I have collected images of S9 phones, added labels with labellmg and trained for a few hours in google colab. I had minimal loss so I thought it is enough. I only selected the rectangles where the phone is displayed and nothing else. What I dont understand is, it draws a lot of rectangles on the phone. I only want 1 or 2 rectangles drawn on the phone itself. Did I do something wrong?
def detect_img(self, img):
blob = cv2.dnn.blobFromImage(img, 0.00392 ,(416,416), (0,0,0), True, crop=False)
input_img = self.net.setInput(blob)
output = self.net.forward(self.output)
height, width, channel = img.shape
boxes = []
trusts = []
class_ids = []
for out in output:
for detect in out:
total_scores = detect[5:]
class_id = np.argmax(total_scores)
trust_factor = total_scores[class_id]
if trust_factor > 0.5:
x_center = int(detect[0] * width)
y_center = int(detect[1] * height)
w = int(detect[2] * width)
h = int(detect[3] * height)
x = int(x_center - w / 2)
y = int(x_center - h / 2)
boxes.append([x,y,w,h])
trusts.append(float(trust_factor))
class_ids.append(class_id)
cv2.rectangle(img, (x_center,y_center), (x + w, y + h), (0,255,0), 2)
When I set the trust_factor to 0.8, a lot of the rectangles are gone but there are still rectangles outside the phone, while I only selected the phone itself in labellmg and not the background.
You can use function "non maximum suppression" that it removes rectangles which have less score. I put a code for NMS
def NMS(boxes, overlapThresh = 0.4):
# Return an empty list, if no boxes given
if len(boxes) == 0:
return []
x1 = boxes[:, 0] # x coordinate of the top-left corner
y1 = boxes[:, 1] # y coordinate of the top-left corner
x2 = boxes[:, 2] # x coordinate of the bottom-right corner
y2 = boxes[:, 3] # y coordinate of the bottom-right corner
# Compute the area of the bounding boxes and sort the bounding
# Boxes by the bottom-right y-coordinate of the bounding box
areas = (x2 - x1 + 1) * (y2 - y1 + 1) # We add 1, because the pixel at the start as well as at the end counts
# The indices of all boxes at start. We will redundant indices one by one.
indices = np.arange(len(x1))
for i,box in enumerate(boxes):
# Create temporary indices
temp_indices = indices[indices!=i]
# Find out the coordinates of the intersection box
xx1 = np.maximum(box[0], boxes[temp_indices,0])
yy1 = np.maximum(box[1], boxes[temp_indices,1])
xx2 = np.minimum(box[2], boxes[temp_indices,2])
yy2 = np.minimum(box[3], boxes[temp_indices,3])
# Find out the width and the height of the intersection box
w = np.maximum(0, xx2 - xx1 + 1)
h = np.maximum(0, yy2 - yy1 + 1)
# compute the ratio of overlap
overlap = (w * h) / areas[temp_indices]
# if the actual boungding box has an overlap bigger than treshold with any other box, remove it's index
if np.any(overlap) > treshold:
indices = indices[indices != i]
#return only the boxes at the remaining indices
return boxes[indices].astype(int)
I used surface.DrawTexturedRectRotated() to make filled circle but it rotates from center and i want to make it rotate from left.
I tried to rotate it but it makes full circle when its 180 degrees
function draw.FilledCircle( x, y, w, h, ang, color )
for i=1,ang do
draw.NoTexture()
surface.SetDrawColor( color or color_white )
surface.DrawTexturedRectRotated( x,y, w, h, i )
end
end
How do i make it to be rotated from left?
If you want a function that allows you to create pie-chart-like filled circle by specifying the ang parameter, your best bet is probably surface.DrawPoly( table vertices ). You should be able to use it like so:
function draw.FilledCircle(x, y, r, ang, color) --x, y being center of the circle, r being radius
local verts = {{x = x, y = y}} --add center point
for i = 0, ang do
local xx = x + math.cos(math.rad(i)) * r
local yy = y - math.sin(math.rad(i)) * r
table.insert(verts, {x = xx, y = yy})
end
--the resulting table is a list of counter-clockwise vertices
--surface.DrawPoly() needs clockwise list
verts = table.Reverse(verts) --should do the job
surface.SetDrawColor(color or color_white)
draw.NoTexture()
surface.DrawPoly(verts)
end
I have put surface.SetDrawColor() before draw.NoTexture() as this example suggests it.
You may want to use for i = 0, ang, angleStep do instead to reduce the number of vertices, therefore reducing hardware load, however that is viable only for small circles (like the one in your example) so the angle step should be some function of radius to account for every situation. Also, additional computing needs to be done to allow for angles that do not divide by the angle step with remainder of zero.
--after the for loop
if ang % angleStep then
local xx = x + math.cos(math.rad(ang)) * r
local yy = y - math.sin(math.rad(ang)) * r
table.insert(verts, {x = xx, y = yy})
end
As for the texturing, this will be very different from rectangle if your texture is anything else than solid color, but a swift look at the library did not reveal any better way to achieve this.
I'm training a YOLO model, I have the bounding boxes in this format:-
x1, y1, x2, y2 => ex (100, 100, 200, 200)
I need to convert it to YOLO format to be something like:-
X, Y, W, H => 0.436262 0.474010 0.383663 0.178218
I already calculated the center point X, Y, the height H, and the weight W.
But still need a away to convert them to floating numbers as mentioned.
for those looking for the reverse of the question (yolo format to normal bbox format)
def yolobbox2bbox(x,y,w,h):
x1, y1 = x-w/2, y-h/2
x2, y2 = x+w/2, y+h/2
return x1, y1, x2, y2
Here's code snipet in python to convert x,y coordinates to yolo format
def convert(size, box):
dw = 1./size[0]
dh = 1./size[1]
x = (box[0] + box[1])/2.0
y = (box[2] + box[3])/2.0
w = box[1] - box[0]
h = box[3] - box[2]
x = x*dw
w = w*dw
y = y*dh
h = h*dh
return (x,y,w,h)
im=Image.open(img_path)
w= int(im.size[0])
h= int(im.size[1])
print(xmin, xmax, ymin, ymax) #define your x,y coordinates
b = (xmin, xmax, ymin, ymax)
bb = convert((w,h), b)
Check my sample program to convert from LabelMe annotation tool format to Yolo format https://github.com/ivder/LabelMeYoloConverter
There is a more straight-forward way to do those stuff with pybboxes. Install with,
pip install pybboxes
use it as below,
import pybboxes as pbx
voc_bbox = (100, 100, 200, 200)
W, H = 1000, 1000 # WxH of the image
pbx.convert_bbox(voc_bbox, from_type="voc", to_type="yolo", image_size=(W,H))
>>> (0.15, 0.15, 0.1, 0.1)
Note that, converting to YOLO format requires the image width and height for scaling.
YOLO normalises the image space to run from 0 to 1 in both x and y directions. To convert between your (x, y) coordinates and yolo (u, v) coordinates you need to transform your data as u = x / XMAX and y = y / YMAX where XMAX, YMAX are the maximum coordinates for the image array you are using.
This all depends on the image arrays being oriented the same way.
Here is a C function to perform the conversion
#include <stdlib.h>
#include <stdio.h>
#include <errno.h>
#include <math.h>
struct yolo {
float u;
float v;
};
struct yolo
convert (unsigned int x, unsigned int y, unsigned int XMAX, unsigned int YMAX)
{
struct yolo point;
if (XMAX && YMAX && (x <= XMAX) && (y <= YMAX))
{
point.u = (float)x / (float)XMAX;
point.v = (float)y / (float)YMAX;
}
else
{
point.u = INFINITY;
point.v = INFINITY;
errno = ERANGE;
}
return point;
}/* convert */
int main()
{
struct yolo P;
P = convert (99, 201, 255, 324);
printf ("Yolo coordinate = <%f, %f>\n", P.u, P.v);
exit (EXIT_SUCCESS);
}/* main */
There are two potential solutions. First of all you have to understand if your first bounding box is in the format of Coco or Pascal_VOC. Otherwise you can't do the right math.
Here is the formatting;
Coco Format: [x_min, y_min, width, height]
Pascal_VOC Format: [x_min, y_min, x_max, y_max]
Here are some Python Code how you can do the conversion:
Converting Coco to Yolo
# Convert Coco bb to Yolo
def coco_to_yolo(x1, y1, w, h, image_w, image_h):
return [((2*x1 + w)/(2*image_w)) , ((2*y1 + h)/(2*image_h)), w/image_w, h/image_h]
Converting Pascal_voc to Yolo
# Convert Pascal_Voc bb to Yolo
def pascal_voc_to_yolo(x1, y1, x2, y2, image_w, image_h):
return [((x2 + x1)/(2*image_w)), ((y2 + y1)/(2*image_h)), (x2 - x1)/image_w, (y2 - y1)/image_h]
If need additional conversions you can check my article at Medium: https://christianbernecker.medium.com/convert-bounding-boxes-from-coco-to-pascal-voc-to-yolo-and-back-660dc6178742
For yolo format to x1,y1, x2,y2 format
def yolobbox2bbox(x,y,w,h):
x1 = int((x - w / 2) * dw)
x2 = int((x + w / 2) * dw)
y1 = int((y - h / 2) * dh)
y2 = int((y + h / 2) * dh)
if x1 < 0:
x1 = 0
if x2 > dw - 1:
x2 = dw - 1
if y1 < 0:
y1 = 0
if y2 > dh - 1:
y2 = dh - 1
return x1, y1, x2, y2
There are two things you need to do:
Divide the coordinates by the image size to normalize them to [0..1] range.
Convert (x1, y1, x2, y2) coordinates to (center_x, center_y, width, height).
If you're using PyTorch, Torchvision provides a function that you can use for the conversion:
from torch import tensor
from torchvision.ops import box_convert
image_size = tensor([608, 608])
boxes = tensor([[100, 100, 200, 200], [300, 300, 400, 400]], dtype=float)
boxes[:, :2] /= image_size
boxes[:, 2:] /= image_size
boxes = box_convert(boxes, "xyxy", "cxcywh")
Just reading the answers I am also looking for this but find this more informative to know what happening at the backend.
Form Here: Source
Assuming x/ymin and x/ymax are your bounding corners, top left and bottom right respectively. Then:
x = xmin
y = ymin
w = xmax - xmin
h = ymax - ymin
You then need to normalize these, which means give them as a proportion of the whole image, so simple divide each value by its respective size from the values above:
x = xmin / width
y = ymin / height
w = (xmax - xmin) / width
h = (ymax - ymin) / height
This assumes a top-left origin, you will have to apply a shift factor if this is not the case.
So the answer
I was wondering how you would go about this assuming you were working with a 2D coordinate frame in pixels. I created some examples of what I mean:
Red dot represents the origin point
Grey circle shows the radius but wouldn't actually be drawn
Green dots have a set amount and get evenly distributed along the
circle
With 3 dots:
http://prntscr.com/5vbj86
With 8 dots:
http://prntscr.com/5vbobd
Spektre answered my question but in C++, here it is in lua for anyone interested:
local x,y
local n = 10
local r = 100.0
local x0 = 250.0
local y0 = 250.0
local da = 2.0 * math.pi/n
local a = 0.0
for i = 0, n - 1 do
x = x0 + r * math.cos(a)
y = y0 + r * math.sin(a)
-- draw here using x,y
a = a + da
end
on circle very easy
for evenly distributed points the angle is increasing with the same step
so for N points the step is da=2.0*M_PI/N;
The code in C++ is like this:
int i,n=10;
double x,y,a,da;
double r=100.0,x0=250.0,y0=250.0; // circle definition
da=2.0*M_PI/double(n);
for (a=0.0,i=0;i<n;i++,a+=da)
{
x=x0+r*cos(a);
y=y0+r*sin(a);
// here draw or do something with (x,y) point
}
I better explain my problem with an Image
I have a contour and a line which is passing through that contour.
At the intersection point of contour and line I want to draw a perpendicular line at the intersection point of a line and contour up to a particular distance.
I know the intersection point as well as slope of the line.
For reference I am attaching this Image.
If the blue line in your picture goes from point A to point B, and you want to draw the red line at point B, you can do the following:
Get the direction vector going from A to B. This would be:
v.x = B.x - A.x; v.y = B.y - A.y;
Normalize the vector:
mag = sqrt (v.x*v.x + v.y*v.y); v.x = v.x / mag; v.y = v.y / mag;
Rotate the vector 90 degrees by swapping x and y, and inverting one of them. Note about the rotation direction: In OpenCV and image processing in general x and y axis on the image are not oriented in the Euclidian way, in particular the y axis points down and not up. In Euclidian, inverting the final x (initial y) would rotate counterclockwise (standard for euclidean), and inverting y would rotate clockwise. In OpenCV it's the opposite. So, for example to get clockwise rotation in OpenCV: temp = v.x; v.x = -v.y; v.y = temp;
Create a new line at B pointing in the direction of v:
C.x = B.x + v.x * length; C.y = B.y + v.y * length;
(Note that you can make it extend in both directions by creating a point D in the opposite direction by simply negating length.)
This is my version of the function :
def getPerpCoord(aX, aY, bX, bY, length):
vX = bX-aX
vY = bY-aY
#print(str(vX)+" "+str(vY))
if(vX == 0 or vY == 0):
return 0, 0, 0, 0
mag = math.sqrt(vX*vX + vY*vY)
vX = vX / mag
vY = vY / mag
temp = vX
vX = 0-vY
vY = temp
cX = bX + vX * length
cY = bY + vY * length
dX = bX - vX * length
dY = bY - vY * length
return int(cX), int(cY), int(dX), int(dY)