Drawing quadrant shape with cv2 in Python - python-3.x

I'm extracting door shapes with Cubicasa5k dataset but I don't know how svg draws quarter rounds.
Extracted some nodes of svg files with xml encoding, I found that it had a dictionary including 'd' key and value under of "Panel" id tree node like :
{'d': 'M825.47,986.05 q0.00,-72.92 72.92,-72.92 l0.00,72.92Z'}
It draws like this quarter round on svg image.
I want to draw that shape on raster image with cv2 in Python with that dictionary value.
Read article about 'd' commands but I'm still confused.
How can I draw it?

Okay, I hope I understood that d command of svg clearly now
That arc shape is Bézier curve, and it is drawn concavely based on the outer point, not on the centripetal point.
According to the example above ('M825.47,986.05 q0.00,-72.92 72.92,-72.92 l0.00,72.92Z'), it means drawing process following this:
The starting point of arc shape(called Bézier curve) is (825.47, 986.05).
The drawn reference point exists at a position shifted only to the y-axis. So the reference point is (825.47+0.00, 986.05-72.92) = (825.47, 913.13)
And the end point of arc shape is set following calculation about starting point like 2 : (825.47+72.92, 986.05-72.92) = (898.39, 913.13)
Drawing door shape finishes with drawing straight line that starts from the end point of the curve and ends with l code, it points (898.39+0.00, 913.13+72.92) = (898.39, 986.05)
Simply speaking, the door shape is drawn with arc and straight line, arc starts from left bottom and ends right top, then straight line starts from that right top and ends right bottom.
So if using cv2, should call cv2.ellipse, and set center point from process 4 above, axes(same value will show circle-like shape, so it's radius) from q command above, startangle and endangle(it differs from command above)

Related

how to get a “thick” poly line from vtkPlaneCutter & vtkPolyData?

I am trying to cut a mesh with a plane and render the contour on top of an image (resliced along using the same plane), much like the image below:
The examples at https://gitlab.kitware.com/vtk/vtk/-/blob/v9.1.0/Filters/Core/Testing/Python/TestPolyDataPlaneCutter.py almost do the slicing part that I need
But when viewing the contour from the direction of the normal the contour / polyline is invisible (see screenshot below left side) as opposed to viewing it from the side (right side of the screenshot below)
So I assume I should try to "dilate" the resulting contour somehow? Any suggestions?

Extract the 3×3image segment from the input image centred around the position (x,y)

I want to ask how to Extract the a n×n image segment from an input image centered around the position (x,y), the image has format like this [[num,num,num],[num,numm,num],[num,num,num]........], size of image is about 10 * 10. Thank you !
I do really recommend using numpy, when working with multidimensional arrays/when you want to manipulate these arrays, but here you go, a numpy solution and a solution without numpy.
import numpy as np
img = np.array([[1,2,3,4,5],[2,3,4,5,6],[3,4,5,6,7],[4,5,6,7,8],[5,6,7,8,9]])
print('img:\n',img,'\n')
partOfImage = img[0:3,0:3]
print('partOfImage:\n',partOfImage)
img = [[1,2,3,4,5],[2,3,4,5,6],[3,4,5,6,7],[4,5,6,7,8],[5,6,7,8,9]]
noNumpy =[[img[i][j] for j in range(0,3)] for i in range(0,3)]
print(noNumpy)
Question. So you wish to find the center of an image, expand 1 pixel left, left down, down, right down, right, right up, up, and left up to get an image from the middle of your sample. Correct? I do not have a code example, but a conceptual comment.
Fixed spatial images normally start their addressing at upper left for 0,0. Because of their even numbers of pixels approach, it's near-impossible to find a "middle". So finding a "middle pixel" means you need to apply a percentage and choose the nearest "whole pixel" to the choice. Then expand to your 3x3 dimensions.

How to get edges of a polygon using pixijs?

I'm using pixi.js to create some editable polygons. So, what I want to achieve is this:
I have one polygon
Then, when I hit the edge a small circle should appear
And next I can drag and drop that part of the edge to creating a new point for the polygon
For now, what I know is the polygon vertices and I'm thinking to use the line function (y=mx+b) to check if the point where the mouse is belongs to the edge. My problem here, is that I have no idea how to obtain that edges. Any Suggestion? Of course, if you have any other idea to do this feel free to share =).
For now, what I know is the polygon vertices
You probably draw your polygon using https://pixijs.download/dev/docs/PIXI.Graphics.html#drawPolygon method by passing to it a list of points - similar as last shape in this example: https://pixijs.io/examples/#/graphics/simple.js
// draw polygon
const path = [600, 370, 700, 460, 780, 420, 730, 570, 590, 520];
graphics.lineStyle(0);
graphics.beginFill(0x3500FA, 1);
graphics.drawPolygon(path);
graphics.endFill();
^ In that example we have 5 points: P (600, 370), Q (700, 460), R (780, 420), S (730, 570), T (590, 520).
It also means that we have 5 edges: PQ, QR, RS, ST, TP
Now, we should have some way to tell if mouse pointer "is hovered over some some edge". By "is hovered" i mean: it lies in some distance from edge - lets say said distance is 10 pixels. So we want to know if mouse pointer is 10 pixels away from some edge.
To know that we can use formula explained in Line defined by two points part in: https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_line#Line_defined_by_two_points
P1=(x1,y1) and P2=(x2,y2) - are the beginning and end vertices of some edge (for example PQ)
(x0,y0) is our "mouse point"
You can iterate over all edges and perform above calculation - if the distance is less that 10 pixel for some edge then you have the answer. If there is more than one edge which meets this requirement then you should pick one with smallest distance (it can happen if for example mouse is placed near some vertice).
Now you have the selected edge. Now lets do following point from your question:
2. Then, when I hit the edge a small circle should appear
To calculate position of this circle we can use equation from same Wikipedia page: https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_line#Line_defined_by_an_equation - the part The point on this line which is closest to (x0,y0) has coordinates:.
Here you need to convert coordinates of vertices from your selected edge to line function.
Then we can proceed to last point from your question:
3. And next I can drag and drop that part of the edge to creating a new point for the polygon
You can do it by adding new vertice to your polygon.
Lets assume that selected edge is PQ - then this new vertice should be added between vertices P and Q in the vertices list which you pass to drawPolygon method. Lets name this new vertice X. Coordinates of vertice X should be equal to current mouse coordinates.
Then you will have following edges: PX, XQ, QR, RS, ST, TP.
You probably want to activate this "mode" after mouse is clicked and when mouse button is down etc - but that is separate issue related to interactivity / GUI etc - not graphics :) .
Note: is good to separate your presentation part of application (graphics / pixi.js related things) from mechanics and interactivity / GUI etc. So for example: do your calculations in separate place (other class, method etc) from where you do your actual drawing (calling pixi.js methods, update canvas etc). Store results of calculations in some place (from above example it would be: list of vertices, position of circle, colors etc), and then when time comes to draw you take those results and just draw polygons using them. Dont mix everything in one place ;)

Matplotlib Text artist - how to get size? (not using pyplot)

Background
I've moved some code to use matplotlib only, instead of pyplot (The reason is it's generating png files in multi-process with futures, and pyplot isn't thread/process safe that way).
So I'm creating my figure and axis via matplotlib.Figure(), not via pyplot.
I've got some code that draw's a text 'table' on a figure, with one side right justified and the other left. In order to keep the spacing between the two sides constant, I was previously using get_window_extent() on my left-side text artist to figure out the location of the right hand side:
# draw left text at figure normalised coordinates, right justified
txt1 = figure.text(x, y, left_str,
ha='right', color=left_color, fontsize=fontsize)
# get the bounding box of that text - this is in display coords
bbox = txt1.get_window_extent(renderer=figure.canvas.get_renderer())
# get x location for right hand side offset from left's bbox
trans = figure.transFigure.inverted()
xr, _ = trans.transform((bbox.x1 + spacing, bbox.y0))
# draw right side text, using lhs's y value
figure.text(xr, y, right_str,
ha='left', color=right_color, fontsize=fontsize)
Problem
My problem is now that I'm not using pyplot, the above code fails to work, because figure.canvas.get_renderer() fails to return a renderer, as I haven't set one, and am simply using Figure.savefig(path) to save my file when I'm done.
So is there a way to find out the bounding box size of an Artist in a Figure without having a renderer set?
From looking at legend, which allows you to use a bounding box line with variable text size, I'm presuming there is, but I can't find how.
I've had a look at https://matplotlib.org/3.1.3/tutorials/intermediate/artists.html, and also tried matplotlib.artist.getp(txt1) on the above, but didn't find any seemingly helpful properties.

SVG path - Add and subtract || lasso selection tool in canvas

I want to combine two SVG paths such that
1.Both paths will be there but, at the area of intersection, there will be no strokes.
2.Second path will be excluded and there will be a complete stroke
See image at http://s18.postimg.org/et4m803rd/shape_combinations.jpg
I want similar function also in HTML5 canvas.
The purpose of this scenario is to implement a lasso selection tool (freehand selection) similar to that of photoshop, with Ctrl and Alt options for adding and subtracting selection [ + some other functions ].
What have you tried? This sounds a little bit like a homework assignment.
The first one is easy to replicate. Just draw the two circles with the stroke, then draw them again in the same place without the stroke.
You can achieve the second shape (the "pacman") by drawing a purple circle on the right that is clipped by a circle that is in the same position as the left (black) circle.

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