Can I get pdist2 results as only one value?

Hi,

I've been using some code I wrote to calculate some distances (Euclidean, Canberra, etc). However, I did some research and I found out that using pdist2 function I could get those distances in an easier way. The problem is that pdist2 returns to me a matrix of distances, and not only one value as my code does. Is it possible to get only one value as a distance result using pdist2 function or another Matlab function? I also tried Jeffrey's distance code I found in file exchange, but I get the same (matrix of results).

Thank you in advance for your help.

Answer 1

[Image Analyst]

pdist2 gets the distances between every possible pair of points. What do you have? Just two points? If so, use hypot() or sqrt(). If you have multiple points then explain how you would like to get just one distance when there are really N*(N-1) distances.

[Walter Roberson]

You are thinking of pdist() not pdist2(). pdist2() has N * M distances

[Image Analyst]

I was thinking of pdist2 - I just gave the wrong formula. If you have a set of N points, then there will be (N * (N-1))/2 distances (not counting the distance of a point to itself), which should have been the formula I'd given. You can get these in a N-by-N array, but of course the distance between the kth point and itself will be zero so along the diagonal they're all zeros. Of course pdist2() can also be used between two different sets of points, with different numbers of points in each of them, rather than within the same set of points.

[Walter Roberson]

If you are going to use the same points for your X and Y in pdist2() then you might as well use pdist() which is designed for the case of each point to each other point.

[Queila Martins]

Hi, Image Analyst. Thank you for answering me too.

Well, imagine that, for example, I have 2 matrices h and h1:

h =  [2894; 0; 0; 186232];
h1 = [3386; 0; 0; 185740];

I'm using a formula to calculate the euclidean distance (or similarity) between then. The formula in matlab is:

 E_distance = sqrt(sum((h-h1).^2));

The result I get is only one value, which is: 695.7930727. I added those values to some websites that have a euclidean distance calculator and the value is the same (example: http://calculator.vhex.net/calculator/distance/euclidean-distance). But when I tried to use pdist2 with those same two matrices, I get a matrix as a result, with all possible distance between all h and h1 values.

So, I was wondering if I'm doing something wrong or if I shouldn't use pdist2 for this kind of calculation.

I've added the code here so you can see it too.

Thank you in advance!

clc;
clear;

h =  [2894; 0; 0; 186232];
h1 = [3386; 0; 0; 185740];

E_distance = sqrt(sum((h-h1).^2));

[Image Analyst]

Do you have two vectors (points, locations) in 4-D space? Or do you have 2 sets of 4 points in 1-D space, like along a number line?

[Queila Martins]

In fact, I have two images and then I get their histograms. So I apply Euclidean distance because I want to get their distance (or similarity) value. Those histograms are 2 sets (1 column, many rows)and both have the same size (1 column, 256 rows). First and last row have values, but others are zero for both sets.

[Image Analyst]

Why can't you just sum the counts differences:

d = sum(abs(counts1-counts2));

Just looking at histograms may or may not be a good way of comparing images. For example the same image turned upside down, or even a completely different image, may have the same histogram. There are other ways of comparing images.

[Queila Martins]

I need to try some distance equations, like Canberra, Euclidean and Jeffrey's, as they are in my master's work. So, I'm gonna try these with the images themselves and not their histograms (the images have the same size too). What other ways of comparing images do you suggest? thanks a lot!

[Image Analyst]

Queila, the "distance" on the histogram, as if the bar tips were (x,y) coordinates like h = [2894; 0; 0; 186232], seems dumb to me and not based on any theory. Better would be what I suggested but I don't even like that - that was before I knew what you really wanted to do - I was just trying to answer your question. It depends on why you want to "compare" the two images and how you do it. For example, the SAD https://en.wikipedia.org/wiki/Sum_of_absolute_differences is a common metric. Also PSNR https://en.wikipedia.org/wiki/Peak_signal-to-noise_ratio and MSE https://en.wikipedia.org/wiki/Mean_squared_error are common and have built-in MATLAB functions for them. The SSIM https://en.wikipedia.org/wiki/Structural_similarity is a very common and popular, somewhat newer, metric.

Here's a File Exchange entry that computes 8 more.

Lots more are listed here: https://en.wikipedia.org/wiki/Image_quality

[Queila Martins]

Thank you SO much for all this information you gave me, Image Analyst. I'll check those links you've sent me and see how they apply to my problem. I'm using LBP function to get textures from an image and after this I need to compare those images to check their grade of similarity and then order them. Well, thank you very much again.

Answer 2

[Walter Roberson]

Pass a single point for each of the first two arguments to pdist2 and the result will be a single distance. pdist2 is optimized to return the distance from each of the points in the first argument to each of the points in the second argument, so if you only want one distance then only pass a single point to each.

[Queila Martins]

So, I understand it's not possible to pass two matrices to pdist2 and get only one result. I guess the best solution is to use the distance formula as I've been doing, right? thanks!

[Walter Roberson]

Please give an example of what you would like the inputs and outputs to be.

[Queila Martins]

Thank you for your answer, Walter. Can you please check the answer I gave to Image Analyst as this answers your questions too. Thanks a lot!

[Walter Roberson]

pdist2(h.', h1.')

http://www.mathworks.com/help/stats/pdist2.html

"Rows of X and Y correspond to observations, columns correspond to variables."

You were sending in columns instead of rows.

Likewise,

pdist( [h, h1].' )

[Queila Martins]

Thanks, Walter, that makes sense. I tried what you sent and it works for a matrix with only one row. However, I tried with matrices with a lot of rows and the result is another matrix. Sorry I've been asking this, I just want to make sure I'm doing the right thing. I need to get the distance value between some images, and then I will check which one is most similar to another.

I've added two samples of images here.

Thank you

[Walter Roberson]

pdist() and pdist2() are not designed to find the distances between matrices: they are designed to find the distance between points.

You could reshape your matrix into a row vector if you wanted to get a "distance" between two of them. But note that if you do that, then the euclidean "distance" would be essentially the same thing as MSE. Euclidean would be square root of (sum of squares of differences), whereas MSE would be mean of (sum of squares of differences)

Answer 3

[sergi torres]

Hey Martins, did you finally find a solution? I need to do the same, find the distance (one number) of two histograms [255, 255]. Thanks!

[Walter Roberson]

What does the "distance" between two histograms mean to you?

Are the histograms 1D histograms or 2D histograms?

Is the number of points the same for the two histograms?

As I wrote above:

You could reshape your matrix into a row vector if you wanted to get a "distance" between two of them. But note that if you do that, then the euclidean "distance" would be essentially the same thing as MSE. Euclidean would be square root of (sum of squares of differences), whereas MSE would be mean of (sum of squares of differences)

[Image Analyst]

Yeah, it could mean virtually nothing. I could take an image, scramble the pixels up so much that it looks just like white noise, then take the histogram and it would be the identical histogram as the original yet the image would nothing like the original. Or I could rearrange/sort the pixels to form a grayscale ramp. Again, the histogram "distance" is zero because the histogram is the same, but the image looks totally different.

[Queila Martins]

Hi, Sergi

I did exactly what Walter Roberson said: I reshaped my matrix (2 histograms) into a row vector and then I used pdist formulas. The results are not the best in the world as I used LBP (Local Binary Pattern), but they are good. Now I'm gonna try GLCM and other descriptors that maybe will bring me better results.