}
}
/**************************************************************************************************/
-//The silhouette width S(i)of individual data points i is calculated using the following formula:
-/*
- s(i) = b(i) - a(i)
- -----------
- max(b(i),a(i))
- where a(i) is the average dissimilarity (or distance) of sample i to all other samples in the same cluster, while b(i) is the average dissimilarity (or distance) to all objects in the closest other cluster.
-
- The formula implies -1 =< S(i) =< 1 . A sample which is much closer to its own cluster than to any other cluster has a high S(i) value, while S(i) close to 0 implies that the given sample lies somewhere between two clusters. Large negative S(i) values indicate that the sample was assigned to the wrong cluster.
- */
-
-vector<double> KMeans::calcSilhouettes(vector<vector<double> > dists) {
- try {
- vector<double> silhouettes; silhouettes.resize(numSamples, 0.0);
- if (numPartitions < 2) { return silhouettes; }
-
- map<int, int> clusterMap; //map sample to partition
- for (int j = 0; j < numSamples; j++) {
- double maxValue = 0.0;
- for (int i = 0; i < numPartitions; i++) {
- if (m->control_pressed) { return silhouettes; }
- if (zMatrix[i][j] > maxValue) { //for kmeans zmatrix contains values for each sample in each partition. partition with highest value for that sample is the partition where the sample should be
- clusterMap[j] = i;
- maxValue = zMatrix[i][j];
- }
- }
- }
-
- vector<int> nextClosestPartition;
- findSecondClosest(nextClosestPartition, dists, clusterMap);
-
- if (m->control_pressed) { return silhouettes; }
-
- vector<double> a, b; a.resize(numSamples, 0.0); b.resize(numSamples, 0.0);
-
- //calc a - all a[i] are the same in the same partition
- for (int k = 0; k < numPartitions; k++) {
- if (m->control_pressed) { break; }
-
- int count = 0;
- double totalA = 0.0;
- for (int i = 0; i < numSamples; i++) {
- for (int j = 0; j < numSamples; j++) {
- if (m->control_pressed) { break; }
- if ((clusterMap[i] == k) && (clusterMap[j] == k)){ //are both samples in the partition, if so add there distance
- totalA += dists[j][i]; //distance from this sample to the other samples in the partition
- count++;
- }
- }
- }
- totalA /= (double) count;
-
- //set a[i] to average for cluster
- for (int i = 0; i < numSamples; i++) {
- if (clusterMap[i] == k) { a[i] = totalA; }
- }
- }
-
- //calc b
- for (int i = 0; i < numSamples; i++) {
- if (m->control_pressed) { break; }
-
- int thisPartition = nextClosestPartition[i];
- int count = 0;
- double totalB = 0.0;
- for (int j = 0; j < numSamples; j++) {
- if (clusterMap[j] == thisPartition) { //this sample is in this partition
- totalB += dists[i][j];
- count++;
- }
- }
- b[i] = totalB / (double) count;
- }
-
- //calc silhouettes
- for (int i = 0; i < numSamples; i++) {
- if (m->control_pressed) { break; }
-
- double denom = a[i];
- if (b[i] > denom) { denom = b[i]; } //max(a[i],b[i])
-
- silhouettes[i] = (b[i] - a[i]) / denom;
-
- //cout << "silhouettes " << i << '\t' << silhouettes[i] << endl;
- }
-
- return silhouettes;
- }
- catch(exception& e) {
- m->errorOut(e, "KMeans", "calcSilhouettes");
- exit(1);
- }
-}
-/**************************************************************************************************/
-int KMeans::findSecondClosest(vector<int>& nextClosestPartition, vector<vector<double> >& dists, map<int, int> clusterMap) {
- try {
- vector<double> minScores; minScores.resize(numSamples, 1e6);
- nextClosestPartition.resize(numSamples, 0);
-
-
- for (int i = 0; i < numSamples; i++) {
- for (int j = 0; j < numPartitions; j++) {
- if (m->control_pressed) { break; }
-
- //is this the one we are assigned to - ie the "best" cluster. We want second best.
- //if numPartitions = 2, then special case??
- if (clusterMap[i] != j) {
- double score = 1e6;
- if (numPartitions == 2) {
- score = 0.0; //choose other option, there are only 2.
- }else{ score = calcScore(i, j, dists, clusterMap); }
-
- if (m->debug) { m->mothurOut("[DEBUG]: sample = " + toString(i) + " partition = " + toString(j) + " score = " + toString(score) + "\n"); }
-
- //is this better than our last find
- if (score < minScores[i]) {
- minScores[i] = score;
- nextClosestPartition[i] = j;
- }
- }else {} //best case, ignore
- }
- }
-
- return 0;
- }
- catch(exception& e) {
- m->errorOut(e, "KMeans", "findSecondClosest");
- exit(1);
- }
-}
-/**************************************************************************************************/
-double KMeans::calcScore(int sample, int partition, vector<vector<double> >& dists, map<int, int> clusterMap) {
- try {
- //square the distances and then for each pair of clusters, calculate the sum of the squraed distances between the clusters
- //then with the sum of hte squared dsitances take the square root and divide by the number of distances in the sum
-
- double sum = 0.0; int count = 0;
- for (int i = 0; i < numSamples; i++) {
- if (m->control_pressed) { break; }
- if (clusterMap[i] == partition) { //samples in this cluster
- sum += (dists[sample][i] * dists[sample][i]);
- count++;
- }
- }
-
- sum = sqrt(sum);
- sum /= (double) count;
-
- return sum;
- }
- catch(exception& e) {
- m->errorOut(e, "KMeans", "calcScore");
- exit(1);
- }
-}
-/**************************************************************************************************/
-/*To assess the optimal number of clusters our dataset was most robustly partitioned into, we used the Calinski-Harabasz (CH) Index that has shown good performance in recovering the number of clusters. It is defined as:
-
- CHk=Bk/(k−1)/Wk/(n−k)
-
- where Bk is the between-cluster sum of squares (i.e. the squared distances between all points i and j, for which i and j are not in the same cluster) and Wk is the within-clusters sum of squares (i.e. the squared distances between all points i and j, for which i and j are in the same cluster). This measure implements the idea that the clustering is more robust when between-cluster distances are substantially larger than within-cluster distances. Consequently, we chose the number of clusters k such that CHk was maximal.*/
-double KMeans::calcCHIndex(vector< vector< double> > dists){
- try {
- double CH = 0.0;
-
- if (numPartitions < 2) { return CH; }
-
- map<int, int> clusterMap; //map sample to partition
- for (int j = 0; j < numSamples; j++) {
- double maxValue = 0.0;
- for (int i = 0; i < numPartitions; i++) {
- if (m->control_pressed) { return 0.0; }
- if (zMatrix[i][j] > maxValue) { //for kmeans zmatrix contains values for each sample in each partition. partition with highest value for that sample is the partition where the sample should be
- clusterMap[j] = i;
- maxValue = zMatrix[i][j];
- }
- }
- }
-
- double sumBetweenCluster = 0.0;
- double sumWithinClusters = 0.0;
-
- for (int i = 0; i < numSamples; i++) { //lt
- for (int j = 0; j < i; j++) {
- if (m->control_pressed) { return 0.0; }
- int partitionI = clusterMap[i];
- int partitionJ = clusterMap[j];
-
- if (partitionI == partitionJ) { //they are from the same cluster so this distance is added to Wk
- sumWithinClusters += (dists[i][j] * dists[i][j]);
- }else { //they are NOT from the same cluster so this distance is added to Bk
- sumBetweenCluster += (dists[i][j] * dists[i][j]);
- }
- }
- }
- //cout << numPartitions << '\t' << sumWithinClusters << '\t' << sumBetweenCluster << '\t' << (numSamples - numPartitions) << endl;
-
- CH = (sumBetweenCluster / (double)(numPartitions - 1)) / (sumWithinClusters / (double) (numSamples - numPartitions));
-
- return CH;
- }
- catch(exception& e){
- m->errorOut(e, "KMeans", "calcCHIndex");
- exit(1);
- }
-}
-/**************************************************************************************************/