Doug's MATLAB Video Tutorials

January 22nd, 2010

Advanced: making a 2d or 3d histogram to visualize data density

 This short video makes a 2d histogram as an alternative to plotting data points and visually estimating where the most data is.
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By Doug Hull

Doug first used MATLAB in 1994, could not figure it out until he got some help in 1995. He is now dedicated to making sure that no one else wastes a year of their life not knowing MATLAB like he did.

18:43 UTC | Posted in Format: Video, Level: Advanced | Permalink | 25 Comments »

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25 Responses to “Advanced: making a 2d or 3d histogram to visualize data density”

  1. Daniel Armyr replied on January 25th, 2010 at 07:58 UTC :

    Good tutorial. This is one type of problem I often find myself trying to solve.

    In fact, I find I try to solve it so often that I took the time to write a properly optimized and documented function for the file exchange:
    http://www.mathworks.com/matlabcentral/fileexchange/23238-cloudplot

  2. John A replied on January 26th, 2010 at 18:39 UTC :

    Doug, I am a little confused…I assume x and y are just random coordinates…you then created xd, yd…what is this actually?? How did you do it? Then in linespace you use x,y again, but in interp you use xd,yd. I think I understand the concept, but I’m lost in the details. Can you post all the code so I can run what you did??

  3. dhull replied on January 26th, 2010 at 19:11 UTC :

    @John,

    XD and YD are the random points. When used in the interp1 function, XD and YD are the raw data that is then interpolated to the nearest nicely spaced data points in XI.

    Make sense?

    Doug

  4. John A replied on January 26th, 2010 at 20:23 UTC :

    Doug, in the linspace functions then you should use xd,yd not x,y…was that just a typo?? If so all makes sense, thx

  5. dhull replied on January 26th, 2010 at 20:52 UTC :

    @John,

    I can see where that would work too. For simplicity, I had hidden where the data came from. X and XD both had the same range, so it did not matter.

    I had originally planned on showing where the data came, but in production the movie got too long, so it ended on the cutting room floor!

    Good catch!

    Doug

  6. Joe Pohedra replied on January 28th, 2010 at 22:04 UTC :

    Sorry, I’m lost. I could not follow what’s happening without the data set to reproduce the example. Fortunately for me, the hist3 function seems to do all the heavy lifting I need.

    randx = randn(100000,1);
randy = randn(100000,1);
z = hist3([randx randy],[40 40]);
surf(z)
  7. Douglas Neal replied on February 1st, 2010 at 20:44 UTC :

    Doug,

    This is a great code! I am also lost though. When I run the line

    >> z = accumarray([xr yr], 1, [n n]);

    I get the following error:

    ??? Error using ==> accumarray
Third input SZ must be a full row vector with one element for each column of SUBS.

    Also, can you comment how this differs from hist3.m?

  8. dhull replied on February 1st, 2010 at 20:55 UTC :

    @Doug

    Please post all the code so I can reproduce it by copying and pasting.

    Doug

  9. Douglas Neal replied on February 2nd, 2010 at 15:59 UTC :

    @dhull

    Here is what I was doing to try and duplicate your code:

    
xd = randn(100000,1);
yd = randn(100000,1);

n=49;
xi = linspace(min(xd(:)),max(xd(:)),n);
yi = linspace(min(yd(:)),max(yd(:)),n);

xr = interp1(xi,1:numel(xi),xd,'nearest')';
yr = interp1(yi,1:numel(yi),yd,'nearest')';

z = accumarray([xr yr], 1, [n n]);

figure(2)
surf(z)

    I copied everything save the declarations of xd, yd out of your example in the video. Thanks!

  10. dhull replied on February 2nd, 2010 at 16:45 UTC :

    @Douglas,

    z = accumarray([xr' yr'], 1, [n n]);

    You needed to transpose xr and yr so they are columns when you put them together. The way it was done made them a long row, not two columns. Works fine now!

    Doug

  11. Douglas Neal replied on February 2nd, 2010 at 18:04 UTC :

    @dhull

    Thanks! Works great now — very nice code!

  12. Doug Carter replied on April 2nd, 2010 at 19:54 UTC :

    I received this error:

    Error in ==> testrun at 22
    z = accumarray([xr' yr'], 1,[n n]);

    Is there anyway to get around this or what is the best way to figure out how to maximize the amount of data without excedding the memory limits?

    Thanks,
    Doug (Ohio State University)

  13. dhull replied on April 2nd, 2010 at 19:57 UTC :

    @Doug,

    Without more context, I can not really answer the question. Please contact http://www.mathworks.com/support with all the relevant files to reproduce, and we will take a look at it.

    Thanks,
    doug

  14. Colin Norris replied on July 28th, 2010 at 15:22 UTC :

    I’m trying to use this function (which I’ve modified below) but I’m getting an error

    ??? Error using ==> accumarray
First input SUBS must contain positive integer subscripts.

    EngSpd & EngTorq are independant 24000 x 1 vectors
    SpdBP is a 1×16 vector
    TrqBP is a 1×21 vector

    Any ideas on how to fix this issue?

    Thanks
    Colin

    
xd = EngSpd;
yd = EngTorq;

xi = SpdBP;
yi = TrqBP;

xr = interp1(xi,1:numel(xi),xd,'nearest')';
yr = interp1(yi,1:numel(yi),yd,'nearest')';

z = accumarray([xr' yr'], 1, [length(SpdBP) length(TrqBP)]);

figure(5)
surf(z)
  15. dhull replied on July 28th, 2010 at 17:17 UTC :

    @Colin,

    I would put a break point on the line with accumarray. Once stopped in the debugger, figure out what the inputs to accumarray are. It looks like the first one is not what is expected.

    Doug

  16. Andrea replied on August 13th, 2010 at 17:31 UTC :

    Thanks for the tutorial. Nice thing, I never came up against accumarray which seems to be very useful!

    But actually there is a mistake when calculating the BIN-number: since interp1 assigns the nearest value in every dimension, all data points say in x-direction belonging to BIN 1 and half of them in BIN 2 will be assigned to (1,:). Correct should be interpolating to

    xr = interp1(xi,0.5:numel(xi)-0.5,xd,'nearest')';
yr = interp1(yi,0.5:numel(yi)-0.5,yd,'nearest')';

    which requires 

    z = accumarray([xr yr] + 0.5, 1, [n n]);

    Best
    Andrea

  17. dhull replied on August 13th, 2010 at 17:51 UTC :

    @Andrea,

    Thank you for catching a subtlety there!

    Doug

  18. Thomas Smith replied on March 23rd, 2011 at 20:34 UTC :

    In the example code, the data from xd will be plotted on the Y axis, and yd on the X axis. To get a properly labeled graph, with the X data along the X axis, you actually need to say:

    surf(xi, yi, Z');

    Why is that? In the example, we have

    Z = accumarray([xr yr], 1, [n n]);

    For instance, if something falls in the 1st X-bin and the 5th Y-bin, it will get counted in Z(1,5). But, the MATLAB plotting commands like SURF and CONTOUR take their inputs in the opposite way: To have something show up at (1,5) in the graph, it should be in Z(5,1).

    Hope this is useful to the next person who’s thinking sideways!
    -Thomas

  19. Thomas Smith replied on March 23rd, 2011 at 21:01 UTC :

    I made the changes suggested by Andrea and myself, and put them into a nice copy-pasteable (and editable!) file on GitHub:

    https://gist.github.com/883933

  20. Jbrand replied on June 22nd, 2011 at 20:31 UTC :

    I am trying to plot out a point spread function, there are positive and negative x and y values and I get this message

    ??? Error using ==> accumarray
    First input SUBS must contain positive integer
    subscripts.

    any help?

  21. dhull replied on June 29th, 2011 at 18:17 UTC :

    @Jbrand,

    You can not pass negative numbers to accumarray. Check your inputs.

    Doug

  22. Megan replied on December 3rd, 2011 at 09:29 UTC :

    Thanks for the video! I was able to use it effectively for 2 vectors (xd & yd).
    I had some other questions about the extent to which I could use this code, or if there are other options available.
    1. Is there an alternative to accumarray that will take negative inputs -or am I able to edit accumarray in a way that would allow it to take negative inputs?
    2. Is there any way to have such a histogram plot for 3 vectors (xd, yd and zd)? What I would like to have is the x and y coordinates with z as the height and color coded by the frequency distribution. I am basically trying to visualize a special euclidean(2) space (since my z vector lies on polar coordinates). Is there any way to adapt your code to work for 3 vectors in this way?

    I know these questions don’t directly apply to the working of the posted tutorial, but any suggestions would be very much appreciated!
    Thanks in advance!

  23. Rod replied on April 11th, 2012 at 17:30 UTC :

    I believe I have implemented the 3d analogy for use with a 3d color histogram. Correct me if this is wrong but it seems to be working. Hope this helps someone.

    colorimage = imread(‘rgb.jpg’);
    nbins = 16;
    redmat = colorimage(:,:,1);
    greenmat = colorimage(:,:,2);
    bluemat = colorimage(:,:,3);
    ri = linspace(0,255,nbins);
    gi = linspace(0,255,nbins);
    bi = linspace(0,255,nbins);
    rtp = interp1(ri,1:numel(ri),double(redmat(:)),’nearest’);
    gtp = interp1(gi,1:numel(gi),double(greenmat(:)),’nearest’);
    btp = interp1(bi,1:numel(bi),double(bluemat(:)),’nearest’);
    Z=accumarray([rtp,gtp,btp],1,[nbins,nbins,nbins]);

    The only issue I have found is displaying this behemoth, since surf() only takes in mxn or mxnx3.

  24. Catherine replied on June 11th, 2012 at 10:38 UTC :

    Hi there, I would like to ask if you know how to label the peak values of the data points in your 3D plot. I can’t seem to find any commands in MATLAB that are able to perform that function. Thank you.

  25. Doug replied on June 11th, 2012 at 12:18 UTC :

    Check out the command >>annotation

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