Transcript Point Pattern Analysis - The University of Texas at Dallas

Point Pattern Analysis
using
Spatial Inferential Statistics
1
Briggs Henan University 2010
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
2
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
3
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 Association using 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.
4
Briggs Henan University 2010
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
=
Modifiable Areal Unit
Problem
A=study area
N= number of points
Briggs Henan University 2010
5
Uniform grid
--used for secondary
data
Random sampling
--useful in field work
Types of Quadrats
Frequency counts by
Quadrat would be:
Census Q = 64
Multiple ways to create quadrats
--and results can differ accordingly!
Quadrats don’t have to be square
--and their size has a big influence
Number
of points
in
Quadrat
0
1
2
3
Count
51
11
2
0
64
Q = # of quadarts
P = # of points =
Sampling Q = 38
Proportion Count
0.797
29
0.172
8
0.031
1
0.000
0
Proportion
0.763
0.211
0.026
0.000
15
6
Briggs Henan University 2010
Quadrat Analysis: Variance/Mean Ratio (VMR)
• Apply uniform or random grid over
area (A) with width of square given by:
Where:
A = area of region
n = # of points
• 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
7
3
5
2
1
3
1
0
1
3
1
2
2
2
2
2
UNIFORM/
DISPERSED
CLUSTERED
Quadrat
#
1
2
3
4
5
6
7
8
9
10
Number of
Points Per
Quadrat
3
1
5
0
2
1
1
3
3
1
20
Variance
Mean
Var/Mean
2.222
2.000
1.111
x^2
9
1
25
0
4
1
1
9
9
1
60
Number
of Points
Quadrat
Per
#
Quadrat
1
2
2
2
3
2
4
2
5
2
6
2
7
2
8
2
9
2
10
2
20
Variance
Mean
Var/Mean
random
2
(
X
i

X
)
i 1
N 1
0.000
2.000
0.000
uniform
Formulae for variance
n
0
0
10
0
0
x
x
RANDOM
2
2
2
2
2


n
i 1
Xi2 [(  X )2 / N ]
N 1
0
0
10
0
0
x
x^2
4
4
4
4
4
4
4
4
4
4
40
Number of
Quadrat Points Per
#
Quadrat
1
0
2
0
3
0
4
0
5
10
6
10
7
0
8
0
9
0
10
0
20
Variance
Mean
Var/Mean
x^2
0
0
0
0
100
100
0
0
0
0
200
17.778
2.000
8.889
Clustered
Note:
N = number of Quadrats = 10
Ratio = Variance/mean
Briggs Henan University 2010
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:
random
60-(202)/10
2
=
10
uniform
40-(202)/10
2
=
0
clustered
200-(202)/10
2
=
80
• 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
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
10
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%) = 1.36
where 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:
λ
p(0) = e- = 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 exampleBriggs
for Henan
cluster
case
University
2010
11
Calculation of Poisson Frequencies for Kolmogorov-Smirnov test
CLUSTERED pattern as used in lecture
A
B
C
D
E
F
G
=ColA * ColB=Col B / q
H
!Col E - Col G
Number of Observed
Cumulative
Cumulative Absolute
Points in Quadrat Total
Observed
Observed Poisson
Poisson
Difference
quadrat
Count
Point Probability
Probability Probability Probability
0
8
0
0.8000
0.8000
0.1353
0.1353
0.6647
1
0
0
0.0000
0.8000
0.2707
0.4060
0.3940
2
0
0
0.0000
0.8000
0.2707
0.6767
0.1233
3
0
0
0.0000
0.8000
0.1804
0.8571
0.0571
4
0
0
0.0000
0.8000
0.0902
0.9473
0.1473
5
0
0
0.0000
0.8000
0.0361
0.9834
0.1834
6
0
0
0.0000
0.8000
0.0120
0.9955
0.1955
7
0
0
0.0000
0.8000
0.0034
0.9989
0.1989
8
0
0
0.0000
0.8000
0.0009
0.9998
0.1998
9
0
0
0.0000
0.8000
0.0002
1.0000
0.2000
10
2
20
0.2000
1.0000
0.0000
1.0000
0.0000
The Kolmogorov-Smirnov D test statistic is the largest Absolute Difference
= largest value in Column h
Critical Value at 5% for one sample given by: 1.36/sqrt(Q)
Critical Value at 5% for two sample given by: 1.36*sqrt((Q1+Q2)/Q1*Q2))
number of quadrats
Q
number of points
P
number of points in a quadrat x
poisson probability
The spreadsheet spatstat.xls
contains worked examples for
the Uniform/ Clustered/
Random data previously used,
as well as for Lee and Wong’s
data
0.6647
0.4301 Significant
10 (sum of column B)
20 (sum of Col C)
p(x) = p(x-1)*(P/Q)/x (Col E, Row 11 onwards)
if x=0 then p(x) = p(0)=2.71828^P/Q
Euler's constant
Row 10
2.7183
12
(Col E, Row 10)
Briggs Henan University 2010
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
For example, quadrat analysis cannot distinguish
between these two, obviously different, patterns
• 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, overall pattern here is dispersed, but
there are some local clusters
Briggs Henan University 2010
13
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
NNI =
Expected Average Distance
For random pattern, NNI = 1
For clustered pattern, NNI = 0
For dispersed pattern, NNI = 2.149
See next slide
for formulae
for calculation
Briggs Henan University 2010
14
Calculating Nearest Neighbor Index
Where:
The average
distance to nearest
neighbor
Area of region:
result very
dependent on
this value
Briggs Henan University 2010
15
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.
• It tests if the observed pattern 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
16
Calculating Test Statistic for Nearest
Neighbor Index
Where:
(Standard error)

0.26136
n2 / A
Briggs Henan University 2010
17
RANDOM
Point
1
2
3
4
5
6
7
8
9
10
Nearest
Neighbor Distance
2
1
3
0.1
2
0.1
5
1
4
1
5
2
6
2.7
10
1
10
1
9
1
10.9
Meanrdistance
Area of
Region
Density
Expected
Mean
R
NNI
Z
1.09
50
0.2
1.118034
0.974926
= -0.1515
CLUSTERED
Nearest
Neighbor Distance
2
0.1
3
0.1
2
0.1
5
0.1
4
0.1
5
0.1
6
0.1
9
0.1
10
0.1
9
0.1
1
Point
1
2
3
4
5
6
7
8
9
10
Mean
r distance
Area of
Region
Density
Expected
Mean
RNNI
Z
0.1
50
0.2
1.118034
0.089443
= 5.508
UNIFORM
Point
1
2
3
4
5
6
7
8
9
10
Nearest
Neighbor Distance
3
2.2
4
2.2
4
2.2
5
2.2
7
2.2
7
2.2
8
2.2
9
2.2
10
2.2
9
2.2
22
Mean
r distance 2.2
Area of
Region
50
Density
0.2
Expected
Mean
1.118034
RNNI
1.96774
18
Z
= 5.855
Briggs Henan University 2010
Source: Lembro
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 19
results
Produced if “Display output
graphically “ box is 
Scroll up the window to see all
the results.
Note: Progress box continues
to run until graphic is closed.
Always close graphic window
20
first.
Briggs Henan University 2010
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
21
Briggs Henan University 2010
Ripley’s K(d) Function
• 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
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
Briggs Henan University 2010
22
Not this simple with real data!!!
Source: O’Sullivan & Unwin, p.
The high end (0.6)
corresponds to
distance between
the clusters
within
between
dispersed
The distance bands are placed
around every point.
Note the big problem of edge
effects from circles outside the
study area.
clustered
The low end (0.2)
corresponds to
distances within the
cluster
Begins flat
Briggs Henan University 2010
23
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
24
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
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
26
Briggs Henan University 2010
Distance bands:
start 10,000 feet
size: 20,000 feet
Also experiment with
different region (study
area) boundaries.
Results for 20,000 feet Bands
27
Briggs Henan University 2010
Y field = Diffk = ObservedK - ExpectedK
Plotting the
Difference Between
Observed and
Expected K, versus
Distance
X field = ExpectedK or HiConfEnv
Distance between clusters
70,000 feet = 13 miles
28
= 20 km
Briggs Henan University 2010
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
29
What have we learned?
How to measure and test if spatial
patterns are clustered or dispersed.
30
Briggs Henan University 2010
Why is this important?
?
We can measure and test
--not just look and guess!
Is it clustered?
That is science.
Briggs Henan University 2010
31
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
32
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
33
34
Briggs Henan University 2010