Tell me more ×
Geographic Information Systems Stack Exchange is a question and answer site for cartographers, geographers and GIS professionals. It's 100% free, no registration required.
up vote 3 down vote favorite

I am trying to determine the level of spatial autocorrelation in a point dataset. The attribute I'm interested in is binary (presence/absence of a species), for which Moran's I is not appropriate. On the other hand, Joint Count statistics, which are typically recommended for binary or categorical data, apparently are not appropriate for point data. In short, the question is thus: what is an appropriate statistic to measure global and/or local spatial autocorrelation of points when the attribute of interest is binary?

share|improve this question

2 Answers

up vote 2 down vote

Your assertion that a Join-Counts statistic is not appropriate for binary data is not correct. It is just a matter of how the spatial weights matrix (Wij) is specified. As in a Morna's-I, you cannot use a distance matrix in this type of analysis, However, an appropriate binary matrix of contingency can be calculated using a distance cutoff. You can create this type of spatial weights matrix as well as conduct a Join-Count analysis in the R spdep library. See the "joincount.test" and joincount.mc (for Monte Carlo permutation test) functions.

share|improve this answer
Thanks, Jeffrey. Joint counts are clearly the way to go for binary data, but I saw a suggestion (can't recall where, now) that joint counts were only appropriate for area (not point) data. It wasn't apparent to me why you couldn't create the weights matrix using a distance threshold, and use a joint count, but I couldn't find examples of this in some cursory searches. Is there a reference you can provide for this type of use? – user13706 Jan 29 at 19:15
This is a large body of literature on Point Pattern Analysis. The Join-Counts statistic is not commonly used and, as such, is not very prevalent in the current literature. I would go back to early work by Diggle or Geits. What is your goal in quantifying spatial dependency in binomial data? You cannot use a Join-Counts coefficient in something like a mixed effects or CAR/SAR model. Here some some interesting background on occupancy pattern scaling (en.wikipedia.org/wiki/Scaling_pattern_of_occupancy) – Jeffrey Evans Jan 29 at 19:48
I'm constructing a species distribution model, and have both presence and reliable absence locations for the species. There is clear clustering of points on the landscape, but perhaps little spatial autocorrelation, since presence and absence points occur in close proximity. In other words, our samples are spatially biased, with points occurring within hundreds of meters of other points, but if spatial autocorrelation is not significant, I'm wondering how much the clustering matters in terms of modeling, inferences, etc. I'm using Random Forest for modeling. – user13706 Jan 29 at 20:08
RandomForest is a nonparametric model and, as such, is not effected by autocorrelation. The concern with this model is correlation within the bootstrap ensemble. Often autocorrelation can create "redundancy" in your data that creates bias in the Bootstrap. I would look at based on the conditional distribution(s) of your covariates. I have R code available "R - Plot Probability Density by Grouping Factor" here: conserveonline.org/workspaces/emt/documents/all.html – Jeffrey Evans Jan 29 at 20:19
(+1) But (re a comment) I don't follow how "nonparametric" implies "not affected by autocorrelation." That's certainly not true in general. But since a random forest isn't really a model--it's just a black box predictor--maybe you meant "nonparametric" in that sense. – whuber♦ Jan 29 at 20:28
show 5 more comments
up vote 0 down vote

Binary data is a normal use case for spatial autocorrelation. I think most of spatial analysis book will talk about it. This document might be of help.

share|improve this answer
1  
The first page of your reference emphasizes that "data locations are regions," so it looks like it does not apply to point data at all. – whuber♦ Jan 29 at 20:28

Your Answer

 
discard

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.