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This question already has an answer here:

Self organizing maps are claimed to be able to visualize/cluster high-dimensional data in a smaller dimensional space. I have some difficulties in understanding this statement.

Consider a six-dimensional data set; the codebook vector/reference vector is also six-dimensional. According to the SOM algorithm, updating these reference vectors is also conducted in the six-dimensional vector space. If we are considering a two dimensional map, how should I understand the map between the six-dimensional data space and two-dimensional map space?

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marked as duplicate by John, jbowman, gung, Nick Cox, Peter Flom Dec 10 '13 at 11:15

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  • $\begingroup$ Not sure exactly what the question is: are you wanting an intuition or an equation, or further algorithm description? Intuition-wise, the SOM, in your example, is a 2D manifold embedded in a 6D problem space, and the SOM training method smooths out the manifold in problem space in such a way that it's weighted by the training data, so that the density of the points on the manifold in problem space reflect the density of the training samples in problem space. SOM (2D) neighbors will be relatively close in problem (6D) space. $\endgroup$ – Wayne Nov 15 '11 at 19:55
  • $\begingroup$ This appears to be a duplicate of stats.stackexchange.com/questions/22774/… . See also: stats.stackexchange.com/questions/64659/… $\endgroup$ – naught101 Dec 10 '13 at 4:24
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You can think of the SOM as a grid, where the reference vectors are just placed. However, while the elements of the reference vector define the vectors orientation in the input space, they have no direct relation to the placement on the SOM.

During the competitive training of the SOM for each input vector, the winning node is determined, as its reference vector has the smallest (typically Euclidian) distance to the input vector. This reference vector is then adjusted towards the input vector. You can imagine that, as if the input vector pulls the reference vector in the input space towards itself. The "strength" of pulling is determined by the training rate, which decreases monotonically over time.

Now comes the trick, which makes the SOM topology preserving: The reference vector of the winning node will adjust all reference vectors on the SOM towards itself. The "strength" of pulling is decreasing over the distance on the SOM. Hence, after having performed one training cycle, the reference vectors close to each other in input space will be also located close to each other on the SOM.

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  • $\begingroup$ Say that you had a 3x3 SOM and an input space consisting of 30 5-Dimensional vectors. When you train, you present each one of the 30 5-Dimensional vectors to the SOM. When training is done, you will have 9 5-Dimensional reference vectors, each attached to one node of the SOM. So, is this where the dimensionality reduction takes place? You started with 30 5-Dimensional vectors, now you have 9. $\endgroup$ – user1231745 Jul 31 '13 at 17:36
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    $\begingroup$ @user1231745 No, you start with 30 5-dimensional vectors and you end up with 30, nondistinct, 2-dimensional vectors (i.e. on a 2-dimensional grid). $\endgroup$ – Beasterfield Aug 1 '13 at 12:07
  • $\begingroup$ @user1231745 You appear to have multiple accounts. Please merge them. Thanks. $\endgroup$ – whuber Aug 1 '13 at 22:37

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