UPGMA (Unweighted Pair Group Method with Arithmetic Mean) is a simple agglomerative (bottom-up) hierarchical clustering method. The method is generally attributed to Sokal and Michener.
The UPGMA method is similar to its weighted variant, the WPGMA method.
Note that the unweighted term indicates that all distances contribute equally to each average that is computed and does not refer to the math by which it is achieved. Thus the simple averaging in WPGMA produces a weighted result and the proportional averaging in UPGMA produces an unweighted result (see the working example).
The UPGMA algorithm constructs a rooted tree (dendrogram) that reflects the structure present in a pairwise similarity matrix (or a dissimilarity matrix). At each step, the nearest two clusters are combined into a higher-level cluster. The distance between any two clusters and , each of size (i.e., cardinality) and , is taken to be the average of all distances between pairs of objects in and in , that is, the mean distance between elements of each cluster:
In other words, at each clustering step, the updated distance between the joined clusters and a new cluster is given by the proportional averaging of the and distances:
The UPGMA algorithm produces rooted dendrograms and requires a constant-rate assumption - that is, it assumes an ultrametric tree in which the distances from the root to every branch tip are equal. When the tips are molecular data (i.e., DNA, RNA and protein), the ultrametricity assumption is called the molecular clock.
Let us assume that we have five elements and the following matrix of pairwise distances between them :
In this example, is the smallest value of , so we join elements and .
- First branch length estimation
Let denote the node to which and are now connected. Setting ensures that elements and are equidistant from . This corresponds to the expectation of the ultrametricity hypothesis. The branches joining and to then have lengths (see the final dendrogram)
- First distance matrix update
We then proceed to update the initial distance matrix into a new distance matrix (see below), reduced in size by one row and one column because of the clustering of with . Bold values in correspond to the new distances, calculated by averaging distances between the first cluster and each of the remaining elements:
Italicized values in are not affected by the matrix update as they correspond to distances between elements not involved in the first cluster.
We now reiterate the three previous steps, starting from the new distance matrix
Here, is the smallest value of , so we join cluster and element .
- Second branch length estimation
Let denote the node to which and are now connected. Because of the ultrametricity constraint, the branches joining or to , and to are equal and have the following length:
We deduce the missing branch length: (see the final dendrogram)
- Second distance matrix update
We then proceed to update into a new distance matrix (see below), reduced in size by one row and one column because of the clustering of with . Bold values in correspond to the new distances, calculated by proportional averaging:
Thanks to this proportional average, the calculation of this new distance accounts for the larger size of the cluster (two elements) with respect to (one element). Similarly:
Proportional averaging therefore gives equal weight to the initial distances of matrix . This is the reason why the method is unweighted, not with respect to the mathematical procedure but with respect to the initial distances.
We again reiterate the three previous steps, starting from the updated distance matrix .
Here, is the smallest value of , so we join elements and .
- Third branch length estimation
Let denote the node to which and are now connected. The branches joining and to then have lengths (see the final dendrogram)
- Third distance matrix update
There is a single entry to update, keeping in mind that the two elements and each have a contribution of in the average computation:
The final matrix is:
So we join clusters and .
Я обязан позвонить в службу безопасности, - решил. - Что еще мне остается? - Он представил Хейла на скамье подсудимых, вываливающего все, что ему известно о Цифровой крепости. - Весь мой план рухнет. Должен быть какой-то другой выход.