2D (Joint) Histogram

 This function creates a joint histogram using two input waves, which must be of equal size. The result will be placed in the wave 'hist', which must be made beforehand and scaled to determine the histogram binning and boundaries. For example,

make /o/n=1000 data1=gnoise(1), data2=gnoise(1)
make /o/n=(25,25) myHist
setscale x,-3,3,myHist
setscale y,-3,3,myHist
JointHistogram(data1,data2,myHist)
NewImage myHist


This is the function, which runs slightly differently depending on how many cores your processor has. In my experience, the MatrixOp version is faster unless you have 4 or more cores, in which case the Multithreaded Make version is faster. You don't have to worry about this, as it is taken care of automatically.

Function JointHistogram(w0,w1,hist)
    wave w0,w1,hist
   
    variable bins0=dimsize(hist,0)
    variable bins1=dimsize(hist,1)
    variable n=numpnts(w0)
    variable left0=dimoffset(hist,0)
    variable left1=dimoffset(hist,1)
    variable right0=left0+bins0*dimdelta(hist,0)
    variable right1=left1+bins1*dimdelta(hist,1)
   
    // Scale between 0 and the number of bins to create an index wave.  
    if(ThreadProcessorCount<4) // For older machines, matrixop is faster.  
        matrixop /free idx=round(bins0*(w0-left0)/(right0-left0))+bins0*round(bins1*(w1-left1)/(right1-left1))
    else // For newer machines with many cores, multithreading with make is faster.  
        make /free/n=(n) idx
        multithread idx=round(bins0*(w0-left0)/(right0-left0))+bins0*round(bins1*(w1-left1)/(right1-left1))
    endif
   
    // Compute the histogram and redimension it.  
    histogram /b={0,1,bins0*bins1} idx,hist
    redimension /n=(bins0,bins1) hist // Redimension to 2D.  
    setscale x,left0,right0,hist // Fix the histogram scaling in the x-dimension.  
    setscale y,left1,right1,hist // Fix the histogram scaling in the y-dimension.  
End
Hello,

this is a very nice example, however there is an error in your code.

The round() should be replaced by floor().
In your example half of the amount in the last histogram class per horizontal line is attributed to the first class of the next line.

Regards,
Michael Huth

August 8, 2014 at 10:08 am - Permalink

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