Hi am using HoughLines Method to detect lines from a camera, i've filtered my image "imgProcessed" using ROI it means getting just the black objects to make the tracking simple, then when i intend to use the HoughLines method it gives me an error that my "CannyEdges" has some invalid arguments, here's my code :
Image<Gray, Byte> gray = imgProcessed.Convert<Gray, Byte>().PyrDown().PyrUp();
Gray cannyThreshold = new Gray(180);
Gray cannyThresholdLinking = new Gray(120);
Gray circleAccumulatorThreshold = new Gray(120);
Image<Gray, Byte> cannyEdges = gray.Canny(cannyThreshold, cannyThresholdLinking);
LineSegment2D[] lines = imgProcessed.cannyEdges.HoughLines(
cannyThreshold,
cannyThresholdLinking,
1, //Distance resolution in pixel-related units
Math.PI / 45.0, //Angle resolution measured in radians.
50, //threshold
100, //min Line width
1 //gap between lines
)[0]; //Get the lines from the first channel
I've edited my code & it works, i've devised it in two parts : the first part where i've detected the canny edges rather then use the HoughLines method to detect it automatically, and the 2nd where i've used the HoughLinesBinary method rather than HoughLines with less arguments, here is the code :
Image<Gray, Byte> gray1 = imgProcessed.Convert<Gray, Byte>().PyrDown().PyrUp();
Image<Gray, Byte> cannyGray = gray1.Canny(120, 180);
imgProcessed = cannyGray;
LineSegment2D[] lines = imgProcessed.HoughLinesBinary(
1, //Distance resolution in pixel-related units
Math.PI / 45.0, //Angle resolution measured in radians.
50, //threshold
100, //min Line width
1 //gap between lines
)[0]; //Get the lines from the first channel
Related
I am using EmguCV(c#) histogram to compare two HSV images. But sometimes I get negative values. I assumed that when I compare 2 histogram values, the value will be in the interval <0 and 1>. However, some of the values of hue or saturation are sometimes negative numbers like -0.145.
Firstly, I get byte image array, which I convert into Image<Hsv, Byte> - img1.
Image<Hsv, Byte> img1 = null;
Mat byteImageMat = new Mat();
Mat hsvMat = new Mat();
CvInvoke.Imdecode(request.ByteImage, Emgu.CV.CvEnum.ImreadModes.AnyColor, byteImageMat);
CvInvoke.CvtColor(byteImageMat, hsvMat, ColorConversion.Bgr2Hsv);
img1 = hsvMat.ToImage<Hsv, Byte>();
Then I create DenseHistogram and spliting individual channels.
DenseHistogram ComparedHistoHue = new DenseHistogram(180, new RangeF(0, 180));
DenseHistogram ComparedHistoSaturation = new DenseHistogram(256, new RangeF(0, 256));
DenseHistogram ComparedHistoBrightness = new DenseHistogram(256, new RangeF(0, 256));
Image<Gray, Byte> hueChannel = img1[0];
Image<Gray, Byte> saturationChannel = img1[1];
Image<Gray, Byte> brightnessChannel = img1[2];
After that I calculate histograms
ComparedHistoHue.Calculate(new Image<Gray, Byte>[] { hueChannel }, false, null);
ComparedHistoSaturation.Calculate(new Image<Gray, Byte>[] { saturationChannel }, false, null);
ComparedHistoBrightness.Calculate(new Image<Gray, Byte>[] { brightnessChannel }, false, null);
At this point, I loaded histogram from file which I created before and assign it into Mat (loadedMatHue, loadedMatSaturation and loadedMatBrightness).
double hue = CvInvoke.CompareHist(loadedMatHue, ComparedHistoHue, Emgu.CV.CvEnum.HistogramCompMethod.Correl);
double satuation = CvInvoke.CompareHist(loadedMatSaturation, ComparedHistoSaturation, Emgu.CV.CvEnum.HistogramCompMethod.Correl);
double brightnes = CvInvoke.CompareHist(loadedMatBrightness, ComparedHistoBrightness, Emgu.CV.CvEnum.HistogramCompMethod.Correl);
Can somebody tell me, why is in hue or saturation variable negative value? In my opinion and tests, there is always only one negative value at one momemnt across the double variables.
For HSV, the idea that the numbers would be between 0 and 1 is incorrect. If you want your image to have values between 0 and 1, then that image would have to be in grayscale.
In HSV, you split it up into three definitions, Hue, Saturation, and Value.
Hue is stored from 0 to 360 degrees, but can become negative if you rotate the hue past 0.
Saturation is considered from 0 to 1, i.e grayscale values. If you have negative values in this channel, disregard them, as the lowest that this value should be is 0. The same can be said for the highest value, which will be 1 since the highest value of a grayscale channel can only be one. Like I said before, its best to think of this channel in terms of grayscale from 0 to 1.
Value is very similar to saturation, the only difference being that value is considered the "lightness of the color, by the given S[saturation]" This value also can only be between 0 and 1, and any values outside of this space should be clipped.
If you want a more in depth explanation, you can check out this Stack post, which is very detailed and I thought it should be credited in this post.
If you do have to clip these values, you can always access the pixel values for each channel using some sample code below.
Image<Hsv,Byte> sampleImage = new Image<Hsv,Byte>("path\to\image");
//X and y are the pixel coordinates on an image
//Hue channel
byte hue = sampleImage.Data[y,x,0];
//Saturation channel
byte sat = sampleImage.Data[y,x,1];
//Value channel
byte val = sampleImage.Data[y,x,2];
You can throw these values inside of a loop and check if a pixel is outside the boundaries, and if it is replace it with the high or low value respectively.
I know that we desaturate an image by decreasing the values in the Saturation channel. I want to acomplish this using c# with emgu
For instance here is c++ code using opencv to do so:
Mat Image = imread("images/any.jpg");
// Specify scaling factor
float saturationScale = 0.01;
Mat hsvImage;
// Convert to HSV color space
cv::cvtColor(Image,hsvImage,COLOR_BGR2HSV);
// Convert to float32
hsvImage.convertTo(hsvImage,CV_32F);
vector<Mat>channels(3);
// Split the channels
split(hsvImage,channels);
// Multiply S channel by scaling factor
channels[1] = channels[1] * saturationScale;
// Clipping operation performed to limit pixel values
// between 0 and 255
min(channels[1],255,channels[1]);
max(channels[1],0,channels[1]);
// Merge the channels
merge(channels,hsvImage);
// Convert back from float32
hsvImage.convertTo(hsvImage,CV_8UC3);
Mat imSat;
// Convert to BGR color space
cv::cvtColor(hsvImage,imSat,COLOR_HSV2BGR);
// Display the images
Mat combined;
cv::hconcat(Image, imSat, combined);
namedWindow("Original Image -- Desaturated Image", CV_WINDOW_AUTOSIZE);
In c# I have:
var img = new Image<Gray, byte>("images/any.jpg");
var imhHsv = img.Convert<Hsv, byte>();
var channels = imhHsv.Split();
// Multiply S channel by scaling factor and clip (limit)
channels[1] = (channels[1] * saturationScale);
I am not sure how to merge modified saturation channel with imHsv, if I do this:
CvInvoke.Merge(channels, imhHsv);
there is error:
cannot convert 'Emgu.CV.Image[]' to
'Emgu.CV.IInputArrayOfArrays'
I put a VectorOfMat into the CvInvoke.Merge and it works.
Mat[] m = new Mat[3];
m[0] = CvInvoke.CvArrToMat(channels[0]);
m[1] = CvInvoke.CvArrToMat(channels[1]);
m[2] = CvInvoke.CvArrToMat(channels[2]);
VectorOfMat vm = new VectorOfMat(m);
CvInvoke.Merge(vm, imhHsv);
After the InRange and GaussianBlur function, I get the following image:
I find the edges and I get:
I need to get from this image three lines, which are represented in the following image (lines that are not red are bulging nodes):
How can I do that?
You should use an Erosion operation with a clever structuring element.
I suggest to use one with the length of the longest horizontal line to remove.
I've made a little example:
Let's take this toErode.png
double longestHorLine = 201;
Image<Gray, byte> toErode = new Image<Gray, byte>(path+ "toErode.png");
for(int i =1;i<4;i++)
{
Mat element = CvInvoke.GetStructuringElement(ElementShape.Rectangle, new Size((int)(longestHorLine / i), 3), new Point(-1, -1));
CvInvoke.Erode(toErode, toErode, element, new Point(-1, -1), i, BorderType.Default, new MCvScalar(0));
toErode.Save(path + "res"+i+".png");
}
Which produces the following outputs: res1.png ,res2.png and
res3.png
I have an image and I want to find the most dominant lowest hue colour and the most dominant highest hue colour from an image.
It is possible that there are several colours/hues that are close to each other in populace to be dominant and if that is the case I would need to take an average of the most popular.
I am using emgu framework here.
I load an image into the HSV colour space.
I split the hue channel away from the main image.
I then use the DenseHistogram to return my ranges of 'buckets'.
Now, I could enumerate through the bin collection to get what I want but I am mindful of conserving memory when and wherever I can.
So, is there a way of getting what I need at all from the DenseHistogram 'object'?
i have tried MinMax (as shown below) and I have consider using linq but not sure if that is expensive to use and/or how to use it with just using a float array.
This is my code so far:
float[] GrayHist;
Image<Hsv, Byte> hsvSample = new Image<Hsv, byte>("An image file somewhere");
DenseHistogram Histo = new DenseHistogram(255, new RangeF(0, 255));
Histo.Calculate(new Image<Gray, Byte>[] { hsvSample[0] }, true, null);
GrayHist = new float[256];
Histo.MatND.ManagedArray.CopyTo(GrayHist, 0);
float mins;
float maxs;
int[] minVals;
int[] maxVals;
Histo.MinMax(out mins, out maxs, out minVals, out maxVals); //only gets lowest and highest and not most popular
List<float> ranges= GrayHist.ToList().OrderBy( //not sure what to put here..
I need to detect a spiral shaped spring and count its coil turns.
I have tried as follows:
Image<Bgr, Byte> ProcessImage(Image<Bgr, Byte> img)
{
Image<Bgr, Byte> imgClone = new Image<Bgr,byte>( img.Width, img.Height);
imgClone = img.Clone();
Bgr bgrRed = new Bgr(System.Drawing.Color.Red);
#region Algorithm 1
imgClone.PyrUp();
imgClone.PyrDown();
imgClone.PyrUp();
imgClone.PyrDown();
imgClone.PyrUp();
imgClone.PyrDown();
imgClone._EqualizeHist();
imgClone._Dilate(20);
imgClone._EqualizeHist();
imgClone._Erode(10);
imgClone.PyrUp();
imgClone.PyrDown();
imgClone.PyrUp();
imgClone.PyrDown();
imgClone.PyrUp();
imgClone.PyrDown();
imgClone._EqualizeHist();
imgClone._Dilate(20);
imgClone._EqualizeHist();
imgClone._Erode(10);
Image<Gray, Byte> imgCloneGray = new Image<Gray, byte>(imgClone.Width, imgClone.Height);
CvInvoke.cvCvtColor(imgClone, imgCloneGray, Emgu.CV.CvEnum.COLOR_CONVERSION.CV_BGR2GRAY);
imgCloneGray = imgCloneGray.Canny(c_thresh, c_threshLink);//, (int)c_threshSize);
Contour<System.Drawing.Point> pts = imgCloneGray.FindContours(Emgu.CV.CvEnum.CHAIN_APPROX_METHOD.CV_CHAIN_APPROX_SIMPLE, Emgu.CV.CvEnum.RETR_TYPE.CV_RETR_EXTERNAL);
CvInvoke.cvCvtColor(imgCloneGray, imgCloneYcc, Emgu.CV.CvEnum.COLOR_CONVERSION.CV_GRAY2BGR);
if (null != pts)
{
imgClone.Draw(pts, bgrRed, 2);
imgClone.Draw(pts.BoundingRectangle, bgrRed, 2);
}
#endregion
return imgClone;
}
I am some how able to get the spring but how to get the counts. I am looking for algorithms.
I am currently not looking for speed optimization.
This is similar like counting fingers. Spring spiral is very thin to get using contour. What else can be done. http://www.luna-arts.de/others/misc/HandsNew.zip
You have a good final binarization over there, but it looks like to be too restricted to this single case. I would do a relatively simpler, but probably more robust, preprocessing to allow a relatively good binarization. From Mathematical Morphology, there is a transform called h-dome, which is used to remove irrelevant minima/maxima by suppressing minima/maxima of height < h. This operation might not be readily available in your image processing library, but it is not hard to implement it. To binarize this preprocessed image I opted for Otsu's method, since it is automatic and statistically optimal.
Here is the input image after h-dome transformations, and the binary image:
Now, to count the number of "spiral turns" I did something very simple: I split the spirals so I can count them as connected components. To split them I did a single morphological opening with a vertical line, followed by a single dilation by an elementary square. This produces the following image:
Counting the components gives 15. Since you have 13 of them that are not too close, this approach counted them all correctly. The groups at left and right were counted as a single one.
The full Matlab code used to do these steps:
f = rgb2gray(imread('http://i.stack.imgur.com/i7x7L.jpg'));
% For this image, the two next lines are optional as they will to lead
% basically the same binary image.
f1 = imhmax(f, 30);
f2 = imhmin(f1, 30);
bin1 = ~im2bw(f2, graythresh(f2));
bin2 = bwmorph(imopen(bin1, strel('line', 15, 90)), 'dilate');