org.scijava.ops.image.threshold.triangle.ComputeTriangleThreshold Maven / Gradle / Ivy
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package org.scijava.ops.image.threshold.triangle;
import org.scijava.ops.image.threshold.AbstractComputeThresholdHistogram;
import net.imglib2.histogram.Histogram1d;
import net.imglib2.type.numeric.RealType;
// NB - this plugin adapted from Gabriel Landini's code of his AutoThreshold
// plugin found in Fiji (version 1.14).
/**
* Implements a Triangle algorithm threshold method from Zack, Rogers,
* {@literal &} Latt.
*
* @author Barry DeZonia
* @author Gabriel Landini
* @implNote op names='threshold.triangle', priority='100.'
*/
public class ComputeTriangleThreshold> extends
AbstractComputeThresholdHistogram
{
/**
* TODO
*
* @param hist the {@link Histogram1d}
* @return the Triangle threshold value
*/
@Override
public long computeBin(final Histogram1d hist) {
final long[] histogram = hist.toLongArray();
return computeBin(histogram);
}
/**
* Zack, G. W., Rogers, W. E. and Latt, S. A., 1977,
* Automatic Measurement of Sister Chromatid Exchange Frequency,
* Journal of Histochemistry and Cytochemistry 25 (7), pp. 741-753
*
* modified from Johannes Schindelin plugin
*/
public static long computeBin(final long[] histogram) {
// find min and max
int min = 0, max = 0, min2 = 0;
long dmax = 0;
for (int i = 0; i < histogram.length; i++) {
if (histogram[i] > 0) {
min = i;
break;
}
}
if (min > 0) min--; // line to the (p==0) point, not to histogram[min]
// The Triangle algorithm cannot tell whether the data is skewed to one
// side
// or another. This causes a problem as there are 2 possible thresholds
// between the max and the 2 extremes of the histogram. Here I propose
// to
// find out to which side of the max point the data is furthest, and use
// that as the other extreme.
for (int i = histogram.length - 1; i > 0; i--) {
if (histogram[i] > 0) {
min2 = i;
break;
}
}
// line to the (p==0) point, not to histogram[min]
if (min2 < histogram.length - 1) min2++;
for (int i = 0; i < histogram.length; i++) {
if (histogram[i] > dmax) {
max = i;
dmax = histogram[i];
}
}
// find which is the furthest side
// IJ.log(""+min+" "+max+" "+min2);
boolean inverted = false;
if ((max - min) < (min2 - max)) {
// reverse the histogram
// IJ.log("Reversing histogram.");
inverted = true;
int left = 0; // index of leftmost element
int right = histogram.length - 1; // index of rightmost element
while (left < right) {
// exchange the left and right elements
final long temp = histogram[left];
histogram[left] = histogram[right];
histogram[right] = temp;
// move the bounds toward the center
left++;
right--;
}
min = histogram.length - 1 - min2;
max = histogram.length - 1 - max;
}
if (min == max) {
// IJ.log("Triangle: min == max.");
return min;
}
// describe line by nx * x + ny * y - d = 0
double nx, ny, d;
// nx is just the max frequency as the other point has freq=0
// lowest value bmin = (p=0)% in the image
nx = histogram[max]; // -min; // histogram[min];
ny = min - max;
d = Math.sqrt(nx * nx + ny * ny);
nx /= d;
ny /= d;
d = nx * min + ny * histogram[min];
// find split point
int split = min;
double splitDistance = 0;
for (int i = min + 1; i <= max; i++) {
final double newDistance = nx * i + ny * histogram[i] - d;
if (newDistance > splitDistance) {
split = i;
splitDistance = newDistance;
}
}
split--;
if (inverted) {
// The histogram might be used for something else, so let's reverse
// it
// back
int left = 0;
int right = histogram.length - 1;
while (left < right) {
final long temp = histogram[left];
histogram[left] = histogram[right];
histogram[right] = temp;
left++;
right--;
}
return (histogram.length - 1 - split);
}
return split;
}
}