13 Geometric operations with vector layers
13.1 Aims
Our aims in this chapter are:
- Learn how to join data to a vector layer, by attributes or by location 
- Learn to make geometric calculations with vector layers 
We will use the following R packages:
- sf
- units
13.2 Join by location
For most of the examples in this Chapter, we are going to use the US counties and three airports layers (Section 7.7):
library(sf) airports = st_read("airports.geojson") county = st_read("USA_2_GADM_fips.shp")transformed to the US National Atlas projection (Section 7.9.2):
airports = st_transform(airports, 2163) county = st_transform(county, 2163)Join by spatial location, or spatial join, is one of the most common operations in spatial analysis. In a spatial join we are "attaching" attributes from one layer to another, based on their spatial relations. In ArcGIS this is done using "Join data from another layer based on spatial location" Figures 8.1–8.2.
Figure 8.1: Join by location in ArcGIS (Step 1)
Figure 8.2: Join by location in ArcGIS (Step 2)
For simplicity, we will create a subset of the counties of New Mexico only, since all three airports are located in that state:
nm = county[county$NAME_1 == "New Mexico", ]Figure 8.3 shows the geometry of the nm and airports layers. Note that the pch parameter determines point shape (see ?points), while cex determines point size (Section 7.8):
plot(st_geometry(nm)) plot(st_geometry(airports), col = "red", pch = 16, cex = 2, add = TRUE)Figure 8.3: The nm and airports layers
Given two layers x and y, a function call of the form st_join(x, y) returns the x layer along with matching attributes from y. The default join type is by intersection, meaning that two geometries are considered matching if they intersect. Other join types can be specified using the join parameter (see ?st_join).
For example, the following expression joins the airports layer with the matching nm attributes. The resulting layer contains all three airports with their original geometry and name attribute, plus the NAME_1, NAME_2, TYPE_2 and FIPS of the nm feature that each airport intersects with:
st_join(airports, nm) ## Simple feature collection with 3 features and 5 fields ## Geometry type: POINT ## Dimension:     XY ## Bounding box:  xmin: -619184.5 ymin: -1081605 xmax: -551689.8 ymax: -1022366 ## Projected CRS: US National Atlas Equal Area ##                        name     NAME_1     NAME_2 TYPE_2  FIPS ## 1 Albuquerque International New Mexico Bernalillo County 35001 ## 2           Double Eagle II New Mexico Bernalillo County 35001 ## 3        Santa Fe Municipal New Mexico   Santa Fe County 35049 ##                     geometry ## 1 POINT (-603868.2 -1081605) ## 2 POINT (-619184.5 -1068431) ## 3 POINT (-551689.8 -1022366)What do you think will happen if we reverse the order of the arguments, as in
st_join(nm, airports)? How many features is the result going to have, and why?
13.3 Subsetting by location
In Section 7.6 we learned how to subset sf vector layers based on their attributes, just like an ordinary data.frame (Section 4.1.5). We can also subset an sf layer based on its intersection with another layer. Subsetting by location uses the same [ operator, only using a layer as the "index." Namely, an expression of the form x[y, ]—where x and y are vector layers—returns a subset of x features that intersect with y.
For example, the following expression returns only those nm counties that intersect with the airports layer:
nm1 = nm[airports, ]The subset is shown in light blue in Figure 8.4:
plot(st_geometry(nm)) plot(st_geometry(nm1), col = "lightblue", add = TRUE) plot(st_geometry(airports), col = "red", pch = 16, cex = 2, add = TRUE)Figure 8.4: Subset of the nm layer based on intersection with airports
What will be the result of
airports[nm, ]?
What will be the result of
nm[nm[20, ], ]?
13.4 Geometric calculations
13.4.1 Introduction
Geometric operations on vector layers can conceptually be divided into three groups, according to their output:
- Numeric values (Section 8.3.2)—Functions that summarize geometrical properties of: 
- Logical values (Section 8.3.3)—Functions that evaluate whether a certain condition holds true, regarding: - A single layer—e.g., geometry is valid 
- A pair of layers—e.g., feature A intersects feature B (Section 8.3.3) 
 
- Spatial layers (Section 8.3.4)—Functions that create a new layer based on: 
The following three sections elaborate and give examples of numeric (Section 8.3.2), logical (Section 8.3.3) and spatial (Section 8.3.4) geometric calculations on vector layers.
13.4.2 Numeric geometric calculations
13.4.2.1 Introduction
There are several functions to calculate numeric geometric properties of vector layers in package sf:
- st_bbox—Section 7.5.2
- st_area—Section 8.3.2.2
- st_distance—Section 8.3.2.3
- st_length—Section 11.4.6
- st_dimension
Functions st_bbox, st_area, st_length and st_dimension are operation applicable for individual geometries. The st_distance function is applicable to pairs of geometries. The next two sections give an example from each category: st_area for individual geometries (Section 8.3.2.2) and st_distance for pairs of geometries (Section 8.3.2.3).
13.4.2.2 Area
The st_area function returns the area of each feature or geometry. For example, we can use st_area to calculate the area of each feature in the county layer (i.e., each county):
county$area = st_area(county) county$area[1:6] ## Units: [m^2] ## [1]  2451875694  1941109814  1077788608  1350475635 16416571545  2626706989The result is an object of class units, which is basically a numeric vector associated with the units of measurement:
class(county$area) ## [1] "units"CRS units (e.g., meters) are used by default in length, area and distance calculations. For example, the units of measurement in the US National Atlas CRS are meters, which is why the area values are returned in m2�2 units.
We can convert units objects to different units, with set_units from package units29. Note that the units package needs to be loaded to do that. For example, here is how we can convert the area column from m2�2 to km2��2:
library(units) county$area = set_units(county$area, "km^2") county$area[1:6] ## Units: [km^2] ## [1]  2451.876  1941.110  1077.789  1350.476 16416.572  2626.707Dealing with units, rather than numeric values has several advantages:
- We don't need to remember the coefficients for conversion between one unit to another, avoiding conversion mistakes. 
- We don't need to remember, or somehow encode, the units of measurement (e.g., rename the area column to - area_m2) since they are already binded with the values.
- We can keep track of the units when making numeric calculations. For example, if we divide a numeric value by km2��2, the result is going to be associated with units of density, km−2��−2 (Section 8.5). 
Nevertheless, if necessary, we can always convert units to an ordinary numeric vector with as.numeric:
as.numeric(county$area[1:6]) ## [1]  2451.876  1941.110  1077.789  1350.476 16416.572  2626.707The calculated county area attribute values are visualized in Figure 8.5:
plot(county[, "area"])Figure 8.5: Calculated area attribute
13.4.2.3 Distance
An example of a numeric operator on a pair of geometries is geographical distance. Distances can be calculated using function st_distance. For example, the following expression calculates the distance between each feature in airports and each feature in nm:
d = st_distance(airports, nm)The result is a matrix of units values:
d[, 1:6] ## Units: [m] ##          [,1]     [,2]      [,3]      [,4]      [,5]      [,6] ## [1,] 266778.9 140470.5 103328.22 140913.62 175353.34 121934.94 ## [2,] 275925.0 120879.9  93753.25 141511.80 177624.40 125327.45 ## [3,] 197540.4 145580.9  34956.53  62231.01  96549.17  43630.68In the distance matrix resulting from st_distance(x,y):
- Rows refer to features of - x, e.g.,- airports
- Columns refer to features of - y, e.g.,- nm
Therefore, the matrix dimensions are equal to c(nrow(x), nrow(y)):
c(nrow(airports), nrow(nm)) ## [1]  3 33dim(d) ## [1]  3 33Just like areas (Section 8.3.2.2), distances can be converted to different units with set_units. For example, the following expression converts the distance matrix d from m� to km��:
d = set_units(d, "km") d[, 1:6] ## Units: [km] ##          [,1]     [,2]      [,3]      [,4]      [,5]      [,6] ## [1,] 266.7789 140.4705 103.32822 140.91362 175.35334 121.93494 ## [2,] 275.9250 120.8799  93.75325 141.51180 177.62440 125.32745 ## [3,] 197.5404 145.5809  34.95653  62.23101  96.54917  43.63068To work with the distance matrix, it can be convenient to set row and column names. We can use any of the unique attributes of x and y, such as airport names and county names:
rownames(d) = airports$name colnames(d) = nm$NAME_2 d[, 1:6] ## Units: [km] ##                              Union San Juan Rio Arriba      Taos    Colfax ## Albuquerque International 266.7789 140.4705  103.32822 140.91362 175.35334 ## Double Eagle II           275.9250 120.8799   93.75325 141.51180 177.62440 ## Santa Fe Municipal        197.5404 145.5809   34.95653  62.23101  96.54917 ##                                Mora ## Albuquerque International 121.93494 ## Double Eagle II           125.32745 ## Santa Fe Municipal         43.63068Once row and column names are set, it is more convenient to find out the distance between a specific airport and a specific county:
d["Santa Fe Municipal", "Santa Fe", drop = FALSE] ## Units: [km] ##                    Santa Fe ## Santa Fe Municipal        013.4.3 Logical geometric calculations
Given two sets of geometries, or layers, x and y, the following logical geometric functions check whether each of the geometries in x maintains the specified relation with each of the geometries in y:
- st_contains_properly
- st_contains
- st_covered_by
- st_covers
- st_crosses
- st_disjoint
- st_equals_exact
- st_equals
- st_intersects
- st_is_within_distance
- st_overlaps
- st_touches
- st_within
In this book we mostly use the first function in the above list, st_intersects. The st_intersects function detects intersection, which is the simplest and most useful type of relation. Two geometries intersect when they share at least one point in common. However, it is important to be aware of the many other relations that can be evaluated using the other 12 functions.
By default, the above functions return a list, which takes less memory but is less convenient to work with. When specifying sparse=FALSE the functions return a logical matrix which is more convenient to work with. Each element i,j in the matrix is TRUE when f(x[i], y[j]) is TRUE. For example, the following expression creates a matrix of intersection relations between counties in nm:
int = st_intersects(nm, nm, sparse = FALSE) int[1:4, 1:4] ##       [,1]  [,2]  [,3]  [,4] ## [1,]  TRUE FALSE FALSE FALSE ## [2,] FALSE  TRUE  TRUE FALSE ## [3,] FALSE  TRUE  TRUE  TRUE ## [4,] FALSE FALSE  TRUE  TRUEFigure 8.6 shows two matrices of spatial relations between counties in New Mexico: st_intersects (left) and st_touches (right).
Figure 8.6: Spatial relations between counties in New Mexico, st_intersects (left) and st_touches (right)
How can we calculate
airportscount per county innm, usingst_intersects(Figure 8.7)?
Figure 8.7: Airport count per county in New Mexico
13.4.4 Spatial geometric calculations
13.4.4.1 Introduction
The sf package provides common geometry-generating functions applicable to individual geometries, such as:
- st_centroid—Section 8.3.4.2
- st_buffer—Section 8.3.4.5
- st_convex_hull—Section 8.3.4.7
- st_voronoi
- st_sample
What each function does is illustrated in Figure 8.8.
Figure 8.8: Geometry-generating operations on individual layers. The input geometries are shown in red, the output geometries are shown in black.
Additionally, the st_cast function is used to transform geometries from one type to another (Figure 7.6), which can also be considred as geometry generation (Section 8.3.4.4). The st_union and st_combine functions are used to combine multiple geometries into one, possibly dissolving their borders (Section 8.3.4.3).
Finally, geometry-generating functions which operate on pairs of geometries (Section 8.3.4.6) include:
- st_intersection
- st_difference
- st_sym_difference
- st_union
We elaborate on these functions in the following Sections 8.3.4.2–8.3.4.7.
13.4.4.2 Centroids
A centroid is the geometric center of a geometry, i.e., its center of mass, represented as a point. For example, the following expression uses st_centroid to create the centroid of each polygon in county:
ctr = st_centroid(nm)The nm layer along with the calculated centroids is shown in Figure 8.9:
plot(st_geometry(nm), border = "grey") plot(st_geometry(ctr), col = "red", pch = 3, add = TRUE)Figure 8.9: Centroids of New Mexico counties
13.4.4.3 Combining and dissolving
The st_combine and st_union functions can be used to combine multiple geometries into a single one. The difference is that st_combine only combines the geometries into one, just like the c function combines sfg objects into one sfg (Section 7.2.2), while st_union also dissolves internal borders.
To understand the difference between st_combine and st_union, let's examine the result of applying each of them on nm:
st_combine(nm) ## Geometry set for 1 feature  ## Geometry type: MULTIPOLYGON ## Dimension:     XY ## Bounding box:  xmin: -863763.9 ymin: -1478601 xmax: -267074.1 ymax: -845634.3 ## Projected CRS: US National Atlas Equal Area ## MULTIPOLYGON (((-267074.1 -884175.1, -267309.2 ...st_union(nm) ## Geometry set for 1 feature  ## Geometry type: POLYGON ## Dimension:     XY ## Bounding box:  xmin: -863763.9 ymin: -1478601 xmax: -267074.1 ymax: -845634.3 ## Projected CRS: US National Atlas Equal Area ## POLYGON ((-852411.9 -1350402, -850603.1 -133127...The results are plotted in Figure 8.10:
plot(st_combine(nm), main = "st_combine(nm)") plot(st_union(nm), main = "st_union(nm)")Figure 8.10: Combining (st_combine) and dissolving (st_union) polygons
What is the class of the resulting objects? Why? What happened to the original
nmattributes?
Why did
st_combinereturn aMULTIPOLYGON, whilest_unionreturned aPOLYGON?
Why do you think there are internal "patches" in the result of
st_union(nm)(Figure 8.10)30?
Let's solve a geometric question to practice some of the functions we learned so far. What is the distance between the centroids of California and New Jersey, in kilometers? Using [, st_union, st_centroid and st_distance, we can calculate the distance as follows:
# Subsetting ca = county[county$NAME_1 == "California", ] nj = county[county$NAME_1 == "New Jersey", ]  # Union ca = st_union(ca) nj = st_union(nj)  # Calculating centroids ca_ctr = st_centroid(ca) nj_ctr = st_centroid(nj)  # Calculating distance d = st_distance(ca_ctr, nj_ctr)  # Converting to km set_units(d, "km") ## Units: [km] ##         [,1] ## [1,] 3846.64Note that we are using st_union to combine the county polygons into a single geometry, so that we can calculate the centroid of the state polygon, as a whole.
To plot the centroids more conveniently, we can use the c function to combine two sfc (geometry columns) into one, long, geometry column. Recall that this is unlike what the c function does when given sfg (geometries), in which case the geometries are combined to one (multi- or collection) geometry (Section 7.2.2).
p = c(ca_ctr, nj_ctr) p ## Geometry set for 2 features  ## Geometry type: POINT ## Dimension:     XY ## Bounding box:  xmin: -1707694 ymin: -667480.7 xmax: 2111204 ymax: -206332.8 ## Projected CRS: US National Atlas Equal Area ## POINT (-1707694 -667480.7) ## POINT (2111204 -206332.8)The centroids we calculated can be displayed as follows (Figure 8.11):
plot(st_geometry(county), border = "grey") plot(p, col = "red", pch = 16, cex = 2, add = TRUE)Figure 8.11: California and New Jersey centroids
How can we draw a line line between the centroids, corresponding to the distance we just calculated (Figure 8.12)?
First, we can use st_combine to transform the points into a single MULTIPOINT geometry. Remember, the st_combine function is similar to st_union, but only combines and does not dissolve.
p = st_combine(p) p ## Geometry set for 1 feature  ## Geometry type: MULTIPOINT ## Dimension:     XY ## Bounding box:  xmin: -1707694 ymin: -667480.7 xmax: 2111204 ymax: -206332.8 ## Projected CRS: US National Atlas Equal Area ## MULTIPOINT ((-1707694 -667480.7), (2111204 -206...Do you think the results of
st_union(p)andst_combine(p)will be different, and if so in what way?
What is left to be done is to transform the MULTIPOINT geometry to a LINESTRING geometry, which is what we learn next (Section 8.3.4.4).
13.4.4.4 Geometry casting
Sometimes we need to convert a given geometry to a different type, to express different aspects of the geometries in our analysis. For example, we would have to convert the county MULTIPOLYGON geometries to MULTILINESTRING geometries to calculate the perimeter of each county, using st_length (Section 8.3.2).
The st_cast function can be used to convert between different geometry types. The st_cast function accepts:
- The input - sfg,- sfcor- sf
- The target geometry type 
For example, we can use st_cast to convert our MULTIPOINT geometry p (Figure 8.11), which we calculated in Section 8.3.4.3, to a LINESTRING geometry l:
l = st_cast(p, "LINESTRING") l ## Geometry set for 1 feature  ## Geometry type: LINESTRING ## Dimension:     XY ## Bounding box:  xmin: -1707694 ymin: -667480.7 xmax: 2111204 ymax: -206332.8 ## Projected CRS: US National Atlas Equal Area ## LINESTRING (-1707694 -667480.7, 2111204 -206332.8)Each MULTIPOINT geometry is replaced with a LINESTRING geometry, comprising a line that goes through all points from first to last. In this case, p contains just one MULTIPOINT geometry that contains two points. Therefore, the resulting geometry column l contains one LINESTRING geometry with a straight line.
The line we calculated is shown in Figure 8.12:
plot(st_geometry(county), border = "grey") plot(st_geometry(ca_ctr), col = "red", pch = 16, cex = 2, add = TRUE) plot(st_geometry(nj_ctr), col = "red", pch = 16, cex = 2, add = TRUE) plot(l, col = "red", add = TRUE)Figure 8.12: California and New Jersey centroids, with a line
It is worth mentioning that not every conversion operation between geometry types is possible. Meaningless st_cast operations will fail with an error message. For example, a LINESTRING composed of just two points cannot be converted to a POLYGON:
st_cast(l, "POLYGON") ## Error in FUN(X[[i]], ...): polygons require at least 4 points13.4.4.5 Buffers
Another example of a function that generates new geometries based on individual input geometries is st_buffer. The st_buffer function calculates buffers, polygons encompassing all point up to the specified distance from a given geometry.
For example, here is how we can calculate 100 km�� buffers around the airports. Note that the st_buffer requires buffer distance (dist), which can be passed as numeric, in which case CRS units are assumed, or as units, in which case we can use any distance unit we wish. The following two expressions are equivalent, in both cases resulting in a layer of 100 km�� buffers named airports_100:
# Buffer distance as 'numeric' airports_100 = st_buffer(airports, dist = 100 * 10^3)# Buffer distance as 'units' airports_100 = st_buffer(airports, dist = set_units(100, "km"))The airports_100 layer with the airport buffers is shown in Figure 8.13:
plot(st_geometry(nm)) plot(st_geometry(airports_100), add = TRUE)Figure 8.13: airports buffered by 100 km
13.4.4.6 Geometry generation from pairs
In addition to geometry-generating functions that operate on individual geometries, such as st_centroid (Section 8.3.4.2) and st_buffer (Section 8.3.4.5), there are four geometry-generating functions that operate on pairs of input geometries. These are:
- st_intersection
- st_difference
- st_sym_difference
- st_union
Figure 8.14 shows what each of these four functions does. Note that st_intersection, st_sym_difference, st_union are symmetrical meaning that the order of the two inputs x and y does not matter. The st_difference function is not symmetrical: it returns the area of x, minus the intersection of x and y.
Figure 8.14: Geometry-generating operations on pairs of layers
For example, suppose that we need to calculate the area that is within 100 km�� range of all three airports at the same time? We can find the area of intersection of the three airports, using st_intersection. Since st_intersection accepts two geometries at a time, this can be done in two steps:
inter1 = st_intersection(airports_100[1, ], airports_100[2, ]) inter2 = st_intersection(inter1, airports_100[3, ])The resulting polygon inter2 is shown, in blue, in Figure 8.15:
plot(st_geometry(nm)) plot(st_geometry(airports_100), add = TRUE) plot(inter2, col = "lightblue", add = TRUE)Figure 8.15: The intersection of three airports buffers
How can we calculate the area that is within 100 km�� range of at least one of the three airports? We can use exactly the same technique, only replacing st_intersection with st_union, as follows:
union1 = st_union(airports_100[1, ], airports_100[2, ]) union2 = st_union(union1, airports_100[3, ])As shown in Section 8.3.4.3, st_union can be applied on an individual layer, in which case it returns the (dissolved) union of all geometries. Therefore we can get to the same result with just one expression, as follows:
union2 = st_union(airports_100)The resulting polygon union2 is shown, in blue, in Figure 8.16:
plot(st_geometry(nm)) plot(union2, col = "lightblue", add = TRUE)Figure 8.16: The union of three airports buffers
13.4.4.7 Convex hull
The Convex Hull of a set X� of points is the smallest convex polygon that contains X�. The convex hull may be visualized as the shape resulting when stretching a rubber band around X�, then releasing it until it shrinks to minimal size (Figure 8.17).
Figure 8.17: Convex Hull: elastic-band analogy (https://en.wikipedia.org/wiki/Convex_hull)
The st_convex_hull function is used to calculate the convex hull of the given geometry or geometries. For example, the following expression calculates the convex hull (per feature) in nm1:
h = st_convex_hull(nm1)The resulting layer is shown in red in Figure 8.18:
plot(st_geometry(nm1)) plot(st_geometry(h), add = TRUE, border = "red")Figure 8.18: Convex hull polygons for two counties in New Mexico
How can we calculate the convex hull of all polygons in
nm1(Figure 8.19)?
Figure 8.19: Convex Hull of multiple polygons
Here is another question to practice the geometric operations we learned in this chapter. Suppose we build a tunnel, 10 km�� wide, between the centroids of "Harding" and "Sierra" counties in New Mexico. Which counties does the tunnel go through? The following code gives the result. Take a moment to go over it and make sure you understand what happens in each step:
w = nm[nm$NAME_2 %in% c("Harding", "Sierra"), ] w_ctr = st_centroid(w) w_ctr_buf = st_buffer(w_ctr, dist = 5000) w_ctr_buf_u = st_combine(w_ctr_buf) w_ctr_buf_u_ch = st_convex_hull(w_ctr_buf_u) nm_w = nm[w_ctr_buf_u_ch, ] nm_w$NAME_2 ## [1] "Harding"    "San Miguel" "Guadalupe"  "Torrance"   "Socorro"    ## [6] "Lincoln"    "Sierra"Figure 8.20 shows the calculated tunnel geometry and highlights the counties that the tunnel goes through. The 4th expression uses the text function to draw text labels. The text function accepts the matrix of coordinates, where labels should be drawn (in this case, county centroid coordinates), and the corresponding character vector of lables. In this case, the labels are county names (nm_w$NAME_2). (Note that in case the labels are numeric, such as in Figure 8.7, then we must convert them to character using as.character for them to be displayed using the text function.)
plot(st_geometry(nm_w), border = NA, col = "lightblue") plot(st_geometry(nm), add = TRUE) plot(st_geometry(w_ctr_buf_u_ch), add = TRUE) text(st_coordinates(st_centroid(nm_w)), nm_w$NAME_2)Figure 8.20: Tunnel between "Sierra" and "Harding" county centroids
How is the area of the "tunnel" divided between the various counties that it crosses? We can use st_intersection (Section 8.3.4.6) to calculate all intersections between the tunnel polygon w_ctr_buf_u_ch, and the county polygons nm_w. Since we already know that the tunnel (w_ctr_buf_u_ch) intersects with each of the seven polygons in nm_w, the result of the intersection is of the same length, and in the same order, as the corresponding geometries in nm_w:
int = st_intersection(w_ctr_buf_u_ch, nm_w) int ## Geometry set for 7 features  ## Geometry type: POLYGON ## Dimension:     XY ## Bounding box:  xmin: -677075.1 ymin: -1293982 xmax: -340138.8 ymax: -1002634 ## Projected CRS: US National Atlas Equal Area ## First 5 geometries: ## POLYGON ((-359510.4 -1013406, -348484.4 -100391... ## POLYGON ((-430575.6 -1074562, -359510.4 -101340... ## POLYGON ((-484643 -1121091, -430575.6 -1074562,... ## POLYGON ((-545452.3 -1173421, -484643 -1121091,... ## POLYGON ((-638440.3 -1253443, -546193.4 -117405...Therefore, we can divide the area (Section 8.3.2.2) of each county intersection, by the total area of the tunnel, to get the proportion of tunnel area in each county:
area = st_area(int) area ## Units: [m^2] ## [1]  199064581  872145769  813577715  702562404 1119738156  108077571  575538789prop = area / sum(area) prop ## Units: [1] ## [1] 0.04533773 0.19863456 0.18529546 0.16001130 0.25502469 0.02461508 0.13108118The proportions can be displayed as text labels, using the text function (Figure 8.21):
plot(st_geometry(nm_w), border = "darkgrey") plot(int, col = hcl.colors(7, "Set3"), border = "grey", add = TRUE) text(st_coordinates(st_centroid(int)), paste0(round(prop, 2)*100, "%"))Figure 8.21: Proportion of tunnel area within each county
13.5 Vector layer aggregation
We don't always want to dissolve all features into a single geometry (Section 8.3.4.3). Sometimes, we need to dissolve each group of geometries, whereas groups are specified by attribute or by location. To demonstrate dissolving by attribute, also known as aggregation, let's take a subset of county with the counties of just Arizona and Utah states:
s = county[county$NAME_1 %in% c("Arizona", "Utah"), ]The selected counties are shown in Figure 8.22:
plot(s[, "NAME_1"], key.width = lcm(3))Figure 8.22: Subset of two states from county
As shown previously (Section 8.3.4.3), we can dissolve all features into a single geometry with st_union:
s1 = st_union(s) s1 ## Geometry set for 1 feature  ## Geometry type: POLYGON ## Dimension:     XY ## Bounding box:  xmin: -1389076 ymin: -1470784 xmax: -757960.3 ymax: -235145.7 ## Projected CRS: US National Atlas Equal Area ## POLYGON ((-852411.9 -1350402, -852611 -1352512,...The dissolved geometry is shown in Figure 8.23:
plot(s1)Figure 8.23: Union of all counties in s
Aggregating/dissolving by attributes can be done with aggregate31, which accepts the following arguments:
- x—The- sflayer being dissolved
- by—A corresponding- data.framedetermining the groups
- FUN—The function to be applied on each attribute in- x
For example, the following expression dissolves the layer s, summing the area attribute, by state:
s2 = aggregate(s[, "area"], st_drop_geometry(s[, "NAME_1"]), sum) s2 ## Simple feature collection with 2 features and 2 fields ## Attribute-geometry relationship: 0 constant, 1 aggregate, 1 identity ## Geometry type: POLYGON ## Dimension:     XY ## Bounding box:  xmin: -1389076 ymin: -1470784 xmax: -757960.3 ymax: -235145.7 ## Projected CRS: US National Atlas Equal Area ##    NAME_1            area                       geometry ## 1 Arizona 295423.1 [km^2] POLYGON ((-852411.9 -135040... ## 2    Utah 219634.4 [km^2] POLYGON ((-1011317 -819521....The result is shown in Figure 8.24:
plot(s2[, "area"])Figure 8.24: Union by state name of s
13.6 Join by attributes
Attaching new attributes to a vector layer from a table, based on common attribute(s), is known as a non-spatial join, or join by attributes. Joining by attributes works exactly the same way as a join between two tables, e.g., using the merge function (Section 4.6). The difference is just that the "left" table, x, is an sf layer, instead of a data.frame. Using merge to join an sf layer with a data.frame is analogous to the "Join attributes from a table" workflow in ArcGIS (Figure 8.25).
Figure 8.25: Join by attributes in ArcGIS
In the next example we will join county-level demographic data, from CO-EST2012-Alldata.csv, with the county layer. The join will be based on the common Federal Information Processing Standards (FIPS) code of each county.
First, let's read the CO-EST2012-Alldata.csv file into a data.frame named dat. This file contains numerous demographic variables, on the county level, in the US:
dat = read.csv("CO-EST2012-Alldata.csv")Next, we subset the columns we are interested in, so that it is easier to display the table:
- STATE—State code
- COUNTY—County code
- CENSUS2010POP—Population size in 2010
The first two columns, STATE and COUNTY, contain the state and county components of the FIPS code, which we will use to join the data. The third column, CENSUS2010POP, contains our variable of interest, which we want to attach to the county layer.
dat = dat[, c("STATE", "COUNTY", "CENSUS2010POP")] head(dat) ##   STATE COUNTY CENSUS2010POP ## 1     1      0       4779736 ## 2     1      1         54571 ## 3     1      3        182265 ## 4     1      5         27457 ## 5     1      7         22915 ## 6     1      9         57322Records where COUNTY code is 0 are state sums, i.e. sub-totals. We need to remove the state sums and keep just the county-level records:
dat = dat[dat$COUNTY != 0, ] head(dat) ##   STATE COUNTY CENSUS2010POP ## 2     1      1         54571 ## 3     1      3        182265 ## 4     1      5         27457 ## 5     1      7         22915 ## 6     1      9         57322 ## 7     1     11         10914The FIPS code in the county layer is a character value with five digits, where the first two digits are the state code and the last three digits are the county code. When necessary, leading zeros are added. For example, state code 9 and county code 5 is encoded as "09005".
county$FIPS[1:6] ## [1] "09005" "09003" "09013" "09015" "06093" "06015"while the dat table contains separate state and county codes as numeric values:
dat$STATE[1:10] ##  [1] 1 1 1 1 1 1 1 1 1 1dat$COUNTY[1:10] ##  [1]  1  3  5  7  9 11 13 15 17 19To get the corresponding county FIPS codes in dat, we need to:
- Standardize the state code to a two-digit code 
- Standardize the county code to three-digit code 
- Paste the state code and the county code, to obtain the five-digit FIPS code 
The formatC function can be used to format numeric values into different formats, using different "scenarios." The "add leading zeros" scenario is specified using width=n, where n is the required number of digits, and flag="0". Therefore:
dat$STATE = formatC(dat$STATE, width = 2, flag = "0") dat$COUNTY = formatC(dat$COUNTY, width = 3, flag = "0") dat$FIPS = paste0(dat$STATE, dat$COUNTY)Now we have a column named FIPS, with FIPS codes in exactly the same format as in the county layer:
head(dat) ##   STATE COUNTY CENSUS2010POP  FIPS ## 2    01    001         54571 01001 ## 3    01    003        182265 01003 ## 4    01    005         27457 01005 ## 5    01    007         22915 01007 ## 6    01    009         57322 01009 ## 7    01    011         10914 01011This means we can join the county layer with the dat table, based on the common column named FIPS:
county = merge(county, dat[, c("FIPS", "CENSUS2010POP")], by = "FIPS", all.x = TRUE) county ## Simple feature collection with 3103 features and 6 fields ## Geometry type: MULTIPOLYGON ## Dimension:     XY ## Bounding box:  xmin: -2031903 ymin: -2116922 xmax: 2516534 ymax: 732304.6 ## Projected CRS: US National Atlas Equal Area ## First 10 features: ##     FIPS  NAME_1   NAME_2 TYPE_2            area CENSUS2010POP ## 1  01001 Alabama  Autauga County 1562.805 [km^2]         54571 ## 2  01003 Alabama  Baldwin County 4247.589 [km^2]        182265 ## 3  01005 Alabama  Barbour County 2330.135 [km^2]         27457 ## 4  01007 Alabama     Bibb County 1621.356 [km^2]         22915 ## 5  01009 Alabama   Blount County 1692.125 [km^2]         57322 ## 6  01011 Alabama  Bullock County 1623.924 [km^2]         10914 ## 7  01013 Alabama   Butler County 2013.066 [km^2]         20947 ## 8  01015 Alabama  Calhoun County 1591.091 [km^2]        118572 ## 9  01017 Alabama Chambers County 1557.054 [km^2]         34215 ## 10 01019 Alabama Cherokee County 1549.770 [km^2]         25989 ##                          geometry ## 1  MULTIPOLYGON (((1225972 -12... ## 2  MULTIPOLYGON (((1200268 -15... ## 3  MULTIPOLYGON (((1408158 -13... ## 4  MULTIPOLYGON (((1207082 -12... ## 5  MULTIPOLYGON (((1260171 -11... ## 6  MULTIPOLYGON (((1370531 -12... ## 7  MULTIPOLYGON (((1280245 -13... ## 8  MULTIPOLYGON (((1313695 -11... ## 9  MULTIPOLYGON (((1374433 -12... ## 10 MULTIPOLYGON (((1333380 -10...Note that we are using a left join, therefore the county layer retains all of its records, whether or not there is a match in the dat table (Section 4.6.2). Features in county that do not have a matching FIPS value in dat get NA in the new CENSUS2010POP column.
A map of the new CENSUS2010POP attribute, using the default equal breaks, is shown in Figure 8.26:
plot(county[, "CENSUS2010POP"])Figure 8.26: Population size per county in the US (equal breaks)
Almost all counties fall into the first category, therefore the map is uniformely colored and not very informative. A map using quantile breaks (Figure 8.27) reveals the spatial pattern of population sizes more clearly:
plot(county[, "CENSUS2010POP"], breaks = "quantile")Figure 8.27: Population size per county in the US (quantile breaks)
Another issue with the map is that county population size also depends on county area, which can make the pattern misleading in case when county area sizes are variable (which they are). To safely compare the counties we need to standardize population size by county area. In other words, we need to calculate population density:
county$density = county$CENSUS2010POP / county$areaNote how the measurement units (km−2��−2, i.e., number of people per km2��2) for the new column are automatically determined based on the inputs:
head(county$density) ## Units: [1/km^2] ## [1] 34.91863 42.91023 11.78344 14.13323 33.87575  6.72076The map of population densities per county is shown in Figure 8.28:
plot(county[, "density"], breaks = "quantile")Figure 8.28: Population density in the US, quantile breaks
How many features in
countydid not have a matching FIPS code indat? What are their names?
How many features in
datdo not have a matching FIPS code incounty?
We can also calculate the average population density in the US, by dividing the total population by the total area:
sum(county$CENSUS2010POP, na.rm = TRUE) / sum(county$area) ## 39.2295 [1/km^2]This is close to the population density in 2010 according to Wikipedia, which is 40.015.
Simply averaging the density column gives an overestimation, because all counties get equal weight while in reality the smaller counties are more dense:
mean(county$density, na.rm = TRUE) ## 93.87977 [1/km^2]13.7 Recap: main data structures
Table 8.1 summarizes the main data structures we learn in this book.
| Table 8.1: Main data structures in the book | ||
| Category | Class | Chapter | 
| Vector | numeric,character,logical | 2 | 
| Date | Date | 3 | 
| Table | data.frame | 4 | 
| Matrix | matrix | 5 | 
| Array | array | 5 | 
| Raster | stars | 5 | 
| Vector layer | sfg,sfc,sf | 7 | 
| Units | units | 8 | 
| List | list | 11 | 
| Geostatistical model | gstat | 12 | 
| Variogram model | variogramModel | 12 |