Hey gang I am trying to make a simple custom triangle shape in draw.io
Unfortunately, there is something very fundamental that I don't understand. Will someone throw me a bone!
The first image shows the XML I used to create the image. Note that a preview shows the offset from (0,0)
The second image shows the triangle after I place it. Note the offset from (0,0) is real, and the box around it is unexpected - I expected a 40x20 box.
How do I remove the 10 pixel offset and also tighten up that bounding box?
Related
Looking for a solution to get the grid lines inside or over the polar area points.
Like the image attached.
I found the code which draws the gridLines and the one draws the point.
I also see it's possible to pass a creationPattern() as background.
But I don't know how to get them working together.
Maybe a solution can be simply show the gridLines over the points and not behind.
(a) what I have, (b) what I get, (c) what I want
I have a simple vector graphic in Inkscape, which consists of a rectangle, filled points and stars. Since the axis ranges are not really nice (the height equals approximatly 3 times the width of the picture) for a publication, I want to rescale the picture. However, I do not have the raw data, such that I can plot it again. How can I rescale my graphic (see figure (a)), such that the x-range is more wide (see figure (c)) without getting distortions (see figure (b))? In the end I want to create a PDF file out of it.
Any ideas on that?
Thanks for your help.
You can try to do it in 2 steps, using the Object -> Transform tool (Shift-Ctrl-M).
First, select everything, and with the transform tool select the Scale tab, and scale horizontally by, say, 300%. All figures will be distorted.
Now, unselect the rectangle, and scale horizontally again by 33.3%, but first click on Apply to each object separately. This will undo the distortion (but not the translation) of each object.
Note that 300% followed by 33.3% should leave the individual objects with the same size.
Documentation here.
I created a flowchart in InkScape in which each shape is filled with a linear gradient.
Some shapes share the same gradient tones but when I created them via copy&past did not know that gradients could be transformed during edit, I thought that copy&pasting a shape simply copy the entire object and each colors.
Anyway, it resulted in shapes that changes background gradient during re-positioning. After several searches I discovered that this is caused by a wrong position of "Gradient Editor Tool" of each one. In other words I moved objects but start/end of each gradient remained in place.
Please, how to quickly reposition gradient direction/start/end of each shape, maybe centering it on each one? I would avoid to manually reposition each one...
Thanks
Inkscape provides a toggle button to move the gradients along with the objects.
Activating this button during any coying or moving carries the gradient with it to the new location. (Actually it copies the gradient.)
I am trying to develop box sorting application in qt and using opencv. I want to measure width and length of box.
As shown in image above i want to detect only outermost lines (ie. box edges), which will give me width and length of box, regardless of whatever printed inside the box.
What i tried:
First i tried using Findcontours() and selected contour with max area, but the contour of outer edge is not enclosed(broken somewhere in canny output) many times and hence not get detected as a contour.
Hough line transform gives me too many lines, i dont know how to get only four lines am interested in out of that.
I tried my algorithm as,
Convert image to gray scale.
Take one column of image, compare every pixel with next successive pixel of that column, if difference in there value is greater than some threshold(say 100) that pixel belongs to edge, so store it in array. Do this for all columns and it will give upper line of box parallel to x axis.
Follow the same procedure, but from last column and last row (ie. from bottom to top), it will give lower line parallel to x axis.
Likewise find lines parallel to y axis as well. Now i have four arrays of points, one for each side.
Now this gives me good results if box is placed in such a way that its sides are exactly parallel to X and Y axis. If box is placed even slightly oriented in some direction, it gives me diagonal lines which is obvious as shown in below image.
As shown in image below i removed first 10 and last 10 points from all four arrays of points (which are responsible for drawing diagonal lines) and drew the lines, which is not going to work when box is tilted more and also measurements will go wrong.
Now my question is,
Is there any simpler way in opencv to get only outer edges(rectangle) of box and get there dimensions, ignoring anything printed on the box and oriented in whatever direction?
I am not necessarily asking to correct/improve my algorithm, but any suggestions on that also welcome. Sorry for such a big post.
I would suggest the following steps:
1: Make a mask image by using cv::inRange() (documentation) to select the background color. Then use cv::not() to invert this mask. This will give you only the box.
2: If you're not concerned about shadow, depth effects making your measurment inaccurate you can proceed right away with trying to use cv::findContours() again. You select the biggest contour and store it's cv::rotatedRect.
3: This cv::rotatedRect will give you a rotatedRect.size that defines the width en the height of your box in pixels
Since the box is placed in a contrasting background, you should be able to use Otsu thresholding.
threshold the image (use Otsu method)
filter out any stray pixels that are outside the box region (let's hope you don't get many such pixels and can easily remove them with a median or a morphological filter)
find contours
combine all contour points and get their convex hull (idea here is to find the convex region that bounds all these contours in the box region regardless of their connectivity)
apply a polygon approximation (approxPolyDP) to this convex hull and check if you get a quadrangle
if there are no perspective distortions, you should get a rectangle, otherwise you will have to correct it
if you get a rectangle, you have its dimensions. You can also find the minimum area rectangle (minAreaRect) of the convexhull, which should directly give you a RotatedRect
I am using open CV and C++. I have a completely dark image which has 3 colored points on it. I need their center coordinates. If I have only one colored point in the dark image, it will automatically display its center coordinate. However,if I take as input the dark image with the 3 colored points,my program will make an average if those 3 coordinates and return the center of the 3 colored points together,which is my exact problem. I need their individual center coordinates.
Can anyone suggest a method to do that please. Thanks
Here is the code http://pastebin.com/RM7chqBE
Found a solution!
load original image to grayscale
convert original image to gray
set range of intensity value depending on color that needs to be detected
vector of contours and hierarchy
findContours
vector of moments and point
iterate through each contour to find coordinates
One of the ways to do this easily is to use the findContours and drawContours function.
In the documentation you have a bit of code that explains how to retrieve the connected components of an image. Which is what you are actually trying to do.
For example you could draw every connected component you will find (that means every dot) on it's own image and use the code you already have on every image.
This may not be the most efficient way to do this however but it's really simple.
Here is how I would do it
http://pastebin.com/y1Ae3e2V
I'm not sure this works however as I don't have time to test it but you can try it.