Doug's MATLAB Video Tutorials

Advanced: making a 2d or 3d histogram to visualize data density 30

Posted by Doug Hull,

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

30 CommentsOldest to Newest

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??

@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

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

@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

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)

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?

@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!

@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

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)

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)

@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

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

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

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?

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!

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.

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.

Hi,

Thank you very much for this post. Its wonderful and exactly what I need to look some 3.5 million data points I have. So I have gone through each comments and details on this page and summarized the code as given below. But I am getting an error “Error using accumarray, First input SUBS must contain positive integer subscripts.” I know there are no negative values in xr and yr, and as suggested I am adding 0.5. So how can I fix it? Please help. Thanks in advance.

x = MAge000(:);
y = DSTS000(:);

% NaN values taken out by
x = x(isfinite(x(:, 1)), :);
y = y(isfinite(x(:, 1)), :);

% Number of bins
n = 49;
xi = linspace(min(x(:)), max(x(:)), n);
yi = linspace(min(y(:)), max(y(:)), n);

xr = interp1(xi, 0.5:numel(xi)-0.5, x, ‘nearest’);
yr = interp1(yi, 0.5:numel(yi)-0.5, y, ‘nearest’);

% [yr xr]: Y then X because the plotting commands take matrices that look like Z(y-coord, x-coord)
Z = accumarray([yr xr] + 0.5, 1, [n n]);

figure(1);
surf(xi, yi, Z);

Hi,
Never mind, I found my mistake, and fixed it. It should be

y = y(isfinite(y(:, 1)), :);

not

y = y(isfinite(x(:, 1)), :);

But I do have a question regarding the selection of n. Why n = 49 and not 50. May be its a stupid question but I dont know/understand. So if you can explain to me will be great.

Thanks again,

Raj

Hi Doug,
This might be a dumb question but before I spend the time to implement this, do you think this will work for an array having 2 million points?
Thanks
-Vipin

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