Raster Analysis
- Raster data
- Raster algebra and statistics
- Conditional and reclassifications
- Exercise A
- Applications in Remote Sensing
- Exercise B
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Raster Data
Recall:
Cells (or pixels) have dimension (height and width), which corresponds to the area on the ground that each cell covers. This is referred to as the raster’s resolution.
Smaller cell size = higher resolution = greater detail
(note: high resolution can lead to long process time because there are a lot of pixels to process)
So, we see generalization of features and loss of detail as the pixel size increases.
That’s because you can only fit one value per pixel.
Recall what you learned about data scale and map scale (same idea).
Rasters are grids of numbers
and we can add, subtract, multiply, and divide numbers.
Raster Analysis
Calculations of cell values from
one or more raster layers
Let’s start with a simple raster with three rows and three columns, given by the following:
\[ \begin{array}{|c|c|c|} \hline 1 & 2 & 5 \\ \hline 6 & 3 & 9 \\ \hline 4 & 6 & 2 \\ \hline \end{array} \]
Basic scalar multiplication
\[ 2 \times \begin{array}{|c|c|c|} \hline 1 & 2 & 5 \\ \hline 6 & 3 & 9 \\ \hline 4 & 6 & 2 \\ \hline \end{array} = \begin{array}{|c|c|c|} \hline 2 & 4 & 10 \\ \hline 12 & 6 & 18 \\ \hline 8 & 12 & 4 \\ \hline \end{array} \]
Each pixel multiplied by one number.
Calculations from two or more raster layers
are performed
pixel by pixel
\[ \begin{array}{|c|c|c|} \hline 1 & 2 & 5 \\ \hline 6 & 3 & 9 \\ \hline 4 & 6 & 2 \\ \hline \end{array} + \begin{array}{|c|c|c|} \hline 3 & 5 & 14 \\ \hline 8 & 6 & 5 \\ \hline 9 & 0 & 2 \\ \hline \end{array} = \begin{array}{|c|c|c|} \hline 4 & 7 & 19 \\ \hline 14 & 9 & 14 \\ \hline 13 & 6 & 4 \\ \hline \end{array} \]
Notice that pixel location is important, especially if the raster layers represent the same area.
These operations can be combined into algorithms
(i.e.,
raster algebra)
\[ 5 \times \begin{array}{|c|c|c|} \hline 1 & 2 & 5 \\ \hline 6 & 3 & 9 \\ \hline 4 & 6 & 2 \\ \hline \end{array} - \begin{array}{|c|c|c|} \hline 3 & 5 & 14 \\ \hline 8 & 6 & 5 \\ \hline 9 & 0 & 2 \\ \hline \end{array} = \begin{array}{|c|c|c|} \hline 2 & 5 & 11 \\ \hline 22 & 9 & 40 \\ \hline 11 & 30 & 8 \\ \hline \end{array} \]
Remember your order of operations from algebra class!
We can even perform statistics!
\[ \begin{array}{|c|c|c|} \hline 3 & 5 & 14 \\ \hline 8 & 6 & 5 \\ \hline 9 & 0 & 2 \\ \hline \end{array} \]
- Maximum: 14
- Minimum: 0
- Average: (3+5+14+8+6+5+9+0+2)/9 = 5.78
We don’t always have to perform analysis
over an entire raster
layer.
Instead, we can perform focal analyses
(sometimes called kernel
or neighborhood statistics).
\[ \begin{array}{|c|c|c|c|c|c|} \hline 4 & 9 & 13 & 6 & 11 & 7 \\ \hline 7 & 2 & 1 & 2 & 8 & 8 \\ \hline 9 & 12 & 6 & 6 & 3 & 9 \\ \hline 2 & 3 & 1 & 7 & 1 & 10 \\ \hline 15 & 13 & 6 & 14 & 9 & 10 \\ \hline 9 & 7 & 20 & 15 & 17 & 9 \\ \hline \end{array} \]
Let’s say we have a raster with six rows and six columns.
Let’s see if we can find focal minimum values
in a \(3 \times 3\) moving window.
\[ \begin{array}{|c|c|c|c|c|c|} \hline 4 & 9 & 13 & 6 & 11 & 7 \\ \hline 7 & 2 & 1 & 2 & 8 & 8 \\ \hline 9 & 12 & 6 & 6 & 3 & 9 \\ \hline 2 & 3 & 1 & 7 & 1 & 10 \\ \hline 15 & 13 & 6 & 14 & 9 & 10 \\ \hline 9 & 7 & 20 & 15 & 17 & 9 \\ \hline \end{array} \]
\[ \begin{array}{|c|c|c|c|c|c|} \hline {\color{blue}4} & 9 & 13 & 6 & 11 & 7 \\ \hline 7 & 2 & 1 & 2 & 8 & 8 \\ \hline 9 & 12 & 6 & 6 & 3 & 9 \\ \hline 2 & 3 & 1 & 7 & 1 & 10 \\ \hline 15 & 13 & 6 & 14 & 9 & 10 \\ \hline 9 & 7 & 20 & 15 & 17 & 9 \\ \hline \end{array} \]
To begin, we look at the first (top-left) value, which is 4.
Then we look at the eight values that surround it.
note
that along the top and left edges there are no values, so we just ignore
them
\[ \begin{array}{|c|c|c|c|c|c|} \hline {\color{blue}4} & {\color{red}9} & 13 & 6 & 11 & 7 \\ \hline {\color{red}7} & {\color{red}2} & 1 & 2 & 8 & 8 \\ \hline 9 & 12 & 6 & 6 & 3 & 9 \\ \hline 2 & 3 & 1 & 7 & 1 & 10 \\ \hline 15 & 13 & 6 & 14 & 9 & 10 \\ \hline 9 & 7 & 20 & 15 & 17 & 9 \\ \hline \end{array} \]
\[ \begin{array}{|c|c|c|c|c|c|} \hline {\color{blue}4} & {\color{red}9} & 13 & 6 & 11 & 7 \\ \hline {\color{red}7} & {\color{red}2} & 1 & 2 & 8 & 8 \\ \hline 9 & 12 & 6 & 6 & 3 & 9 \\ \hline 2 & 3 & 1 & 7 & 1 & 10 \\ \hline 15 & 13 & 6 & 14 & 9 & 10 \\ \hline 9 & 7 & 20 & 15 & 17 & 9 \\ \hline \end{array} \]
Then we check these nine values (four in this edge case) and calculate their minimum (it’s 2), which will replace the 4 in our output.
Let’s skip down to the 2
\[ \begin{array}{|c|c|c|c|c|c|} \hline 4 & 9 & 13 & 6 & 11 & 7 \\ \hline 7 & {\color{blue}2} & 1 & 2 & 8 & 8 \\ \hline 9 & 12 & 6 & 6 & 3 & 9 \\ \hline 2 & 3 & 1 & 7 & 1 & 10 \\ \hline 15 & 13 & 6 & 14 & 9 & 10 \\ \hline 9 & 7 & 20 & 15 & 17 & 9 \\ \hline \end{array} \]
\[ \begin{array}{|c|c|c|c|c|c|} \hline {\color{red}4} & {\color{red}9} & {\color{red}{13}} & 6 & 11 & 7 \\ \hline {\color{red}7} & {\color{blue}2} & {\color{red}1} & 2 & 8 & 8 \\ \hline {\color{red}9} & {\color{red}{12}} & {\color{red}6} & 6 & 3 & 9 \\ \hline 2 & 3 & 1 & 7 & 1 & 10 \\ \hline 15 & 13 & 6 & 14 & 9 & 10 \\ \hline 9 & 7 & 20 & 15 & 17 & 9 \\ \hline \end{array} \]
and look at the eight surrounding values
and we see the minimum of these 9 values is 1
1 will
replace the 2 in our output
\[ \begin{array}{|c|c|c|c|c|c|} \hline {\color{red}4} & {\color{red}9} & {\color{red}{13}} & 6 & 11 & 7 \\ \hline {\color{red}7} & {\color{blue}2} & {\color{red}1} & 2 & 8 & 8 \\ \hline {\color{red}9} & {\color{red}{12}} & {\color{red}6} & 6 & 3 & 9 \\ \hline 2 & 3 & 1 & 7 & 1 & 10 \\ \hline 15 & 13 & 6 & 14 & 9 & 10 \\ \hline 9 & 7 & 20 & 15 & 17 & 9 \\ \hline \end{array} \]
Repeat this process for each cell.
The minimum around 1 is 1.
\[ \begin{array}{|c|c|c|c|c|c|} \hline 4 & {\color{red}9} & {\color{red}{13}} & {\color{red}6} & 11 & 7 \\ \hline 7 & {\color{red}2} & {\color{blue}1} & {\color{red}2} & 8 & 8 \\ \hline 9 & {\color{red}{12}} & {\color{red}6} & {\color{red}6} & 3 & 9 \\ \hline 2 & 3 & 1 & 7 & 1 & 10 \\ \hline 15 & 13 & 6 & 14 & 9 & 10 \\ \hline 9 & 7 & 20 & 15 & 17 & 9 \\ \hline \end{array} \]
The minimum around 2 is 1.
\[ \begin{array}{|c|c|c|c|c|c|} \hline 4 & 9 & {\color{red}{13}} & {\color{red}6} & {\color{red}{11}} & 7 \\ \hline 7 & 2 & {\color{red}1} & {\color{blue}2} & {\color{red}8} & 8 \\ \hline 9 & 12 & {\color{red}6} & {\color{red}6} & {\color{red}3} & 9 \\ \hline 2 & 3 & 1 & 7 & 1 & 10 \\ \hline 15 & 13 & 6 & 14 & 9 & 10 \\ \hline 9 & 7 & 20 & 15 & 17 & 9 \\ \hline \end{array} \]
The minimum around 8 is 2.
\[ \begin{array}{|c|c|c|c|c|c|} \hline 4 & 9 & 13 & {\color{red}6} & {\color{red}{11}} & {\color{red}7} \\ \hline 7 & 2 & 1 & {\color{red}2} & {\color{blue}8} & {\color{red}8} \\ \hline 9 & 12 & 6 & {\color{red}6} & {\color{red}3} & {\color{red}9} \\ \hline 2 & 3 & 1 & 7 & 1 & 10 \\ \hline 15 & 13 & 6 & 14 & 9 & 10 \\ \hline 9 & 7 & 20 & 15 & 17 & 9 \\ \hline \end{array} \]
and so we go until we find
the focal minimum for each pixel
writing down the new values as we go
until we’ve searched the \(3 \times
3\) windows
around all our original values
and get a result like this
\[ \begin{array}{|c|c|c|c|c|c|} \hline 2 & 1 & 1 & 1 & 2 & 7 \\ \hline 2 & 1 & 1 & 1 & 2 & 3 \\ \hline 2 & 1 & 1 & 1 & 1 & 1 \\ \hline 2 & 1 & 1 & 1 & 1 & 1 \\ \hline 2 & 1 & 1 & 1 & 1 & 1 \\ \hline 7 & 6 & 6 & 6 & 9 & 9 \\ \hline \end{array} \]
The neighborhood analysis is important for understanding local
influences
(e.g., crime statistics or pollution
levels)
The window size and shape may be adjusted as well as the statistics type (e.g., average, maximum, sum) to accommodate a wide variety of analyses.
For more on focal statistics, see here
Another important analysis that we can perform is the conditional (i.e., testing whether our pixel values meet certain conditions).
The output of conditional analysis are boolean rasters (i.e, a raster with pixel values that are either true or false or, because it has to be numeric, 1 or 0, respectively).
Example conditional statement:
“Where pixel value is greater
than 5”
\[ \mathrm{Input} = \begin{array}{|c|c|c|} \hline 3 & 5 & 14 \\ \hline 8 & 6 & 5 \\ \hline 9 & 0 & 2 \\ \hline \end{array} \]
\[ \mathrm{Output} = \begin{array}{|c|c|c|} \hline 0 & 0 & 1 \\ \hline 1 & 1 & 0 \\ \hline 1 & 0 & 0 \\ \hline \end{array} \]
Notice that false is equal to zero and true is equal to one.
This makes it easy to mix conditionals with raster algebra because multiplying by zero always returns zero and multiplying by one always returns the same value.
| 0 | False |
| 1 | True |
note that, depending on your GIS, true and false
values
may be assigned any value of your choice
This can be represented in a VAT (value attribute table)
an attribute table that associates raster values to a class,
group, category, or membership
Example conditional used with raster algebra.
\[ A = \begin{array}{|c|c|c|} \hline 12 & 5 & 14 \\ \hline 8 & 16 & 5 \\ \hline 9 & 10 & 3 \\ \hline \end{array}, B = \begin{array}{|c|c|c|} \hline 1 & 2 & 5 \\ \hline 6 & 3 & 9 \\ \hline 4 & 6 & 2 \\ \hline \end{array} \]
Raster \(B\) where raster \(A\) is less than 10.
\[ \mathrm{Output} = \begin{array}{|c|c|c|} \hline 0 & 2 & 0 \\ \hline 6 & 0 & 9 \\ \hline 4 & 0 & 2 \\ \hline \end{array} \]
(note: this example assumes 0 represents no
data)
You can also change pixel values based on conditional statements in what is called reclassification.
Reclassifications typically have an input raster and a reclassification table that describes conditional statements along with their associated new pixel values.
| Condition* | New Value |
|---|---|
| \(p < 5\) | 1 |
| \(5 \le p < 10\) | 2 |
| \(p \ge 10\) | 3 |
*where \(p\) is pixel value
\[ \mathrm{Input} = \begin{array}{|c|c|c|} \hline 12 & 5 & 14 \\ \hline 8 & 16 & 5 \\ \hline 9 & 10 & 3 \\ \hline \end{array}, \mathrm{Output} = \begin{array}{|c|c|c|} \hline 3 & 2 & 3 \\ \hline 2 & 3 & 2 \\ \hline 2 & 3 & 1 \\ \hline \end{array} \]
Please follow along with the video
demonstration.
Note the correction at 11:36 where the maximum
should be 11.
Exercise A
For this assignment, you’ll want to grab a pen and paper.
Here are your raster sets:
\[ A = \begin{array}{|c|c|c|c|} \hline 16&18&24&28\\ \hline 16&20&26&30\\ \hline 14&18&20&26\\ \hline 14&16&18&22\\ \hline 12&14&14&18\\ \hline \end{array},\;B = \begin{array}{|c|c|c|c|} \hline 30&33&30&27\\ \hline 27&27&24&21\\ \hline 27&24&21&18\\ \hline 24&24&21&15\\ \hline 18&15&15&12\\ \hline \end{array} \]
Perform the following tasks in the given order:
- Multiply A by one half (i.e., 1/2 \(\times\) A)
- Multiply B by one third (i.e., 1/3 \(\times\) B)
- Add the two results together (i.e., 1/2 \(\times\) A + 1/3 \(\times\) B)
- Subtract the result from B (i.e., B - 1/2 \(\times\) A - 1/3 \(\times\) B)
- Apply the following reclassification on the result
| Condition* | New Value |
|---|---|
| \(p < 0\) | 4 |
| \(0 \le p < 4\) | 3 |
| \(4 \le p < 8\) | 2 |
| \(8 \le p < 12\) | 1 |
| \(p \ge 12\) | 0 |
*where \(p\) is pixel value
- Apply the following conditional to the result, setting true values equal to 1 and false values equal to 0.
“Where the result is less than 2.”
- What is the sum of all the resulting pixel values?
- Perform a 3 by 3 focal minimum on A
- Perform a 3 by 3 focal maximum on B
- Subtract what you get in step (8) from what you got in step (9)
- What is the average pixel value of the results you got from step (10)?
Applications
There are numerous applications where raster analysis, raster algebra, and neighborhood statistics are useful. One such application is in the field of remote sensing or the science of obtaining information about objects or areas from a distance, typically from aircraft or satellites.
Please watch this brief introduction to the
field of remote sensing.
Please find the deliverables worksheet for
Lab 9 Exercises A and
B on Blackboard