Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am not sure if this question makes sense or is even ideal, but here is my problem: Many proposed distance formulas only give a distance in terms of 1-dimensions. A common distance formula is the Euclidean distance (the most obvious one). Euclidean distance $d_{ecdn}$ of two points located at $(x_1, x_2, x_3, ... , x_n)$ and $(y_1, y_2, y_3, ... , y_n)$, respectively, is given by

$d_{ecdn} = \sqrt{\displaystyle\sum_{i=1}^{n} (x_i-y_i)^2}$.

Geometrically, in any number of dimensions (but imagine 3 dimensions as an example), if you were to draw a line between the two points, whose length expresses the Euclidean distance, the line itself exists in 1 dimension. Why is this a problem? I am doing a project involve data mining and information retrieval. Basically, each document in a corpus is mapped to a location in an $n$-dimensional space where $n$ is the total number of terms within the whole set of documents. They say that if document A is geometrically closer to document B than document C, then document A is more similar to B than C. Euclidean distance gives the wrong impression of similarity since each dimension matters.

I am wondering if anyone has any good algorithms that will calculate the "distance" between two points, but with a weight on each dimensions. An example would be the Pearson Product-Moment Correlation Coefficient. This formula is given by


share|improve this question

migrated from Aug 22 '12 at 5:53

This question came from our site for people studying math at any level and professionals in related fields.

Note, lines are in 1 dimension. – Karolis Juodelė Aug 22 '12 at 5:52
Fixed! Stupid error. Thank you. – Sidd Aug 22 '12 at 5:57
I think you are looking for Mahalanobis Distance. – TenaliRaman Aug 22 '12 at 6:27
1) Weighted version of euclidean distance is straightforward, see the formula. 2) Why do you think Pearson r incorporates weights? I don't see any weighting. – ttnphns Aug 22 '12 at 7:34
@TenaliRaman +1. That probably would have been appropriate enough for an answer, but regardless, thank you for that! – Sidd Aug 22 '12 at 13:30

Just to create an official answer for this topic (based on @TenaliRaman's comment), you want the Mahalanobis distance.

This is computed similarly to the Euclidian distance, (in fact, the Euclidian distance formula is a special case of the Mahalanobis distance formula) except each squared difference term is 'corrected' based on an inverted covariance matrix.

From the Wikipedia page:

 Mahalanobis distance is thus unitless and scale-invariant, 
 and takes into account the correlations of the data set.
share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.