jenks natural breaks vs k-means I am new to this topic. As far as I know both are data clustering methods. Then my question is when is Jenks prefered over k-means?
I read on this website that jenks is particularly suited for 1-dimensional data, while k-means are for multi-dimensional. Is this true? If so can you point me to some references about this?
I also read from another source that says jenks is usually used to spot gaps in ranged data. Again I cannot find a reference to explain this.
Any pointers to any particular work that uses Jenks natural break or even compares it with k-means would be great. 
 A: The Jenks natural breaks algorithm, just like K-means, assigns data to one of K groups such that the within group distances are minimized. Also just like K-means, one must select K prior to running the algorithm. 
However, Jenks and K-means are different in how they minimize within group distances. Jenks takes advantage of the fact that 1-dimensional data is sortable which makes it a faster algorithm for 1-dimensional data. K-means is more general in that it can handle data in any dimension; including dimensions greater than 1 where the data is not sortable.
A: Previous answers essentially present Jenks as a special case of K-means. However, this source makes an important distinction: K-means solely "searches for minimum distance between data points and the centers of clusters they belong to". Jenks takes this objective and adds a penalty for the proximity between the centers of clusters, and thus it also searches "for maximum difference between cluster centers themselves".
The logic is that, even if two clusters are internally very compact, they may be hard to distinguish when their centers are very close.
Thus, for $n$ data points and $k$ clusters, K-means would minimize $C$:
$$ C = \sum_{i=1}^n  \sum_{j=1}^k dist(d_i, c_j) $$
where $d_i$ is the value of a data point and $c_j$ is the value of its associated cluster center.
In contrast, the Jenks algorithm would minimize $J$:
$$ J = C - \sum_{j=1}^{k-1} dist(c_{j+1}, c_j)$$
Two things to note, however:

*

*I am really no expert in clustering algorithms, so confirmations, comments, corrections and edits are welcome.

*The source I reference states that $dist()$ computes the Euclidean distance (so, $\sqrt{(d_i - c_j)^2}$), but from everything else I read on K-means it seems that the squared Euclidean distance ($(d_i - c_j)^2$) is what is actually minimized.

Full reference:
Khan, F. (2012). An initial seed selection algorithm for k-means clustering of georeferenced data to improve replicability of cluster assignments for mapping application. Applied Soft Computing Journal, 12(11), 3698–3700. https://doi.org/10.1016/j.asoc.2012.07.021
