1 / 34

Point Pattern Analysis

Point Pattern Analysis. using Spatial Inferential Statistics. Last time. Concept of statistical inference Drawing conclusions about populations from samples Null Hypothesis of no difference Alternative hypotheses (which we really want to accept) Random point pattern

ziya
Download Presentation

Point Pattern Analysis

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Point Pattern Analysis using Spatial Inferential Statistics Briggs Henan University 2010

  2. Last time • Concept of statistical inference • Drawing conclusions about populations from samples • Null Hypothesis of no difference • Alternative hypotheses (which we really want to accept) • Random point pattern • Is our observed point pattern “significantly different from random” Briggs Henan University 2010

  3. How Point Pattern Analysis (PPA) is different From Centrographic Statistics (previously): Centrographic Statistics calculates single, summary measures PPA analyses the complete set of points From Spatial Autocorrelation (discussed later): with PPA, the points have location only; there is no “magnitude” value With Spatial Autocorrelation points have different magnitudes; there is an attribute variable. Briggs Henan University 2010

  4. Approaches to Point Pattern Analysis Two primary approaches: • Point Density using Quadrat Analysis • Based on polygons • Analyze points using polygons! • Uses the frequency distribution or density of points within a set of grid squares. • Point Associationusing Nearest Neighbor Analysis • Based on points • Uses distances between the points Although the above would suggest that the first approach examines first order effects and the second approach examines second order effects, in practice the two cannot be separated. Briggs Henan University 2010

  5. Quadrat Analysis:The problem of selecting quadrat size Too small: many quadrats with zero points Too big: many quadrats have similar number of points O.K. Length of Quadrat edge = A=study area N= number of points Modifiable Areal Unit Problem Briggs Henan University 2010

  6. Uniform grid --used for secondary data Types of Quadrats Random sampling --useful in field work Frequency counts by Quadrat would be: Multiple ways to create quadrats --and results can differ accordingly! Quadrats don’t have to be square --and their size has a big influence Briggs Henan University 2010

  7. Quadrat Analysis: Variance/Mean Ratio (VMR) Where: A = area of region n = # of points • Apply uniform or random grid over area (A) with width of square given by: • Treat each cell as an observation and count the number of points within it, to create the variable X • Calculate variance and mean of X, and create the variance to mean ratio: variance / mean • For an uniform distribution, the variance is zero. • Therefore, we expect a variance-mean ratio close to 0 • For a random distribution, the variance and mean are the same. • Therefore, we expect a variance-mean ratio around 1 • For a clustered distribution, the variance is relatively large • Therefore, we expect a variance-mean ratio above 1 See following slide for example. See O&U p 98-100 for another example Briggs Henan University 2010

  8. x x x UNIFORM/ DISPERSED CLUSTERED random uniform Clustered Formulae for variance 2 RANDOM Note: N = number of Quadrats = 10 Ratio = Variance/mean Briggs Henan University 2010

  9. = Significance Test for VMR • A significance test can be conducted based upon the chi-square frequency distribution • The test statistic is given by: (sum of squared differences)/Mean • The test will ascertain if a pattern is significantly more clustered than would be expected by chance (but does not test for a uniformity) • The values of the test statistics in our cases would be: • For degrees of freedom: N - 1 = 10 - 1 = 9, the value of chi-square at the 1% level is 21.666. • Thus, there is only a 1% chance of obtaining a value of 21.666 or greater if the points had been allocated randomly. Since our test statistic for the clustered pattern is 80, we conclude that there is (considerably) less than a 1% chance that the clustered pattern could have resulted from a random process random 60-(202)/10=10 2 uniform 40-(202)/10=0 2 clustered 200-(202)/10=80 2

  10. Quadrat Analysis: Frequency Distribution Comparison • Rather than base conclusion on variance/mean ratio, we can compare observed frequencies in the quadrats (Q= number of quadrats) with expected frequencies that would be generated by • a random process (modeled by the Poisson frequency distribution) • a clustered process (e.g. one cell with P points, Q-1 cells with 0 points) • a uniform process (e.g. each cell has P/Q points) • The standard Kolmogorov-Smirnov test for comparing two frequency distributions can then be applied – see next slide • See Lee and Wong pp. 62-68 for another example and further discussion. Briggs Henan University 2010

  11. Kolmogorov-Smirnov (K-S) Test • The test statistic “D” is simply given by: D = max [ Cum Obser. Freq – Cum Expect. Freq] The largest difference (irrespective of sign) between observed cumulative frequency and expected cumulative frequency • The critical value at the 5% level is given by: D (at 5%) = 1.36where Q is the number of quadrats Q • Expected frequencies for a random spatial distribution are derived from the Poisson frequency distribution and can be calculated with: p(0) = e-λ= 1 / (2.71828P/Q) and p(x) = p(x - 1) * λ/x Where x = number of points in a quadrat and p(x) = the probability of x points P = total number of points Q = number of quadrats λ = P/Q (the average number of points per quadrat) See next slide for worked example for cluster case Briggs Henan University 2010

  12. Row 10 The spreadsheet spatstat.xlscontains worked examples for the Uniform/ Clustered/ Random data previously used, as well as for Lee and Wong’s data Briggs Henan University 2010

  13. Weakness of Quadrat Analysis • Results may depend on quadrat size and orientation (Modifiable areal unit problem) • test different sizes (or orientations) to determine the effects of each test on the results • Is a measure of dispersion, and not really pattern, because it is based primarily on the density of points, and not their arrangement in relation to one another • Results in a single measure for the entire distribution, so variations within the region are not recognized (could have clustering locally in some areas, but not overall) For example, quadrat analysis cannot distinguish between these two, obviously different, patterns For example, overall pattern here is dispersed, but there are some local clusters Briggs Henan University 2010

  14. Nearest-Neighbor Index (NNI) (O&U p. 100) • Uses distances between points • It compares: • the mean of the distance observed between each point and its nearest neighbor • with the expected mean distance if the distribution was random: Observed Average Distance Expected Average Distance For random pattern, NNI = 1 For clustered pattern, NNI = 0 For dispersed pattern, NNI = 2.149 NNI = See next slide for formulae for calculation Briggs Henan University 2010

  15. Calculating Nearest Neighbor Index Where: The average distance to nearest neighbor Area of region: result very dependent on this value Briggs Henan University 2010

  16. Significance Test for NNI • The test statistic is calculated as follows: Z = Av. Distance Observed - Av. Distance Expected. Standard Error • It has a Normal Frequency Distribution. • Ittests if the observedpattern is significantly different from random. • if Z is below –1.96 or above +1.96, we are “95% confident that the distribution is not randomly distributed.” • or can say: If the observed pattern was random, there are less than 5 chances in 100 we would have observed a z value this large. Note: in the example on the next slide, the fact that the NNI for uniform is 1.96 is coincidence! Briggs Henan University 2010

  17. (Standard error) Calculating Test Statistic for Nearest Neighbor Index Where: Briggs Henan University 2010

  18. Mean distance Mean distance Mean distance NNI NNI NNI RANDOM CLUSTERED UNIFORM Z = 5.508 Z = -0.1515 Z = 5.855 Briggs Henan University 2010 Source: Lembro

  19. Running in ArcGIS Telecom and Software Companies Result is very dependent on area of the region. There is an option to insert your own value. Default value is the “minimum enclosed rectangle that encompasses all features. Briggs Henan University 2010

  20. results Scroll up the window to see all the results. Note: Progress box continues to run until graphic is closed. Always close graphic window first. Produced if “Display output graphically “ box is  Briggs Henan University 2010

  21. Evaluating the Nearest Neighbor Index • Advantages • Unlike quadrats, the NNI considers distances between points • No quadrat size problem • However, NNI has problems • Very dependent on the value of A, the area of the study region. What boundary do we use for the study area? • Minimum enclosing rectangle? (highly affected by a few outliers) • Convex hull • Convex hull with buffer. What size buffer? • There is an “adjustment for edge effects” but problems remain • Based on only the mean distance to the nearest neighbor • Doesn’t incorporate local variations, or clustering scale • could have clustering locally in some areas, but not overall • Based on point location only and does not incorporate magnitude of phenomena (quantity) at that point Briggs Henan University 2010

  22. Ripley’s K(d) Function O&U p. 135-137 Where S is a point, and C(si, d) is a circle of radius d, centered at si Ripley B.D. 1976. The second –order analysis of stationary point processes. Journal of Applied Probability 13: 255-266 • Ripley’s K is calculated multiple times, each for a different distance band, • So it is represented as K(d): K is a function of distance, d • The distance bands are placed around every point • K (d) is the average density of points at each distance (d), divided by the average density of points in the entire area (n/a) • If the density is high for a particular band, then clustering is occurring at that distance Briggs Henan University 2010

  23. Not this simple with real data!!! The low end (0.2) corresponds to distances within the cluster The high end (0.6) corresponds to distance between the clusters within clustered between The distance bands are placed around every point. Note the big problem of edge effects from circles outside the study area. dispersed Source: O’Sullivan & Unwin, p. Begins flat Briggs Henan University 2010

  24. Running in ArcGIS Telecom and Software Companies use 9 for tests--99 takes a long time! Weight field—number of points at that location Distance bands Result is very dependent on area of the region. Can insert your own value. Again—study area has big effect so there are several options for this Briggs Henan University 2010

  25. Interpreting the Results Not this simple with real data!!! Clustered, since observed is above expected Observed Expected Dispersed, if observed was below expected Pattern is clustered! Expected based on random pattern

  26. Distance bands: start 5,000 feet size: 10,000 feet Expected assumes random pattern Confidence band—9 iterations (takes long time for 99!) Results for 10,000 feet Bands Briggs Henan University 2010

  27. Distance bands: start 10,000 feet size: 20,000 feet Also experiment with different region (study area) boundaries. Results for 20,000 feet Bands Briggs Henan University 2010

  28. Plotting the Difference Between Observed and Expected K, versus Distance Y field = Diffk = ObservedK - ExpectedK X field = ExpectedK or HiConfEnv Distance between clusters 70,000 feet = 13 miles = 20 km Briggs Henan University 2010

  29. Problems with Ripley K(d) • Dependent on study area boundary (edge effect) • Circles go outside study area • Special adjustments are available (see O&U p. 148) • Try different options for boundary in ArcGIS • Affected by circle radii selected • Try different values • Each point has unit value—no magnitude or quantity • Weight field assumes “X” points at that location • e.g. X = 3, then 3 points at that location Briggs Henan University 2010

  30. What have we learned? How to measure and test if spatial patterns are clustered or dispersed. Briggs Henan University 2010

  31. Why is this important? ? We can measure and test --not just look and guess! Is it clustered? That is science. Briggs Henan University 2010

  32. Not just GIS! I taught these tools to senior undergraduate geography students. They are also used in Earth Management. A former Henan University student and faculty member (now at UT-Dallas) is using Ripley’s K function for research on urban forests. Briggs Henan University 2010

  33. Next Time • No classes next week • Next class will be Wednesday November 17 Topic • Spatial Autocorrelation • Unlike PPA, in Spatial Autocorrelation points have different magnitudes; there is an attribute variable. Briggs Henan University 2010

  34. Briggs Henan University 2010

More Related