An undistorted image typically have lower resolution than the original image due to non-uniform distribution of pixels and (usually) the cropping of the black edges. (See below as an example)
So given the camera calibration parameters, e.g. in ROS format
image_width: 1600
image_height: 1200
camera_name: camera1
camera_matrix:
rows: 3
cols: 3
data: [1384.355466887268, 0, 849.4355708515795, 0, 1398.17734010913, 604.5570699746268, 0, 0, 1]
distortion_model: plumb_bob
distortion_coefficients:
rows: 1
cols: 5
data: [0.0425049914802741, -0.1347528158561486, -0.0002287009852930437, 0.00641133892300999, 0]
rectification_matrix:
rows: 3
cols: 3
data: [1, 0, 0, 0, 1, 0, 0, 0, 1]
projection_matrix:
rows: 3
cols: 4
data: [1379.868041992188, 0, 860.3000889574832, 0, 0, 1405.926879882812, 604.3997819099422, 0, 0, 0, 1, 0]
How would one calculate the final resolution of the undistorted rectified image?
From Fruchtzwerg's comment, the following will give the effective ROI of the undistorted image
import cv2
import numpy as np
mtx = np.array([
[1384.355466887268, 0, 849.4355708515795],
[ 0, 1398.17734010913, 604.5570699746268],
[0, 0, 1]])
dist = np.array([0.0425049914802741, -0.1347528158561486, -0.0002287009852930437, 0.00641133892300999, 0])
cv2.getOptimalNewCameraMatrix(mtx, dist, (1600, 1200), 1)
Related
Suppose I have an image with mask, valid pixels are masked as 1 and others 0, how to perform bilinear interpolation to fill all the invalid pixels?
for example, image:
1, 0, 0, 4
mask:
1, 0, 0, 1
interpolation result should be:
1, 2, 3, 4
The valid pixels are not regularly arranged, a more complicated sample, image:
4, 0, 6, 0,
0, 8, 5, 0
5, 3, 0, 0
mask:
1, 0, 1, 0,
0, 1, 1, 0
1, 1, 0, 0
interpolate with scipy.interpolate.interp2d and the result has many holes and noise
I have some BGR image:
cv::Mat image;
I want to subtract from all the pixels in the image the vector:
[10, 103, 196]
Meaning that the blue channel for all the pixels will be reduced by 10, the green by 103 and the red by 196.
Is there a standard way to do that, or should I run for loops over all the channels and all the pixels?
suppose we have image that all channels filled with zero and for instance it's dimension is 2x3
cv::Mat image = cv::Mat::zeros(2,3,CV_32SC3)
output will be:
[0, 0, 0, 0, 0, 0, 0, 0, 0;
0, 0, 0, 0, 0, 0, 0, 0, 0]
then if we want to add or subtract a singleton variable, then we can use cv::Scalar
1- suppose we want to add 3 in blue channel:
image = image + Scalar(3,0,0); // the result will be same as image=image+3;
with above code our matrix is now:
[3, 0, 0, 3, 0, 0, 3, 0, 0;
3, 0, 0, 3, 0, 0, 3, 0, 0]
2- if you want to add to another channel you can use second or third argument(or forth) of cv::Scalar like below
image = image +Scalar(3,2,-3);
output will be
[3, 2, -3, 3, 2, -3, 3, 2, -3;
3, 2, -3, 3, 2, -3, 3, 2, -3]
Using cv::subtract
cv::Mat image = cv::Mat::zeros(2,3,CV_32SC3);
subtract(image,Scalar(2,3,1),image);
output
[-2, -3, -1, -2, -3, -1, -2, -3, -1;
-2, -3, -1, -2, -3, -1, -2, -3, -1]
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I'm using Genetic Algorithms (GA) on an image processing problem (an image segmentation to be more precise). In this case, an individual represent a block of pixels (i.e. a set of pixel coordinates). I need to encourage individuals with contiguous pixels.
To encourage contiguous blocks of pixels:
The "contiguousness" of an individual need to be considered in the fitness function to encourage individuals having adjacent pixels (best-fit). Hence during the evolution, the contiguousness of a set of coordinates (i.e. an individual) will influence the fitness of this individual.
The problem I'm facing is how to measure this feature (how much contiguous) on a set of pixel coordinates (x, y)?
As can be shown on the image below, the individual (set of pixels in black) on the right is clearly more "contiguous" (and therefore fitter) than the individual on the left:
I think I understand what you are asking, and my suggestion would be to count the number of shared "walls" between your pixels:
I would argue that from left to right the individuals are decreasing in continuity.
Counting the number of walls is not difficult to code, but might be slow the way I've implemented it here.
import random
width = 5
height = 5
image = [[0 for x in range(width)] for y in range(height)]
num_pts_in_individual = 4
#I realize this may give replicate points
individual = [[int(random.uniform(0,height)),int(random.uniform(0,width))] for x in range(num_pts_in_individual)]
#Fill up the image
for point in individual:
image[point[0]][point[1]] = 1
#Print out the image
for row in image:
print row
def count_shared_walls(image):
num_shared = 0
height = len(image)
width = len(image[0])
for h in range(height):
for w in range(width):
if image[h][w] == 1:
if h > 0 and image[h-1][w] == 1:
num_shared += 1
if w > 0 and image[h][w-1] == 1:
num_shared += 1
if h < height-1 and image[h+1][w] == 1:
num_shared += 1
if w < width-1 and image[h][w+1] == 1:
num_shared += 1
return num_shared
shared_walls = count_shared_walls(image)
print shared_walls
Different images and counts of shared walls:
[0, 0, 0, 0, 0]
[0, 1, 0, 0, 0]
[0, 0, 0, 0, 0]
[1, 0, 0, 1, 1]
[0, 0, 0, 0, 0]
2
[1, 0, 0, 0, 0]
[0, 0, 0, 1, 0]
[0, 0, 0, 0, 0]
[0, 0, 0, 0, 0]
[1, 0, 1, 0, 0]
0
[0, 0, 0, 1, 1]
[0, 0, 0, 1, 0]
[0, 0, 0, 0, 0]
[0, 0, 0, 0, 0]
[1, 0, 0, 0, 0]
4
One major problem with this, is that if a change in pixel locations occurs that does not change the number of shared walls, it will not affect the score. Maybe a combination of the distance method you described and the shared walls approach would be best.
I have an image of a planar surface, and I want to compute an image warping that gives me a synthetic view of the same planar surface seen from a virtual camera located at another point in the 3d space.
So, given an image I1 I want to compute an image I2 that represents the image I1 seen from a virtual camera.
In theory, there exists an homography that relates these two images.
How do I compute this homography given the camera pose of the virtual camera, as well as it's matrix of internal parameters?
I'm using opencv's warpPerspective() function to apply this homography and generate the image warped.
Thanks in advance.
Ok, found this post (Opencv virtually camera rotating/translating for bird's eye view), where I found some code doing what I needed.
However, I noticed that the rotation in Y had a sign error (-sin instead of sin) . Here's my solution adapted for python. I'm new to python, sorry if I'm doing something ugly.
import cv2
import numpy as np
rotXdeg = 90
rotYdeg = 90
rotZdeg = 90
f = 500
dist = 500
def onRotXChange(val):
global rotXdeg
rotXdeg = val
def onRotYChange(val):
global rotYdeg
rotYdeg = val
def onRotZChange(val):
global rotZdeg
rotZdeg = val
def onFchange(val):
global f
f=val
def onDistChange(val):
global dist
dist=val
if __name__ == '__main__':
#Read input image, and create output image
src = cv2.imread('/home/miquel/image.jpeg')
dst = np.ndarray(shape=src.shape,dtype=src.dtype)
#Create user interface with trackbars that will allow to modify the parameters of the transformation
wndname1 = "Source:"
wndname2 = "WarpPerspective: "
cv2.namedWindow(wndname1, 1)
cv2.namedWindow(wndname2, 1)
cv2.createTrackbar("Rotation X", wndname2, rotXdeg, 180, onRotXChange)
cv2.createTrackbar("Rotation Y", wndname2, rotYdeg, 180, onRotYChange)
cv2.createTrackbar("Rotation Z", wndname2, rotZdeg, 180, onRotZChange)
cv2.createTrackbar("f", wndname2, f, 2000, onFchange)
cv2.createTrackbar("Distance", wndname2, dist, 2000, onDistChange)
#Show original image
cv2.imshow(wndname1, src)
h , w = src.shape[:2]
while True:
rotX = (rotXdeg - 90)*np.pi/180
rotY = (rotYdeg - 90)*np.pi/180
rotZ = (rotZdeg - 90)*np.pi/180
#Projection 2D -> 3D matrix
A1= np.matrix([[1, 0, -w/2],
[0, 1, -h/2],
[0, 0, 0 ],
[0, 0, 1 ]])
# Rotation matrices around the X,Y,Z axis
RX = np.matrix([[1, 0, 0, 0],
[0,np.cos(rotX),-np.sin(rotX), 0],
[0,np.sin(rotX),np.cos(rotX) , 0],
[0, 0, 0, 1]])
RY = np.matrix([[ np.cos(rotY), 0, np.sin(rotY), 0],
[ 0, 1, 0, 0],
[ -np.sin(rotY), 0, np.cos(rotY), 0],
[ 0, 0, 0, 1]])
RZ = np.matrix([[ np.cos(rotZ), -np.sin(rotZ), 0, 0],
[ np.sin(rotZ), np.cos(rotZ), 0, 0],
[ 0, 0, 1, 0],
[ 0, 0, 0, 1]])
#Composed rotation matrix with (RX,RY,RZ)
R = RX * RY * RZ
#Translation matrix on the Z axis change dist will change the height
T = np.matrix([[1,0,0,0],
[0,1,0,0],
[0,0,1,dist],
[0,0,0,1]])
#Camera Intrisecs matrix 3D -> 2D
A2= np.matrix([[f, 0, w/2,0],
[0, f, h/2,0],
[0, 0, 1,0]])
# Final and overall transformation matrix
H = A2 * (T * (R * A1))
# Apply matrix transformation
cv2.warpPerspective(src, H, (w, h), dst, cv2.INTER_CUBIC)
#Show the image
cv2.imshow(wndname2, dst)
cv2.waitKey(1)
Given a 3 x 3 rotation matrix,R, and a 3 x 1 translation matrix,T, I am wondering how to multiply the T and R matrices to an image?
Lets say the Iplimage img is 640 x 480.
What I want to do is R*(T*img).
I was thinking of using cvGemm, but that didn't work.
The function you are searching for is probably warpPerspective() : this is a use case...
// Projection 2D -> 3D matrix
Mat A1 = (Mat_<double>(4,3) <<
1, 0, -w/2,
0, 1, -h/2,
0, 0, 0,
0, 0, 1);
// Rotation matrices around the X axis
Mat R = (Mat_<double>(4, 4) <<
1, 0, 0, 0,
0, cos(alpha), -sin(alpha), 0,
0, sin(alpha), cos(alpha), 0,
0, 0, 0, 1);
// Translation matrix on the Z axis
Mat T = (Mat_<double>(4, 4) <<
1, 0, 0, 0,
0, 1, 0, 0,
0, 0, 1, dist,
0, 0, 0, 1);
// Camera Intrisecs matrix 3D -> 2D
Mat A2 = (Mat_<double>(3,4) <<
f, 0, w/2, 0,
0, f, h/2, 0,
0, 0, 1, 0);
Mat transfo = A2 * (T * (R * A1));
Mat source;
Mat destination;
warpPerspective(source, destination, transfo, source.size(), INTER_CUBIC | WARP_INVERSE_MAP);
I hope it could help you,
Julien
PS : I gave the example with a projection from 2D to 3D but you can use directly transfo = T* R;