineIterator class: http://docs.opencv.org/modules/core/doc/drawing_functions.html?highlight=lineiterator#LineIterator

E.g. if you want to get pixel values failing on that line:

LineIterator it(img, pt1, pt2, 8);
LineIterator it2 = it;
vector<Vec3b> buf(it.count);

for(int i = 0; i < it.count; i++, ++it)
    buf[i] = *(const Vec3b)*it;

it.count in this sample is pixels count for the line between pt1 and pt2 if we use 8 neighbors.

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Did not know about this function, but I assume internally it does something like I suggested :)

StevenPuttemans gravatar imageStevenPuttemans ( 2013-04-05 14:31:09 -0600 )edit

Just wondering. The explanation says Class for iterating pixels on a raster line Does this mean that you can only use this on horizontal and vertical lines?

StevenPuttemans gravatar imageStevenPuttemans ( 2013-04-05 14:34:20 -0600 )edit

No, It doesn't. You can use it for any types of lines.

AlexanderShishkov gravatar imageAlexanderShishkov ( 2013-04-05 14:50:43 -0600 )edit
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answered 2013-04-05 03:05:59 -0600

I do not think he wants to calculate the euclidean distance, but he wants to define the pixel values falling on that line.

You should first consider how a line defined by 2 points can be represented mathematically. Given point 1 and point 2, the line equation equals:

Point1 = Point(x1,y1); Point2 = Point(x2,y2);   
y = y1 + [(y2 - y1) / (x2 - x1)]·(x - x1);

This gives you the ability to know exactly which pixels correspond to that line. Loop over your image, enter the x coördinate in the equation and look if the output is still the index your are currently in y position. If this check holds, than add the current value of the pixel to a storage container, for example an int that increases in value each time.

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And basically you do not even want the values, then just increase a counter if the location suits the equation :)

StevenPuttemans gravatar imageStevenPuttemans ( 2013-04-05 03:14:12 -0600 )edit

YES, I want to define the pixel values falling on that line. And Do you same when i use type of line (straight horizontal, vertical, diagonal, slanted..) Thanks you very much

Vuong Bui gravatar imageVuong Bui ( 2013-04-05 04:51:40 -0600 )edit

This equation is only for a straight line, but it can have any possible orientation. If you want curved lines, then defining an equation will be alot harder. Accept answers if they suit you :)

StevenPuttemans gravatar imageStevenPuttemans ( 2013-04-05 04:54:55 -0600 )edit
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answered 2013-04-05 01:13:20 -0600

berak gravatar image

for a straight line, the euklidean distance is:

sqrt( (p1.x-p2.x)*(p1.x-p2.x) + (p1.y-p2.y)*(p1.y-p2.y) );
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Asked: 2013-04-04 21:50:05 -0600

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Last updated: Apr 05 '13