Identifying Peaks from Line Profile - dm-script

Question

I would like to ask if it is possible to locate the position of every maxima and minima of an intensity profile on DM.
How do I come up with a simple script that identifies the positions of the peaks in the example below?
Here's a screenshot of line intensity profile of a STEM image along the Y-direction:

Answer 1

If you want to filter for "strict" local maxima, then you can easily do this with image expressions and the conditional "tert" operator. The following example illustrates this:
image CreateTestSpec( number nChannels, number nPeaks, number amp, number back )
{
image testImg := RealImage( "TestSpec", 4, nChannels )
testImg = amp * cos( PI() + 2*PI() * nPeaks * icol/(iwidth-1) )
testImg += back
testImg = PoissonRandom( testImg )
return testImg
}
image FilterLocalMaxima1D( image spectrumIn, number range )
{
image spectrumOut := spectrumIn.ImageClone()
for( number dx = -range; dx<=range; dx++ )
spectrumout *= ( spectrumIn >= offset(spectrumIn,dx,0) ? 1 : 0 )
spectrumout.SetName("Local maxima ("+range+") filtered")
return spectrumOut
}
image test1 := CreateTestSpec(256,10,1000,5000)
image test2 := FilterLocalMaxima1D(test1,3)
test1.ShowImage()
test2.ShowImage()
However, considering noise (also in your example image), you might want to perform fits around these "local maxima" to ensure you're really getting what you want. The data from above is then only the starting point for that.
Also: Sometimes you can get away with first averaging your data and then finding the local maxima, instead of doing real data fitting in each peak. This works in particular, if you "know" the width of your real peaks rather well.
This would be the example:
image CreateTestSpec( number nChannels, number nPeaks, number amp, number back )
{
image testImg := RealImage( "TestSpec", 4, nChannels )
testImg = amp * cos( PI() + 2*PI() * nPeaks * icol/(iwidth-1) )
testImg += back
testImg = PoissonRandom( testImg )
return testImg
}
image FilterLocalMaxima1D( image spectrumIn, number range )
{
image spectrumOut := spectrumIn.ImageClone()
for( number dx = -range; dx<=range; dx++ )
spectrumout *= ( spectrumIn >= offset(spectrumIn,dx,0) ? 1 : 0 )
spectrumout.SetName("Local maxima ("+range+") filtered")
return spectrumOut
}
image FilterLocalMaxima1DAveraged( image spectrumIn, number range )
{
image avSpectrum := spectrumIn.ImageClone()
avSpectrum = 0
for( number dx = -range; dx<=range; dx++ )
avSpectrum += offset(spectrumIn,dx,0)
avSpectrum *= 1/(2*range+1)
image spectrumOut := spectrumIn.ImageClone()
for( number dx = -range; dx<=range; dx++ )
spectrumout *= ( avSpectrum >= offset(avSpectrum,dx,0) ? 1 : 0 )
spectrumout.SetName("Local maxima ("+range+") filtered, average")
return spectrumOut
}
image test1 := CreateTestSpec(256,10,1000,5000)
image maxPeaks := FilterLocalMaxima1D(test1,3)
image maxPeaksAv := FilterLocalMaxima1DAveraged(test1,3)
test1.ShowImage()
test1.ImageGetImageDisplay(0).ImageDisplayAddImage( maxPeaks, "local max" )
test1.ImageGetImageDisplay(0).ImageDisplayAddImage( maxPeaksAv, "local max from Average" )
test1.ImageGetImageDisplay(0).LinePlotImageDisplaySetSliceComponentColor( 0, 1, 0.7, 0.7, 0.7 )
test1.ImageGetImageDisplay(0).LinePlotImageDisplaySetSliceDrawingStyle( 1, 2)
test1.ImageGetImageDisplay(0).LinePlotImageDisplaySetSliceComponentColor( 1, 1, 1, 0, 0 )
test1.ImageGetImageDisplay(0).LinePlotImageDisplaySetSliceTransparency( 1, 1, 0.7 )
test1.ImageGetImageDisplay(0).LinePlotImageDisplaySetSliceDrawingStyle( 2, 2)
test1.ImageGetImageDisplay(0).LinePlotImageDisplaySetSliceComponentColor( 2, 1, 0, 1, 0 )
test1.ImageGetImageDisplay(0).LinePlotImageDisplaySetSliceTransparency( 2, 1, 0.7 )

Answer 2

The easiest way is to use 1-point (or 2-point) neighborhood to decide, whether center is minimum (maximum). See pseudo code below:
// assume 0 <= x <= maxX, y(x) is value at point x, radius is 1
x = 1;
while (x + 1 <= maxX)
{
if (y(x) > y(x-1) and y(x) > y(x+1))
// we found a maximum at x
if (y(x) < y(x-1) and y(x) < y(x+1))
// we found a minimum at x
x = x + 1
}
For 2-point neighborhood maximum condition could be
if (y(x) > y(x-1) and y(x-1) >= y(x-2) and y(x) > y(x+1) and y(x+1) >= y(x+2))
Note >= here. You may use > instead.
Note that above approach won't find maximum in case two consecutive x have the same value y e.g. for y(0) = 0, y(1) = 1, y(2) = 1, y(3) = 0 it won't find maximum neither for x = 1, nor for x = 2.