I am trying to get a simple face mask, but I cannot pin point the cause and get rid of tiny artifacts in the textured output
Example image has a few artifacts marked:
Code:
import cv2
import mediapipe as mp
import triangulation_media_pipe as tmp
import numpy as np
mp_drawing = mp.solutions.drawing_utils
mp_face_mesh = mp.solutions.face_mesh
face = "face_textures/yash.jpg"
def load_base_img(face_mesh, image_file_name, ):
image = cv2.imread(image_file_name)
results = face_mesh.process(cv2.cvtColor(image, cv2.COLOR_BGR2RGB))
return {"img": image, "landmarks": results}
def transform_landmarks_from_tf_to_ocv(keypoints, face_width, face_height):
landmark_list = []
if (keypoints.multi_face_landmarks != None):
for face_landmarks in keypoints.multi_face_landmarks:
for l in face_landmarks.landmark:
pt = mp_drawing._normalized_to_pixel_coordinates(l.x, l.y,
face_width, face_height)
landmark_list.append(pt)
return landmark_list
def main():
# For webcam input:
face_mesh = mp_face_mesh.FaceMesh()
base_face_handler, landmark_base_ocv, base_input_image = process_base_face_mesh(face_mesh, face)
cap = cv2.VideoCapture(2)
while cap.isOpened():
_, webcam_img = cap.read()
image_rows, image_cols, _ = webcam_img.shape
results = face_mesh.process(webcam_img)
landmark_target_ocv = transform_landmarks_from_tf_to_ocv(results, image_cols, image_rows)
# Draw the face mesh annotations on the image.
image = webcam_img.copy()
img2_new_face = np.zeros_like(image)
if results.multi_face_landmarks:
if True:
for i in range(0, int(len(tmp.TRIANGULATION) / 3)):
triangle_index = [tmp.TRIANGULATION[i * 3],
tmp.TRIANGULATION[i * 3 + 1],
tmp.TRIANGULATION[i * 3 + 2]]
tbas1 = landmark_base_ocv[triangle_index[0]]
tbas2 = landmark_base_ocv[triangle_index[1]]
tbas3 = landmark_base_ocv[triangle_index[2]]
triangle1 = np.array([tbas1, tbas2, tbas3])
rect1 = cv2.boundingRect(triangle1)
(x1, y1, w1, h1) = rect1
cropped_triangle = base_input_image[y1: y1 + h1, x1: x1 + w1]
cropped_tr1_mask = np.zeros((h1, w1), np.uint8)
points = np.array([[tbas1[0] - x1, tbas1[1] - y1],
[tbas2[0] - x1, tbas2[1] - y1],
[tbas3[0] - x1, tbas3[1] - y1]])
cv2.fillConvexPoly(cropped_tr1_mask, points, 255)
ttar1 = landmark_target_ocv[triangle_index[0]]
ttar2 = landmark_target_ocv[triangle_index[1]]
ttar3 = landmark_target_ocv[triangle_index[2]]
triangle2 = np.array([ttar1, ttar2, ttar3])
rect2 = cv2.boundingRect(triangle2)
(x2, y2, w2, h2) = rect2
cropped_tr2_mask = np.zeros((h2, w2), np.uint8)
points2 = np.array([[ttar1[0] - x2, ttar1[1] - y2],
[ttar2[0] - x2, ttar2[1] - y2],
[ttar3[0] - x2, ttar3[1] - y2]])
cv2.fillConvexPoly(cropped_tr2_mask, points2, 255)
# Warp triangles
points = np.float32(points)
points2 = np.float32(points2)
M = cv2.getAffineTransform(points, points2)
warped_triangle = cv2.warpAffine(cropped_triangle, M, (w2, h2))
warped_triangle = cv2.bitwise_and(warped_triangle, warped_triangle, mask=cropped_tr2_mask)
# Reconstructing destination face
img2_new_face_rect_area = img2_new_face[y2: y2 + h2, x2: x2 + w2]
img2_new_face_rect_area_gray = cv2.cvtColor(img2_new_face_rect_area, cv2.COLOR_BGR2GRAY)
_, mask_triangles_designed = cv2.threshold(img2_new_face_rect_area_gray, 0, 255,
cv2.THRESH_BINARY_INV)
warped_triangle = cv2.bitwise_and(warped_triangle, warped_triangle,
mask=mask_triangles_designed)
img2_new_face_rect_area = cv2.add(img2_new_face_rect_area, warped_triangle)
img2_new_face[y2: y2 + h2, x2: x2 + w2] = img2_new_face_rect_area
cv2.imshow('mask', img2_new_face)
key = cv2.waitKey(5)
face_mesh.close()
cap.release()
def process_base_face_mesh(face_mesh,
image_file):
base_face_handler = load_base_img(face_mesh, image_file)
base_input_image = base_face_handler["img"].copy()
image_rows, image_cols, _ = base_face_handler["img"].shape
landmark_base_ocv = transform_landmarks_from_tf_to_ocv(base_face_handler["landmarks"], image_cols, image_rows)
return base_face_handler, landmark_base_ocv, base_input_image
if __name__ == "__main__":
main()
triangulation_media_pipe.py
The code is a slightly modified version of this
A possible solution I added was to get rid of the artifacts is to just mask those areas with a blurred version of the same image, but I want the pixel values to be as close as possible to the real image values.
Related
I am working on tutorial which use Isomap for image reconginzation, the code is as following:
the major error is that reshape function in def Plot2D , ValueError: cannot reshape array of size 72 into shape (8,8).
function for 2d plot :
def Plot2D(T, title, x, y, num_to_plot=40):
# This method picks a bunch of random samples (images in your case)
# to plot onto the chart:
fig = plt.figure()
ax = fig.add_subplot(111)
ax.set_title(title)
ax.set_xlabel('Component: {0}'.format(x))
ax.set_ylabel('Component: {0}'.format(y))
x_size = (max(T[:,x]) - min(T[:,x])) * 0.08
y_size = (max(T[:,y]) - min(T[:,y])) * 0.08
for i in range(num_to_plot):
img_num = int(random.random() * num_images)
x0, y0 = T[img_num,x]-x_size/2., T[img_num,y]-y_size/2.
x1, y1 = T[img_num,x]+x_size/2., T[img_num,y]+y_size/2.
img = df.iloc[img_num,:].reshape(num_pixels, num_pixels)
ax.imshow(img, aspect='auto', cmap=plt.cm.gray, interpolation='nearest', zorder=100000, extent=(x0, x1, y0, y1))
function for image uploading and processing:
df = []
for image_path in glob.glob("path/*.png"):
image= misc.imread(image_path)
df.append((image[::2, ::2] / 255.0).reshape(-1))
df = pd.DataFrame(df).T
iso = Isomap(n_neighbors=3,n_components=3).fit(df)
T = iso.transform(df)
function for Plotting :
num_images, num_pixels = df.shape
num_pixels = int(math.sqrt(num_pixels))
Plot2D(T, "test", 0, 1, num_to_plot=40)
error message:
<ipython-input-30-e9aeee7b57c9> in Plot2D(T, title, x, y, num_to_plot)
16 x0, y0 = T[img_num,x]-x_size/2., T[img_num,y]-y_size/2.
17 x1, y1 = T[img_num,x]+x_size/2., T[img_num,y]+y_size/2.
-> 18 img = df.iloc[img_num,:].reshape(num_pixels, num_pixels)
19 ax.imshow(img, aspect='auto', cmap=plt.cm.gray,
interpolation='nearest', zorder=100000, extent=(x0, x1, y0, y1))
ValueError: cannot reshape array of size 72 into shape (8,8)
I know it might be too late for you, but hopefully whoever is using this course won't waste their time on something so simple (like I did this morning). The course doesn't get updated anymore and some pandas and numpy function changed since its first release I believe.
img = df.iloc[img_num,:].reshape(num_pixels, num_pixels)
Needs to be changed into:
img = df.iloc[img_num,:].values.reshape(num_pixels, num_pixels)
Same with the code where they transform the MATLAB file:
for i in range(num_images):
df.loc[i,:] = df.loc[i,:].reshape(num_pixels, num_pixels).T.reshape(-1)
change to:
for i in range(num_images):
df.loc[i,:] = df.loc[i,:].values.reshape(num_pixels, num_pixels).T.reshape(-1)
I have lena image in Matlab. First I need to find the centroid C, and then divide the image into equal number of parts. I can calculate the centroid of the image but from that how can I divide the image into equal number of parts as shown below. Please anyone help me.
Thanks
Using poly2mask to create binary sectors and using the resulting sectors for indexing
Code:
im = imread('peppers.png');
r = 300;
out1 = ones(max(size(im,1),r*2)+2,max(size(im,2),r*2)+2,3).*255;
xoffset = floor((size(out1,2)-size(im,2))/2);
yoffset = floor((size(out1,1)-size(im,1))/2);
out1(yoffset:yoffset+size(im,1)-1,xoffset:xoffset+size(im,2)-1,:) = im(:,:,:);
im = out1;
cy = floor(size(im,1)/2);
cx = floor(size(im,2)/2);
figure;
imshow(uint8(im));
hold on
pos = [cx-r+1 cy-r+1 r*2 r*2];
rectangle('Position',pos,'Curvature',[1 1]);
x1 = [-r, 0, -r*cosd(45), -r*cosd(45); r, 0, r*cosd(45), r*cosd(45)]+cx+1;
y1 = [0, -r, -r*sind(45), r*sind(45); 0, r, r*sind(45), -r*sind(45)]+cy+1;
plot(x1,y1);
hold off
figure;
for i = 0:45:315
t = linspace(-i,-i-45,128);
x = [cx, cx+r*cosd(t), cx];
y = [cy, cy+r*sind(t), cy];
bw = poly2mask( x, y, size(im,1),size(im,2));
bw = repmat(bw,1,1,3);
out = ones(size(im,1),size(im,2),size(im,3)).*155;
out(bw) = im(bw);
subplot(2,4,(i/45)+1); imshow(uint8(out));
end;
Results:
Original Image
Partitions drawn over Original Image
Segments of the image
Update
for getting pixel values of the lines, by using Bresenham function from here
figure;
bw1 = zeros(size(im,1),size(im,2));
outmat = zeros(size(bw1));
[X,Y] = bresenham(cx+1-r,cy+1,cx+1+r,cy+1);
ind = sub2ind(size(outmat), Y, X);
outmat(ind) = 1;
[X,Y] = bresenham(cx+1,cy+1-r,cx+1,cy+1+r);
ind = sub2ind(size(outmat), Y, X);
outmat(ind) = 1;
[X,Y] = bresenham(cx+1-r*cosd(45),cy+1-r*sind(45),cx+1+r*cosd(45),cy+1+r*sind(45));
ind = sub2ind(size(outmat), Y, X);
outmat(ind) = 1;
[X,Y] = bresenham(cx+1-r*cosd(45),cy+1+r*sind(45),cx+1+r*cosd(45),cy+1-r*sind(45));
ind = sub2ind(size(outmat), Y, X);
outmat(ind) = 1;
se = strel('disk',5); %// change the '5' value to affect thickness of the line
outmat = imdilate(outmat,se);
outmat = repmat(boolean(outmat),1,1,3);
outmat1 = zeros(size(outmat));
outmat1(outmat) = im(outmat);
imshow(uint8(outmat1));
Pixel values under each lines
Check the following code. I just did it for a grayscale image. You can now change it to a color image as well. Check and pls confirm this is what you wanted.
clear all;
i = rgb2gray(imread('hestain.png'));
imshow(i);
cr = floor(size(i,1)/2);
cl = floor(size(i,2)/2);
r = min(cr, cl);
a = 90;
r1 = cr;
c1 = size(i,2);
v1=[c1 r1]-[cl cr];
i2 = zeros(size(i,1),size(i,2),ceil(360/a));
for ri = 1:size(i,1)
for ci = 1:size(i,2)
v2=[ci ri]-[cl cr];
a2 = mod(-atan2(v1(1)*v2(2)-v1(2)*v2(1), v1*v2'), 2*pi) * 180/pi;
d2 = pdist([ci ri; cl cr],'euclidean');
if d2<=r
if ceil(a2/a)==0
a2 =1;
end
i2(ri,ci,ceil(a2/a)) = i(ri,ci);
end
end
end
figure;
for i=1:360/a
subplot(2,180/a,i);
imshow(mat2gray(i2(:,:,i)));
end
Sample output:
Is there a way (using something like OpenCV) to detect text skew and correct it by rotating the image? Pretty much like this?
Rotating an image seems easy enough if you know the angle, but for the images I'm processing, I wont...it will need to be detected somehow.
Based on your above comment, here is the code based on the tutorial here, working fine for the above image,
Source
Rotated
Mat src=imread("text.png",0);
Mat thr,dst;
threshold(src,thr,200,255,THRESH_BINARY_INV);
imshow("thr",thr);
std::vector<cv::Point> points;
cv::Mat_<uchar>::iterator it = thr.begin<uchar>();
cv::Mat_<uchar>::iterator end = thr.end<uchar>();
for (; it != end; ++it)
if (*it)
points.push_back(it.pos());
cv::RotatedRect box = cv::minAreaRect(cv::Mat(points));
cv::Mat rot_mat = cv::getRotationMatrix2D(box.center, box.angle, 1);
//cv::Mat rotated(src.size(),src.type(),Scalar(255,255,255));
Mat rotated;
cv::warpAffine(src, rotated, rot_mat, src.size(), cv::INTER_CUBIC);
imshow("rotated",rotated);
Edit:
Also see the answer here , might be helpful.
Here's an implementation of the Projection Profile Method algorithm for skew angle estimation. Various angle points are projected into an accumulator array where the skew angle can be defined as the angle of projection within a search interval that maximizes alignment. The idea is to rotate the image at various angles and generate a histogram of pixels for each iteration. To determine the skew angle, we compare the maximum difference between peaks and using this skew angle, rotate the image to correct the skew.
Input
Result
Skew angle: -5
import cv2
import numpy as np
from scipy.ndimage import interpolation as inter
def correct_skew(image, delta=1, limit=5):
def determine_score(arr, angle):
data = inter.rotate(arr, angle, reshape=False, order=0)
histogram = np.sum(data, axis=1, dtype=float)
score = np.sum((histogram[1:] - histogram[:-1]) ** 2, dtype=float)
return histogram, score
gray = cv2.cvtColor(image, cv2.COLOR_BGR2GRAY)
thresh = cv2.threshold(gray, 0, 255, cv2.THRESH_BINARY_INV + cv2.THRESH_OTSU)[1]
scores = []
angles = np.arange(-limit, limit + delta, delta)
for angle in angles:
histogram, score = determine_score(thresh, angle)
scores.append(score)
best_angle = angles[scores.index(max(scores))]
(h, w) = image.shape[:2]
center = (w // 2, h // 2)
M = cv2.getRotationMatrix2D(center, best_angle, 1.0)
corrected = cv2.warpAffine(image, M, (w, h), flags=cv2.INTER_CUBIC, \
borderMode=cv2.BORDER_REPLICATE)
return best_angle, corrected
if __name__ == '__main__':
image = cv2.imread('1.png')
angle, corrected = correct_skew(image)
print('Skew angle:', angle)
cv2.imshow('corrected', corrected)
cv2.waitKey()
Note: You may have to adjust the delta or limit values depending on the image. The delta value controls iteration step, it will iterate up until the limit which controls the maximum angle. This method is straightforward by iteratively checking each angle + delta and currently only works to correct skew in the range of +/- 5 degrees. If you need to correct at a larger angle, adjust the limit value.
I would provide javacv for your reference.
package com.test13;
import org.opencv.core.*;
import org.opencv.imgproc.Imgproc;
import org.opencv.imgcodecs.Imgcodecs;
public class EdgeDetection {
static{ System.loadLibrary(Core.NATIVE_LIBRARY_NAME); }
public static void main( String[] args ) throws Exception{
Mat src = Imgcodecs.imread("src//data//inclined_text.jpg");
Mat src_gray = new Mat();
Imgproc.cvtColor(src, src_gray, Imgproc.COLOR_BGR2GRAY);
Imgcodecs.imwrite("src//data//inclined_text_src_gray.jpg", src_gray);
Mat output = new Mat();
Core.bitwise_not(src_gray, output);
Imgcodecs.imwrite("src//data//inclined_text_output.jpg", output);
Mat points = Mat.zeros(output.size(),output.type());
Core.findNonZero(output, points);
MatOfPoint mpoints = new MatOfPoint(points);
MatOfPoint2f points2f = new MatOfPoint2f(mpoints.toArray());
RotatedRect box = Imgproc.minAreaRect(points2f);
Mat src_squares = src.clone();
Mat rot_mat = Imgproc.getRotationMatrix2D(box.center, box.angle, 1);
Mat rotated = new Mat();
Imgproc.warpAffine(src_squares, rotated, rot_mat, src_squares.size(), Imgproc.INTER_CUBIC);
Imgcodecs.imwrite("src//data//inclined_text_squares_rotated.jpg",rotated);
}
}
private fun main(){
val bmp:Bitmap? = null //Any bitmap (if you are working with bitmap)
var mRgba = Mat() // else you can direct use MAT on onCameraFrame
val mGray = Mat()
val bmp32: Bitmap = bmp.copy(Bitmap.Config.ARGB_8888, true)
Utils.bitmapToMat(bmp32, mRgba)
Imgproc.cvtColor(mRgba, mGray, Imgproc.COLOR_BGR2GRAY)
mRgba = makeOrientationCorrection(mRgba,mGray)// here actual magic starts
Imgproc.cvtColor(mRgba, mGray, Imgproc.COLOR_BGR2GRAY)
val bmpOutX = Bitmap.createBitmap(
mRgba.cols(),
mRgba.rows(),
Bitmap.Config.ARGB_8888
)
Utils.matToBitmap(mRgba, bmpOutX)
binding.imagePreview.setImageBitmap(bmpOutX!!)
}
private fun makeOrientationCorrection(mRGBA:Mat, mGRAY:Mat):Mat{
val dst = Mat()
val cdst = Mat()
val cdstP: Mat
Imgproc.Canny(mGRAY, dst, 50.0, 200.0, 3, false)
Imgproc.cvtColor(dst, cdst, Imgproc.COLOR_GRAY2BGR)
cdstP = cdst.clone()
val linesP = Mat()
Imgproc.HoughLinesP(dst, linesP, 1.0, Math.PI/180, 50, 50.0, 10.0)
var biggestLineX1 = 0.0
var biggestLineY1 = 0.0
var biggestLineX2 = 0.0
var biggestLineY2 = 0.0
var biggestLine = 0.0
for (x in 0 until linesP.rows()) {
val l = linesP[x, 0]
Imgproc.line(
cdstP, org.opencv.core.Point(l[0], l[1]),
org.opencv.core.Point(l[2], l[3]),
Scalar(0.0, 0.0, 255.0), 3, Imgproc.LINE_AA, 0)
}
for (x in 0 until linesP.rows()) {
val l = linesP[x, 0]
val x1 = l[0]
val y1 = l[1]
val x2 = l[2]
val y2 = l[3]
val lineHeight = sqrt(((x2 - x1).pow(2.0)) + ((y2 - y1).pow(2.0)))
if(biggestLine<lineHeight){
val angleOfRotationX1 = angleOf(PointF(x1.toFloat(),y1.toFloat()),PointF(x2.toFloat(),y2.toFloat()))
Log.e("angleOfRotationX1","$angleOfRotationX1")
if(angleOfRotationX1<45.0 || angleOfRotationX1>270.0){
biggestLine = lineHeight
if(angleOfRotationX1<45.0){
biggestLineX1 = x1
biggestLineY1 = y1
biggestLineX2 = x2
biggestLineY2 = y2
}
if(angleOfRotationX1>270.0){
biggestLineX1 = x2
biggestLineY1 = y2
biggestLineX2 = x1
biggestLineY2 = y1
}
}
}
if(x==linesP.rows()-1){
Imgproc.line(
cdstP, org.opencv.core.Point(biggestLineX1, biggestLineY1),
org.opencv.core.Point(biggestLineX2, biggestLineY2),
Scalar(255.0, 0.0, 0.0), 3, Imgproc.LINE_AA, 0)
}
}
var angle = angleOf(PointF(biggestLineX1.toFloat(),biggestLineY1.toFloat()),PointF(biggestLineX2.toFloat(),biggestLineY2.toFloat()))
Log.e("angleOfRotationX2","$angle")
angle -= (angle * 2)
return deskew(mRGBA,angle)
}
fun angleOf(p1: PointF, p2: PointF): Double {
val deltaY = (p1.y - p2.y).toDouble()
val deltaX = (p2.x - p1.x).toDouble()
val result = Math.toDegrees(Math.atan2(deltaY, deltaX))
return if (result < 0) 360.0 + result else result
}
private fun deskew(src:Mat, angle:Double):Mat{
val center = org.opencv.core.Point((src.width() / 2).toDouble(), (src.height() / 2).toDouble())
val scaleBy = if(angle<0){
1.0+((0.5*angle)/45)//max scale down by 0.50(50%) based on angle
}else{
1.0-((0.3*angle)/45)//max scale down by 0.50(50%) based on angle
}
Log.e("scaleBy",""+scaleBy)
val rotImage = Imgproc.getRotationMatrix2D(center, angle, scaleBy)
val size = Size(src.width().toDouble(), src.height().toDouble())
Imgproc.warpAffine(src, src, rotImage, size, Imgproc.INTER_LINEAR + Imgproc.CV_WARP_FILL_OUTLIERS)
return src
}
Make sure you run this "makeOrientationCorrection()" method on another thread. otherwise, UI won't update for 2-5 sec.
I have a polygon made of successive edges on a plane, and would like to subdivide it in sub-polygons being triangles or rectangles.
Where can I find an algorithm to do this ?
Thanks !
In computational geometry, the problem you want to solve is called triangulation.
There are algorithms to solve this problem, giving triangulations with different properties. You will need to decide which one is the best fit.
I was looking for an answer for this myself but couldn't find one. Tried to stitch together several pieces and here's the result.
This is not necessarily the most optimal routine but it did the job for me. If you want to increase performance, try experimenting with the code.
A brief description of the algorithm:
Using the boundaries of the original geometry itself, and the boundaries of its convex hull, and its minimum rotated rectangle, derive all possible rectangles.
Divide all rectangles into smaller squares of specified side length.
Drop duplicates using a rounded off centroid. (r: round off param)
Retain either those squares 'within' the geometry, or those that 'intersect' the geometry, depending on whichever is closer to the total number of required squares.
EDITED
New Solution
#### Python script for dividing any shapely polygon into smaller equal sized polygons
import numpy as np
from shapely.ops import split
import geopandas
from shapely.geometry import MultiPolygon, Polygon
def rhombus(square):
"""
Naively transform the square into a Rhombus at a 45 degree angle
"""
coords = square.boundary.coords.xy
xx = list(coords[0])
yy = list(coords[1])
radians = 1
points = list(zip(xx, yy))
Rhombus = Polygon(
[
points[0],
points[1],
points[3],
((2 * points[3][0]) - points[2][0], (2 * points[3][1]) - points[2][1]),
points[4],
]
)
return Rhombus
def get_squares_from_rect(RectangularPolygon, side_length=0.0025):
"""
Divide a Rectangle (Shapely Polygon) into squares of equal area.
`side_length` : required side of square
"""
rect_coords = np.array(RectangularPolygon.boundary.coords.xy)
y_list = rect_coords[1]
x_list = rect_coords[0]
y1 = min(y_list)
y2 = max(y_list)
x1 = min(x_list)
x2 = max(x_list)
width = x2 - x1
height = y2 - y1
xcells = int(np.round(width / side_length))
ycells = int(np.round(height / side_length))
yindices = np.linspace(y1, y2, ycells + 1)
xindices = np.linspace(x1, x2, xcells + 1)
horizontal_splitters = [
LineString([(x, yindices[0]), (x, yindices[-1])]) for x in xindices
]
vertical_splitters = [
LineString([(xindices[0], y), (xindices[-1], y)]) for y in yindices
]
result = RectangularPolygon
for splitter in vertical_splitters:
result = MultiPolygon(split(result, splitter))
for splitter in horizontal_splitters:
result = MultiPolygon(split(result, splitter))
square_polygons = list(result)
return square_polygons
def split_polygon(G, side_length=0.025, shape="square", thresh=0.9):
"""
Using a rectangular envelope around `G`, creates a mesh of squares of required length.
Removes non-intersecting polygons.
Args:
- `thresh` : Range - [0,1]
This controls - the number of smaller polygons at the boundaries.
A thresh == 1 will only create (or retain) smaller polygons that are
completely enclosed (area of intersection=area of smaller polygon)
by the original Geometry - `G`.
A thresh == 0 will create (or retain) smaller polygons that
have a non-zero intersection (area of intersection>0) with the
original geometry - `G`
- `side_length` : Range - (0,infinity)
side_length must be such that the resultant geometries are smaller
than the original geometry - `G`, for a useful result.
side_length should be >0 (non-zero positive)
- `shape` : {square/rhombus}
Desired shape of subset geometries.
"""
assert side_length>0, "side_length must be a float>0"
Rectangle = G.envelope
squares = get_squares_from_rect(Rectangle, side_length=side_length)
SquareGeoDF = geopandas.GeoDataFrame(squares).rename(columns={0: "geometry"})
Geoms = SquareGeoDF[SquareGeoDF.intersects(G)].geometry.values
if shape == "rhombus":
Geoms = [rhombus(g) for g in Geoms]
geoms = [g for g in Geoms if ((g.intersection(G)).area / g.area) >= thresh]
elif shape == "square":
geoms = [g for g in Geoms if ((g.intersection(G)).area / g.area) >= thresh]
return geoms
# Reading geometric data
geo_filepath = "/data/geojson/pc_14.geojson"
GeoDF = geopandas.read_file(geo_filepath)
# Selecting random shapely-geometry
G = np.random.choice(GeoDF.geometry.values)
squares = split_polygon(G,shape='square',thresh=0.5,side_length=0.025)
rhombuses = split_polygon(G,shape='rhombus',thresh=0.5,side_length=0.025)
Previous Solution:
import numpy as np
import geopandas
from shapely.ops import split
from shapely.geometry import MultiPolygon, Polygon, Point, MultiPoint
def get_rect_from_geom(G, r=2):
"""
Get rectangles from a geometry.
r = rounding factor.
small r ==> more rounding off ==> more rectangles
"""
coordinate_arrays = G.exterior.coords.xy
coordinates = list(
zip(
[np.round(c, r) for c in coordinate_arrays[0]],
[np.round(c, r) for c in coordinate_arrays[1]],
)
)
Rectangles = []
for c1 in coordinates:
Coords1 = [a for a in coordinates if a != c1]
for c2 in Coords1:
Coords2 = [b for b in Coords1 if b != c2]
x1, y1 = c1[0], c1[1]
x2, y2 = c2[0], c2[1]
K1 = [k for k in Coords2 if k == (x1, y2)]
K2 = [k for k in Coords2 if k == (x2, y1)]
if (len(K1) > 0) & (len(K2) > 0):
rect = [list(c1), list(K1[0]), list(c2), list(K2[0])]
Rectangles.append(rect)
return Rectangles
def get_squares_from_rect(rect, side_length=0.0025):
"""
Divide a rectangle into equal area squares
side_length = required side of square
"""
y_list = [r[1] for r in rect]
x_list = [r[0] for r in rect]
y1 = min(y_list)
y2 = max(y_list)
x1 = min(x_list)
x2 = max(x_list)
width = x2 - x1
height = y2 - y1
xcells, ycells = int(np.round(width / side_length)), int(
np.round(height / side_length)
)
yindices = np.linspace(y1, y2, ycells + 1)
xindices = np.linspace(x1, x2, xcells + 1)
horizontal_splitters = [
LineString([(x, yindices[0]), (x, yindices[-1])]) for x in xindices
]
vertical_splitters = [
LineString([(xindices[0], y), (xindices[-1], y)]) for y in yindices
]
result = Polygon(rect)
for splitter in vertical_splitters:
result = MultiPolygon(split(result, splitter))
for splitter in horizontal_splitters:
result = MultiPolygon(split(result, splitter))
square_polygons = list(result)
return [np.stack(SQPOLY.exterior.coords.xy, axis=1) for SQPOLY in square_polygons]
def round_centroid(g, r=10):
"""
Get Centroids.
Round off centroid coordinates to `r` decimal points.
"""
C = g.centroid.coords.xy
return (np.round(C[0][0], r), np.round(C[1][0], r))
def subdivide_polygon(g, side_length=0.0025, r=10):
"""
1. Create all possible rectangles coordinates from the geometry, its minimum rotated rectangle, and its convex hull.
2. Divide all rectangles into smaller squares.
small r ==> more rounding off ==> fewer overlapping squares. (these are dropped as duplicates)
large r ==> may lead to a few overlapping squares.
"""
# Number of squares required.
num_squares_reqd = g.area // (side_length ** 2)
# Some of these combinations can be dropped to improve performance.
Rectangles = []
Rectangles.extend(get_rect_from_geom(g))
Rectangles.extend(get_rect_from_geom(g.minimum_rotated_rectangle))
Rectangles.extend(get_rect_from_geom(g.convex_hull))
Squares = []
for r in range(len(Rectangles)):
rect = Rectangles[r]
Squares.extend(get_squares_from_rect(rect, side_length=side_length))
SquarePolygons = [Polygon(square) for square in Squares]
GDF = geopandas.GeoDataFrame(SquarePolygons).rename(columns={0: "geometry"})
GDF.loc[:, "centroid"] = GDF.geometry.apply(round_centroid, r=r)
GDF = GDF.drop_duplicates(subset=["centroid"])
wgeoms = GDF[GDF.within(g)].geometry.values
igeoms = GDF[GDF.intersects(g)].geometry.values
w = abs(num_squares_reqd - len(wgeoms))
i = abs(num_squares_reqd - len(igeoms))
print(w, i)
if w <= i:
return wgeoms
else:
return igeoms
geoms = subdivide(g)
Stumbled across this after many searches.
Thanks #Aditya Chhabra for your submission, it works great but get_squares_from_rect is very slow for small side lengths due to iterative clips.
We can do this instantaneously if we combine all LineStrings into a single collection, then clip and polygonize in one step, which I found in in this question.
Previously side lengths of 0.0001 (EPSG:4326) took > 1 minute, now it takes no time.
from shapely.ops import unary_union, polygonize, linemerge
from shapely.geometry import LineString
import numpy as np
def get_squares_from_rect_faster(RectangularPolygon, side_length=0.0025):
rect_coords = np.array(RectangularPolygon.boundary.coords.xy)
y_list = rect_coords[1]
x_list = rect_coords[0]
y1 = min(y_list)
y2 = max(y_list)
x1 = min(x_list)
x2 = max(x_list)
width = x2 - x1
height = y2 - y1
xcells = int(np.round(width / side_length))
ycells = int(np.round(height / side_length))
yindices = np.linspace(y1, y2, ycells + 1)
xindices = np.linspace(x1, x2, xcells + 1)
horizontal_splitters = [
LineString([(x, yindices[0]), (x, yindices[-1])]) for x in xindices
]
vertical_splitters = [
LineString([(xindices[0], y), (xindices[-1], y)]) for y in yindices
]
lines = horizontal_splitters + vertical_splitters
lines.append(RectangularPolygon.boundary)
lines = unary_union(lines)
lines = linemerge(lines)
return list(polygonize(lines))
How do I calculate the intersection points of two circles. I would expect there to be either two, one or no intersection points in all cases.
I have the x and y coordinates of the centre-point, and the radius for each circle.
An answer in python would be preferred, but any working algorithm would be acceptable.
Intersection of two circles
Written by Paul Bourke
The following note describes how to find the intersection point(s)
between two circles on a plane, the following notation is used. The
aim is to find the two points P3 = (x3,
y3) if they exist.
First calculate the distance d between the center
of the circles. d = ||P1 - P0||.
If d > r0 + r1 then there are no solutions,
the circles are separate. If d < |r0 -
r1| then there are no solutions because one circle is
contained within the other. If d = 0 and r0 =
r1 then the circles are coincident and there are an
infinite number of solutions.
Considering the two triangles P0P2P3
and P1P2P3 we can write
a2 + h2 = r02 and
b2 + h2 = r12
Using d = a + b we can solve for a, a =
(r02 - r12 +
d2 ) / (2 d)
It can be readily shown that this reduces to
r0 when the two circles touch at one point, ie: d =
r0 + r1
Solve for h by substituting a into the first
equation, h2 = r02 - a2
So P2 = P0 + a ( P1 -
P0 ) / d And finally, P3 =
(x3,y3) in terms of P0 =
(x0,y0), P1 =
(x1,y1) and P2 =
(x2,y2), is x3 =
x2 +- h ( y1 - y0 ) / d
y3 = y2 -+ h ( x1 - x0 ) /
d
Source: http://paulbourke.net/geometry/circlesphere/
Here is my C++ implementation based on Paul Bourke's article. It only works if there are two intersections, otherwise it probably returns NaN NAN NAN NAN.
class Point{
public:
float x, y;
Point(float px, float py) {
x = px;
y = py;
}
Point sub(Point p2) {
return Point(x - p2.x, y - p2.y);
}
Point add(Point p2) {
return Point(x + p2.x, y + p2.y);
}
float distance(Point p2) {
return sqrt((x - p2.x)*(x - p2.x) + (y - p2.y)*(y - p2.y));
}
Point normal() {
float length = sqrt(x*x + y*y);
return Point(x/length, y/length);
}
Point scale(float s) {
return Point(x*s, y*s);
}
};
class Circle {
public:
float x, y, r, left;
Circle(float cx, float cy, float cr) {
x = cx;
y = cy;
r = cr;
left = x - r;
}
pair<Point, Point> intersections(Circle c) {
Point P0(x, y);
Point P1(c.x, c.y);
float d, a, h;
d = P0.distance(P1);
a = (r*r - c.r*c.r + d*d)/(2*d);
h = sqrt(r*r - a*a);
Point P2 = P1.sub(P0).scale(a/d).add(P0);
float x3, y3, x4, y4;
x3 = P2.x + h*(P1.y - P0.y)/d;
y3 = P2.y - h*(P1.x - P0.x)/d;
x4 = P2.x - h*(P1.y - P0.y)/d;
y4 = P2.y + h*(P1.x - P0.x)/d;
return pair<Point, Point>(Point(x3, y3), Point(x4, y4));
}
};
Why not just use 7 lines of your favorite procedural language (or programmable calculator!) as below.
Assuming you are given P0 coords (x0,y0), P1 coords (x1,y1), r0 and r1 and you want to find P3 coords (x3,y3):
d=sqr((x1-x0)^2 + (y1-y0)^2)
a=(r0^2-r1^2+d^2)/(2*d)
h=sqr(r0^2-a^2)
x2=x0+a*(x1-x0)/d
y2=y0+a*(y1-y0)/d
x3=x2+h*(y1-y0)/d // also x3=x2-h*(y1-y0)/d
y3=y2-h*(x1-x0)/d // also y3=y2+h*(x1-x0)/d
Here's an implementation in Javascript using vectors. The code is well documented, you should be able to follow it. Here's the original source
See live demo here:
// Let EPS (epsilon) be a small value
var EPS = 0.0000001;
// Let a point be a pair: (x, y)
function Point(x, y) {
this.x = x;
this.y = y;
}
// Define a circle centered at (x,y) with radius r
function Circle(x,y,r) {
this.x = x;
this.y = y;
this.r = r;
}
// Due to double rounding precision the value passed into the Math.acos
// function may be outside its domain of [-1, +1] which would return
// the value NaN which we do not want.
function acossafe(x) {
if (x >= +1.0) return 0;
if (x <= -1.0) return Math.PI;
return Math.acos(x);
}
// Rotates a point about a fixed point at some angle 'a'
function rotatePoint(fp, pt, a) {
var x = pt.x - fp.x;
var y = pt.y - fp.y;
var xRot = x * Math.cos(a) + y * Math.sin(a);
var yRot = y * Math.cos(a) - x * Math.sin(a);
return new Point(fp.x+xRot,fp.y+yRot);
}
// Given two circles this method finds the intersection
// point(s) of the two circles (if any exists)
function circleCircleIntersectionPoints(c1, c2) {
var r, R, d, dx, dy, cx, cy, Cx, Cy;
if (c1.r < c2.r) {
r = c1.r; R = c2.r;
cx = c1.x; cy = c1.y;
Cx = c2.x; Cy = c2.y;
} else {
r = c2.r; R = c1.r;
Cx = c1.x; Cy = c1.y;
cx = c2.x; cy = c2.y;
}
// Compute the vector <dx, dy>
dx = cx - Cx;
dy = cy - Cy;
// Find the distance between two points.
d = Math.sqrt( dx*dx + dy*dy );
// There are an infinite number of solutions
// Seems appropriate to also return null
if (d < EPS && Math.abs(R-r) < EPS) return [];
// No intersection (circles centered at the
// same place with different size)
else if (d < EPS) return [];
var x = (dx / d) * R + Cx;
var y = (dy / d) * R + Cy;
var P = new Point(x, y);
// Single intersection (kissing circles)
if (Math.abs((R+r)-d) < EPS || Math.abs(R-(r+d)) < EPS) return [P];
// No intersection. Either the small circle contained within
// big circle or circles are simply disjoint.
if ( (d+r) < R || (R+r < d) ) return [];
var C = new Point(Cx, Cy);
var angle = acossafe((r*r-d*d-R*R)/(-2.0*d*R));
var pt1 = rotatePoint(C, P, +angle);
var pt2 = rotatePoint(C, P, -angle);
return [pt1, pt2];
}
Try this;
def ri(cr1,cr2,cp1,cp2):
int1=[]
int2=[]
ori=0
if cp1[0]<cp2[0] and cp1[1]!=cp2[1]:
p1=cp1
p2=cp2
r1=cr1
r2=cr2
if cp1[1]<cp2[1]:
ori+=1
elif cp1[1]>cp2[1]:
ori+=2
elif cp1[0]>cp2[0] and cp1[1]!=cp2[1]:
p1=cp2
p2=cp1
r1=cr2
r2=cr1
if p1[1]<p2[1]:
ori+=1
elif p1[1]>p2[1]:
ori+=2
elif cp1[0]==cp2[0]:
ori+=4
if cp1[1]>cp2[1]:
p1=cp1
p2=cp2
r1=cr1
r2=cr2
elif cp1[1]<cp2[1]:
p1=cp2
p2=cp1
r1=cr2
r2=cr1
elif cp1[1]==cp2[1]:
ori+=3
if cp1[0]>cp2[0]:
p1=cp2
p2=cp1
r1=cr2
r2=cr1
elif cp1[0]<cp2[0]:
p1=cp1
p2=cp2
r1=cr1
r2=cr2
if ori==1:#+
D=calc_dist(p1,p2)
tr=r1+r2
el=tr-D
a=r1-el
b=r2-el
A=a+(el/2)
B=b+(el/2)
thta=math.degrees(math.acos(A/r1))
rs=p2[1]-p1[1]
rn=p2[0]-p1[0]
gd=rs/rn
yint=p1[1]-((gd)*p1[0])
dty=calc_dist(p1,[0,yint])
aa=p1[1]-yint
bb=math.degrees(math.asin(aa/dty))
d=90-bb
e=180-d-thta
g=(dty/math.sin(math.radians(e)))*math.sin(math.radians(thta))
f=(g/math.sin(math.radians(thta)))*math.sin(math.radians(d))
oty=yint+g
h=f+r1
i=90-e
j=180-90-i
l=math.sin(math.radians(i))*h
k=math.cos(math.radians(i))*h
iy2=oty-l
ix2=k
int2.append(ix2)
int2.append(iy2)
m=90+bb
n=180-m-thta
p=(dty/math.sin(math.radians(n)))*math.sin(math.radians(m))
o=(p/math.sin(math.radians(m)))*math.sin(math.radians(thta))
q=p+r1
r=90-n
s=math.sin(math.radians(r))*q
t=math.cos(math.radians(r))*q
otty=yint-o
iy1=otty+s
ix1=t
int1.append(ix1)
int1.append(iy1)
elif ori==2:#-
D=calc_dist(p1,p2)
tr=r1+r2
el=tr-D
a=r1-el
b=r2-el
A=a+(el/2)
B=b+(el/2)
thta=math.degrees(math.acos(A/r1))
rs=p2[1]-p1[1]
rn=p2[0]-p1[0]
gd=rs/rn
yint=p1[1]-((gd)*p1[0])
dty=calc_dist(p1,[0,yint])
aa=yint-p1[1]
bb=math.degrees(math.asin(aa/dty))
c=180-90-bb
d=180-c-thta
e=180-90-d
f=math.tan(math.radians(e))*p1[0]
g=math.sqrt(p1[0]**2+f**2)
h=g+r1
i=180-90-e
j=math.sin(math.radians(e))*h
jj=math.cos(math.radians(i))*h
k=math.cos(math.radians(e))*h
kk=math.sin(math.radians(i))*h
l=90-bb
m=90-e
tt=l+m+thta
n=(dty/math.sin(math.radians(m)))*math.sin(math.radians(thta))
nn=(g/math.sin(math.radians(l)))*math.sin(math.radians(thta))
oty=yint-n
iy1=oty+j
ix1=k
int1.append(ix1)
int1.append(iy1)
o=bb+90
p=180-o-thta
q=90-p
r=180-90-q
s=(dty/math.sin(math.radians(p)))*math.sin(math.radians(o))
t=(s/math.sin(math.radians(o)))*math.sin(math.radians(thta))
u=s+r1
v=math.sin(math.radians(r))*u
vv=math.cos(math.radians(q))*u
w=math.cos(math.radians(r))*u
ww=math.sin(math.radians(q))*u
ix2=v
otty=yint+t
iy2=otty-w
int2.append(ix2)
int2.append(iy2)
elif ori==3:#y
D=calc_dist(p1,p2)
tr=r1+r2
el=tr-D
a=r1-el
b=r2-el
A=a+(el/2)
B=b+(el/2)
b=math.sqrt(r1**2-A**2)
int1.append(p1[0]+A)
int1.append(p1[1]+b)
int2.append(p1[0]+A)
int2.append(p1[1]-b)
elif ori==4:#x
D=calc_dist(p1,p2)
tr=r1+r2
el=tr-D
a=r1-el
b=r2-el
A=a+(el/2)
B=b+(el/2)
b=math.sqrt(r1**2-A**2)
int1.append(p1[0]+b)
int1.append(p1[1]-A)
int2.append(p1[0]-b)
int2.append(p1[1]-A)
return [int1,int2]
def calc_dist(p1,p2):
return math.sqrt((p2[0] - p1[0]) ** 2 +
(p2[1] - p1[1]) ** 2)