2
$\begingroup$

Thank you for reading my question. I have an archaeological case-study, that we can call "Site1", that I want to compare with 9 others "Sites" studied by other scholars. For all of them I have 8 economic (indipendent) variables based on the frecuency of certain artifacts. How can I do to explore the 'economnic' variabiliy of this sites, and more specifically to see to which sites my case-study is more similar to?

This are the steps I followed:

  • I changed from frecuency to percentages in order to avoid the among sites variability. There are site with a lot of observations and sites with only twenty observations. But I am not interested in absolute frecuency, but in relative ones.

  • I calculated Z-values for the entire sample from the percentage values.

  • I run a HCA with Ward method, excluding my 'Site1' from the analysis. I obtain a three-cluster solution, with three Sites each one.

  • I save the cluster centers.

  • I run a K-means analysis using the Ward's cluster-centers, but this time including also my 'Site1'. As result, I see that the cluster-composition is not changed and that my 'Site1' is added to cluster2. So, I can run a One-way Anova and a Tukey to see which variables contribuite at most to each cluster.

Is it a right procedure? Would you suggest some different strategy?

Thanks,

Mark

$\endgroup$
0
$\begingroup$

Why don't you just assign the new observation to the nearest cluster center by least SSQ? Running k-means on a Ward clustering result is like doing a nice steak on the BBQ and then mincing it to make a frozen hamburger.

Technically, you use the cluster centers as new data set, then do a 1NN classification to classify your new observation.

$\endgroup$
  • $\begingroup$ It sound interesting, but no idea on how to perform an K-nearest neighbour... I don't have in my SPSS Statistic v22, and no idea on how do it with R. Why it is not correct to run a 'canonic' K-means? Just one more question... do you think it is right to peform Z-values on percentages? $\endgroup$ – Mark Jan 22 '15 at 20:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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