5
$\begingroup$

I'm trying to find the curvature of the features in an image and I was advised to calculate the gradient vector of pixels. So if the matrix below are the values from a grayscale image, how would I go about calculating the gradient vector for the pixel with the value '99'?

21 20 22 24 18 11 23
21 20 22 24 18 11 23
21 20 22 24 18 11 23
21 20 22 99 18 11 23
21 20 22 24 18 11 23
21 20 22 24 18 11 23
21 20 22 24 18 11 23

Apologies for asking such an open ended question, I've never done much maths and am not sure how to start tackling this.

$\endgroup$
4
  • $\begingroup$ you might try dsp.stackexchange for these type of questions $\endgroup$ Aug 12, 2015 at 13:58
  • $\begingroup$ My first reaction would be that pixel luminosity is not a differentiable function at that point, i.e. there is no gradient. 99 is a rather extreme value in comparison to its neighbors. If anything, the gradient could vanish, because this is a local extremum. $\endgroup$ Aug 12, 2015 at 21:04
  • $\begingroup$ Matlab has a function called "gradient" that will compute the discrete gradient for you. Just one line of code. $\endgroup$
    – littleO
    Jun 18, 2016 at 0:56
  • $\begingroup$ You can read up on convolution on wikipedia. en.wikipedia.org/wiki/Convolution#Discrete_convolution <-- this is what many methods mentioned use as long as the image is sampled on any evenly spaced and relatively nice grid. $\endgroup$ Jul 18, 2016 at 21:45

3 Answers 3

3
$\begingroup$

In Python you can use the numpy.gradient function to do this. This said function uses central differences for the computation, like so: \begin{eqnarray} \nabla_x I (i, j) = \frac{I(i + 1, j) - I(i - 1, j)}{2}, \hspace{.5em}\nabla_y I (i, j) = \frac{I(i, j+1) - I(i, j-1)}{2}. \end{eqnarray}

Here is a code snippet for your specific image:

import numpy as np
import matplotlib.pyplot as plt

# load image
img = np.array([[21.0, 20.0, 22.0, 24.0, 18.0, 11.0, 23.0],
                [21.0, 20.0, 22.0, 24.0, 18.0, 11.0, 23.0],
                [21.0, 20.0, 22.0, 24.0, 18.0, 11.0, 23.0],
                [21.0, 20.0, 22.0, 99.0, 18.0, 11.0, 23.0],
                [21.0, 20.0, 22.0, 24.0, 18.0, 11.0, 23.0],
                [21.0, 20.0, 22.0, 24.0, 18.0, 11.0, 23.0],
                [21.0, 20.0, 22.0, 24.0, 18.0, 11.0, 23.0]])
print "image =", img

# compute gradient of image
gx, gy = np.gradient(img)
print "gx =", gx
print "gy =", gy

# plotting
plt.close("all")
plt.figure()
plt.suptitle("Image, and it gradient along each axis")
ax = plt.subplot("131")
ax.axis("off")
ax.imshow(img)
ax.set_title("image")

ax = plt.subplot("132")
ax.axis("off")
ax.imshow(gx)
ax.set_title("gx")

ax = plt.subplot("133")
ax.axis("off")
ax.imshow(gy)
ax.set_title("gy")
plt.show()

enter image description here

To answer your specific question, the gradient (via central differences!) of the image at pixel with value $99$ is $0$ along the $x$ axis and $-2$ along the $y$ axis.

$\endgroup$
2
  • $\begingroup$ Just jumping in here. But what happens at the ends of the matrices then? You wrote gradient x at i,j = I(i + 1) - i(i - 1) ... But what happens when i does not have i + 1 or i - 1? (at the end of each row/column) $\endgroup$
    – AlanSTACK
    May 15, 2016 at 6:13
  • $\begingroup$ At the boundaries, you can clip the value of the gradients to zero, for example. There are other kinds of boundary conditions though (wrapping, etc.). $\endgroup$
    – dohmatob
    May 15, 2016 at 9:07
1
$\begingroup$

Suppose the image is continuous and differentiable in $x$ and $y$. Then $I(x,y)$ is the value of the pixel at each $(x,y)$, i.e. $I: \mathbb{R}^2 \mapsto \mathbb{R}$. Recall that the gradient at a point $(u,v)$ is:

$$ \nabla I(u,v) = \begin{bmatrix} \frac{\partial I}{\partial x}(u,v) \\ \frac{\partial I}{\partial y}(u,v) \end{bmatrix} $$

Given a discrete grid, you should approximate the partial derivative in $x$ and $y$ directions using finite difference approximations at the point of interest.

Assume your function $I$ is sampled over points $\{1, \ldots, 7 \} \times \{1, \ldots, 7 \}$ in image-coordinates, i.e. $I(1,1) = 21$, $I(1,7) = 23$, etc... So you're looking for the gradient at $(4,4)$. If you assume the resolution between points is 1, then the forward difference approximation in the $x$ direction gives:

$$ \frac{\partial I}{\partial x}(4,4) \approx I(5,4) - I(4,4) = 24 - 99 $$

Do the same in $y$ to obtain the full gradient at the point.

$\endgroup$
0
$\begingroup$

The theoretical thing you may want to read up on is convolution and especially discrete convolution.


Since there already are practical approaches in a few computer languages I might as well add one that will work in Matlab & GNU Octave. We start out by feeding in our image. It is typed in as a matrix:

lImage = [21,20,22,24,18,11,23;
          21,20,22,24,18,11,23;
          21,20,22,24,18,11,23;
          21,20,22,99,18,11,23;
          21,20,22,24,18,11,23;
          21,20,22,24,18,11,23;
          21,20,22,24,18,11,23];

Then the command "conv2" creates a 2 dimensional convolution, first we can do one in the x direction by feeding it a row vector with the [1,0,-1] filter:

dx = conv2(lImage,[1,0,-1],'valid')

which gives us the output:

dx =

   1    4   -4  -13    5
   1    4   -4  -13    5
   1    4   -4  -13    5
   1   79   -4  -88    5
   1    4   -4  -13    5
   1    4   -4  -13    5
   1    4   -4  -13    5 

and the y direction we can easily do by just applying a transpose (denoted by ') on the x filter row vector:

dy = conv2(lImage,[1,0,-1]','valid')

gives us the output:

dy= 

   0    0    0    0    0    0    0
   0    0    0   75    0    0    0
   0    0    0    0    0    0    0
   0    0    0  -75    0    0    0
   0    0    0    0    0    0    0
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.