Spatial Analysis Part 1

Spatial Analysis• What is spatial analysis?

– It is the means by which we turn raw geographic data into useful information

– It does so by adding greater informative content and value

• Spatial analysis reveals patterns, trends, and anomalies that might otherwise be missed– It provides a check on human intuition– It allows for analysis of data that could

never be done by humans

Spatial Analysis

• Analysis is considered spatial if the results depend on the locations of the objects being analyzed.

– Thus if you move the objects, the results of spatial analyses will change.

– Spatial analyses generally requires both attributes and locations of objects.

Steps in Spatial Analysis

1. Frame the question we wish to ask.

2. Find appropriate data to answer the question.

3. Choose an analytical method appropriate to answer question.

4. Process the data using the chosen method.

5. Interpret the results of the analysis.

Spatial Relationships are at the

Core of Spatial Analysis• Most spatial analyses are based on topological

relationships:– How near is Feature A to Feature B– What features contain other features?– What features are adjacent to other features?– What features are connected to other features?

• From these topological building blocks, we can develop all sorts of spatial analysis approaches to answer many complex questions

Types of Spatial Analysis

• We will consider six categories of spatial analyses:

1. Queries (today)2. Measurements (today)3. Transformations

(today)4. Descriptive summaries (next lecture)5. Optimization (next lecture)6. Hypothesis testing (next lecture)

1. Queries

• Queries – Attribute based

• Example: show me all pixels in a raster image with BV > 80.

– Location based• List all the block groups that fall within Orange County

• A GIS can respond to queries by selecting the appropriate data in:– A map view– A table– Both

The Map View

• Queries can be performed through interaction with a GIS on-screen map– Identify objects– Query data objects based on specific criteria of attributes – Find coordinates of objects

The Table View

• Queries can be performed through interaction with a table– Attribute based queries can be performed in the table.– When objects are selected in a table, a GIS can automatically

highlight the selected data objects in the map view, and vice versa.

2. Measurements

• Measure:– Distance between two points

• Distances can be summed– Example: a truck makes multiple stops on a route. What

is the total distance traveled on the route?

• Other mathematical operations can be applied to distances:

– We can square a set of distances, add them up, divide by the amount of distances calculated in the set, and take the square root. What is an example of when this operation is used?

– Area of a polygon– Example: What is the area of a preserved forest tract?

Measurement of Length

• Types of length measurements– Euclidean Distance: straight-line distance

between two points on a flat plane (as the crow flies)– Manhattan Distance: limits movement to

orthogonal directions– Great Circle Distance: the shortest distance

between two points on the globe– Network Distance:

• Along roads • Along pipe network• Along electric grid• Along phone grid• By river channels

Euclidean Distance• Distances can be calculated between points, along lines, or in a

variety of fashions with areas• Euclidean Distance – is calculated in a Cartesian frame of

reference:

(x1 – x2)2 + (y1 – y2)2C=

C

P1 (x1,y1)

P2 (x2,y2)

• On what scales is this valid?

• Can we use this with latitude and longitude?

Manhattan Distance

• Manhattan Distance is useful in some urban environments with orthogonal road networks.

• Movement is limited to city streets:

P1 (x1,y1)

P2 (x2,y2)

dm = | x1 – x2 | + | y1 – y2 |

a reminder – the | symbols denote absolute value

Great Circle Distance

• The Great Circle distance is the shortest distance between two points on the globe.

• The two points must be specified using geographic coordinates (i.e., latitude & longitude positions).

Great Circle Distance

• Calculating the great circle distance is actually pretty complicated– A = Latitude of point A– B = Latitude of point B– C = Difference in longitude between the points (i.e.,

Longitude of point A – Longitude of point B)– D = angular distance

• Simple version (spherical distance)– cos(D) = sin(A) sin(B) + cos(A) cos(B) cos(C)

Great Circle Distance

• ONE more complicated version (also more accurate)

To use these equations:1. Convert latitude and longitude (degree, minute, second) to decimal degrees (if necessary)2. Convert degrees to into radians3. Solve the equation for D4. Great Circle Distance = D * the radius of the earth (6372.795 km)

OR

1. Convert latitude and longitude (degree, minute, second) to decimal degrees (if necessary)2. Convert degrees to into radians3. Solve the equation for D4. Convert D into degrees5. Great Circle Distance = D * length of 1 degree at the equator (111.32 km)

Starting from: Carrboro, NC 27510 Save AddressArriving at: Washington, DC Save Addressistance:272.5 miles Approximate Travel Time:5 hours 23 mins

Network DistanceYahoo maps:

Issues with Length Measurement

• The length of a true curve is longer than the length of its polyline or polygon representation:

Issues with Length Measurement• Length measurements in GIS are usually calculated in

2 dimensions. But changes in elevation increase distances.

X

Z

3. Transformations• Spatial transformations includes

many analytical approaches, applicable to:– Vector data– Raster data– Both

• Transformations can create new:– Attributes– Data objects

Buffering (Proximity Analysis)

• Buffering operations create new objects consisting of areas within a user-defined distance of existing objects.

• Examples of uses:– to determine areas impacted by a proposed

highway– to determine the service area of a proposed

hospital

• Buffering can be performed in both the vector and raster spatial data models

Buffering: The delineation of a zone around the feature of interest within a given distance. For a point feature, it is simply a circle with its radius equal to the buffer distance:

Buffering (Proximity Analysis)

Variable Distance Buffering

• The buffer zone constructed around each feature can be based on a variable distance according to some feature attribute(s)

• Suppose we have a point pollution source, such as a power plant. We want to zone residential areas some distance away from each plant, based on the amount of pollution that power plant produces

– For smaller power plants, the distance might be shorter.

– For larger power plants that generate a lot of pollutant, we choose longer distances

•Buffering higher order objects involves moving a circle of specified radius along the line (or the lines forming polygon)

Buffering Points, Lines, and Polygons

Buffer lines

Buffer polygons

Line and Polygon Buffer Examples

Raster Buffering• Buffering operations also can be performed using the raster data

model• In the raster model, we can perform a simple distance buffer, or in

this case, a distance buffered according to values in a friction layer (e.g. travel time for a bear through different landcover):

lake

Areas reachable in 5 minutesAreas reachable in 10 minutesOther areas

Feature in Feature Transformations

• These transformations determine whether a feature lies inside or outside of another feature

• The most basic of these transformations is point in polygon analysis, which can be applied in various situations:

The Point in Polygon Algorithm

• Draw a line from the point to infinity in any direction, and then count the number of intersections between this line and each polygon’s boundary

• The polygon with an odd number of intersections is the containing polygon; all other polygons have an even number of intersections

How do GIS programs calculate this?

Point in Polygon Algorithm

• For the point to be inside the polygon, there must be an odd number of intersections on either side of the point .

Habitat Area(km2) Frequency Density . A 150 4 0.027 nests/km2

B 320 6 0.019 nests/km2

C 350 3 0.009 nests/km2

D 180 3 0.017 nests/km2

Bird’s NestsA B

DC

Habitat TypesA B

DC

Analysis Results

Point Frequency/Density Analysis

• We can use point in polygon results to calculate frequencies or densities of points per area

• For example, given a point layer of bird’s nests and polygon layer of habitats, we can calculate densities:

• Overlay line layer (A) with polygon layer (B)– In which B polygons are A lines located?– Assign polygon attributes from B to lines in

AA BExample

: Assign land use attributes (polygons) to streams (lines):

Line in Polygon Analysis

David Tenenbaum – GEOG 070 – UNC-CH Spring 2005

Polygon Overlay, Discrete Object Case

•In this example, two polygons are intersected to form nine new polygons.

•One is formed from both input polygons (1);

•four are formed by Polygon A and not Polygon B (2-5);

•four are formed by Polygon B and not Polygon A (6-9)

A B

1

2

3

4

5

6 7

8

9

•Boolean operations of OR & AND correspond to UNION & INTERSECTION, used in vector-based analyses

UNION

INTERSECTION

A BOR

A BAND

Boolean Operations

•We can apply these concepts in the raster spatial data model, when two input layers contain true/false or 1/0 data:

Polygon Combination

Common ways to combine polygons:

1. Show all new polygons as in diagram.2. UNION (Boolean OR)3. INTERSECTION (Boolean AND)

A B

1

2

3

4

5

6 7

8

9

0 1 1

0 0 1

1 0 1

0 0 0

1 1 1

0 0 1

AND =

Boolean Operations with Raster Layers

0 1 1

0 0 1

1 0 1

0 0 0

1 1 1

0 0 1

OR =

•The AND operation requires that the value of cells in both input layers be equal to 1 for the output to have a value of 1:

•The OR operation requires that the value of a cells in either input layer be equal to 1 for the output to have a value of 1:

Problems with Vector Overlay Analysis (esp. Polygon)

• There is a tradeoff between the complexity and interpretability of results

• Complex input layers with many polygons can result in many more polygon combinations… can we make sense of all those combinations?

Problems with Vector Overlay Analysis (esp. Polygon)

• Overlay analysis using the vector spatial data model is highly computationally intensive

• Complicated input layers can tax even current processors

• For example– With 2 partially overlapping polygons (U & V) we can

have 8 options: • Neither• U• V• U AND V• U OR V• U NOT V• V NOT U• V XOR U

Problems with Vector Overlay Analysis (esp. Polygon)

• For example– With 3 (X, Y, & Z) partially overlapping polygons we can have many

more options– Any guesses for how many?– There are 121 possible combinations!– Just think of what this means for a dataset with even a few hundred

polygons… this is what we mean by “computationally intensive”– The graphic shows just a few simple ones

• There are often spatial mismatches between input layers– Overlay can result in spurious sliver

polygons– We can “filter” out spurious slivers by

querying to select all polygons with AREA less than some minimum threshold

• It is difficult to choose a threshold to avoid deleting ‘real’ polygons

Problems with Vector Overlay Analysis (esp. Polygon)

Algebraic Operations w/ Raster Layers

• We can extend this concept from Boolean logic to algebra

• Map algebra:– Each cell is a number– Mathematical operations are using the raster layers as input– Calculations are done on a cell-by-cell basis– The result for each cell is placed in a new raster layer.

• Suitability analysis example: – Multiple raster input layers determine suitable sites:

• The cell values (attributes) in each raster layer represent ‘scores’.• Raster layers are weighted based on their importance.• Output scores are the sum of the input raster layers.

101

100

110

100

111

000

+ =201

211

110

Summation

101

100

110

100

111

000

=100

100

000

Multiplication

101

100

110

100

111

000

+ =301

322

110

100

111

000

+

Summation of more than two layers

Simple Arithmetic Operations

Near the mall Near friend’s houseNear work Good place to live?

Raster Difference (Subtraction)

• One application for subtraction between layers is a simple image change detection:

– Example: Imagine you have 2 images of the same forest, 10 years apart.

– Question: How can the locations where substantial changes have occurred (i.e., logging or regrowth) be identified using the two images?

– Answer: To measure forest growth or logging, you can take the difference in reflected Near InfraRed (NIR) light between image dates.

• More infrared light = more chlorophyll and more vegetation

517

656

345

723

541

653

- =-2-14

115

-3-32

The difference between two layers:

=

Raster Division

• Questions: – Can we perform the following operation?

– Are there any circumstances where we cannot perform this operation?

235

123

421

102

115

001

+ =100

111

000

*a b c

More Complex Operations

• As with other algebra, where you put the parentheses makes a difference

Applying a Model to Our Data• Map algebra can also be applied in the context of computing a

statistical linear regression

• What you probably learned in geometry: Y = a + bX, – X is the explanatory (independent) variable– Y is the dependent variable. – The slope of the line is b and a is the intercept (the value of y when x = 0)

• Linear regression uses the same basic idea: – Y = B0 + B1X1 + B2X2 + random error– The B values are the coefficients– The X values are the independent variables (2 in this case)

• For example maybe we sampled some forested field sites and generated the following equation: LAI = B0 + B1NDVI + B2TMI

• Using the coefficients we derived (the B values) we can apply the equation to our entire study area

Making Inferences from Samples

• Imagine that you have point location data, but you want data for your whole site/region.– For example:

• Air temperature maps created from point data.• Water pollution levels measured at points.

• Use models to predict values between sampling points

• Extrapolation: Predicting missing values using existing values that exist only on one side of the point in question

• Interpolation: Predicting unknown values using known values occurring at locations around the unknown value

Spatial Interpolation:Inverse Distance Weighting (IDW)

• The unknown value at a point is estimated by taking a weighted average of known values

– Those known points closer to the unknown point have higher weights.

– Those known points farther from the unknown point have lower weights.

point iknown value zidistance di

weight wi

unknown value (to be interpolated) atlocation x

i

ii

iix wzwz

21 ii dw

The estimate of the unknown value is a weighted average

Sample weighting function

Spatial Interpolation:Inverse Distance Weighting (IDW)

Issues with IDW• Weighted average estimates are always between the

min and max known values.– If the known (sampled) points did not include the

minima and maxima (e.g., mountain peaks and valleys), your data will be less extreme than reality

– It is thus important to position sample points to include the extremes whenever possible

Issues with IDW

The dashed line is a hill.

The x’s are sampled elevation points.

The black line is the interpolated (estimated) hill elevation.

With IDW, the unknown points tend towards the overall mean.

Triangle Irregular Networks (TIN)

Most often used for elevation surfaces.

Points are known data values (of elevation).

Lines are drawn between nearby points to create a set of irregular triangles.

The value of the variable (e.g. elevation) moves along the lines evenly from one point to the next.

The triangle between 3 lines represents a flat surface with slope and aspect.

With the TIN, the feature’s value every point on the TIN is estimated.