从线串形状文件中提取坐标并找到它们之间的距离
Extract coordinates from linestring shapefiles and find distance between them
我有两个 LINESTRING 类型的 shapefile。
一个是对象A和对象B之间的9行列表。
另一个是对象B和对象C之间的18行列表
对于共享相同对象 B 坐标的线串，我想找到相应的对象 A 坐标和对象 C 坐标之间的距离。
例如；
对象 B 到对象 C 线串：
线绳 (21 15, 40.5 10.5)
线串(23 15,40.5 10.5)
对象 B 到对象 A 线串：
线串(53 11, 40.5 10.5)
那我想计算坐标(21, 15)和(53,11)之间的距离，计算(23, 15)和(53,11)之间的距离
要重现我的 LINESTRING shapefile，您可以运行以下代码：
对象 B 到对象 A
ObjectBtoA <-structure(list(structure(c(53, 40.5, 11, 10.5), .Dim = c(2L,
2L), class = c("XY", "LINESTRING", "sfg")), structure(c(53, 41.5,
11, 10.5), .Dim = c(2L, 2L), class = c("XY", "LINESTRING", "sfg"
)), structure(c(53, 42.5, 11, 10.5), .Dim = c(2L, 2L), class = c("XY",
"LINESTRING", "sfg")), structure(c(53, 40.5, 11.5, 11.5), .Dim = c(2L,
2L), class = c("XY", "LINESTRING", "sfg")), structure(c(53, 41.5,
11.5, 11.5), .Dim = c(2L, 2L), class = c("XY", "LINESTRING",
"sfg")), structure(c(53, 42.5, 11.5, 11.5), .Dim = c(2L, 2L), class = c("XY",
"LINESTRING", "sfg")), structure(c(53, 40.5, 12.5, 12.5), .Dim = c(2L,
2L), class = c("XY", "LINESTRING", "sfg")), structure(c(53, 41.5,
12.5, 12.5), .Dim = c(2L, 2L), class = c("XY", "LINESTRING",
"sfg")), structure(c(53, 42.5, 12.5, 12.5), .Dim = c(2L, 2L), class = c("XY",
"LINESTRING", "sfg"))), n_empty = 0L, precision = 0, crs = structure(list(
input = "EPSG:4326", wkt = "GEOGCS[\"WGS 84\",\n DATUM[\"WGS_1984\",\n
SPHEROID[\"WGS 84\",6378137,298.257223563,\n AUTHORITY[\"EPSG\",\"7030\"]],\n
AUTHORITY[\"EPSG\",\"6326\"]],\n PRIMEM[\"Greenwich\",0,\n
AUTHORITY[\"EPSG\",\"8901\"]],\n UNIT[\"degree\",0.0174532925199433,\n
AUTHORITY[\"EPSG\",\"9122\"]],\n AUTHORITY[\"EPSG\",\"4326\"]]"), class = "crs"), class =
c("sfc_LINESTRING",
"sfc"), bbox = structure(c(xmin = 40.5, ymin = 10.5, xmax = 53,
ymax = 12.5), class = "bbox"))
对象 B 到对象 C：
structure(list(structure(c(21, 40.5, 15, 10.5), .Dim = c(2L,
2L), class = c("XY", "LINESTRING", "sfg")), structure(c(21, 41.5,
15, 10.5), .Dim = c(2L, 2L), class = c("XY", "LINESTRING", "sfg"
)), structure(c(21, 42.5, 15, 10.5), .Dim = c(2L, 2L), class = c("XY",
"LINESTRING", "sfg")), structure(c(21, 40.5, 15, 11.5), .Dim = c(2L,
2L), class = c("XY", "LINESTRING", "sfg")), structure(c(21, 41.5,
15, 11.5), .Dim = c(2L, 2L), class = c("XY", "LINESTRING", "sfg"
)), structure(c(21, 42.5, 15, 11.5), .Dim = c(2L, 2L), class = c("XY", "LINESTRING",
"sfg")), structure(c(21, 40.5, 15, 12.5), .Dim = c(2L,
2L), class = c("XY", "LINESTRING", "sfg")), structure(c(21, 41.5,
15, 12.5), .Dim = c(2L, 2L), class = c("XY", "LINESTRING", "sfg"
)), structure(c(21, 42.5, 15, 12.5), .Dim = c(2L, 2L), class = c("XY",
"LINESTRING", "sfg")), structure(c(23, 40.5, 15, 10.5), .Dim = c(2L,
2L), class = c("XY", "LINESTRING", "sfg")), structure(c(23, 41.5,
15, 10.5), .Dim = c(2L, 2L), class = c("XY", "LINESTRING", "sfg"
)), structure(c(23, 42.5, 15, 10.5), .Dim = c(2L, 2L), class = c("XY",
"LINESTRING", "sfg")), structure(c(23, 40.5, 15, 11.5), .Dim = c(2L,
2L), class = c("XY", "LINESTRING", "sfg")), structure(c(23, 41.5,
15, 11.5), .Dim = c(2L, 2L), class = c("XY", "LINESTRING", "sfg"
)), structure(c(23, 42.5, 15, 11.5), .Dim = c(2L, 2L), class = c("XY",
"LINESTRING", "sfg")), structure(c(23, 40.5, 15, 12.5), .Dim = c(2L,
2L), class = c("XY", "LINESTRING", "sfg")), structure(c(23, 41.5,
15, 12.5), .Dim = c(2L, 2L), class = c("XY", "LINESTRING", "sfg"
)), structure(c(23, 42.5, 15, 12.5), .Dim = c(2L, 2L), class = c("XY",
"LINESTRING", "sfg"))), n_empty = 0L, precision = 0, crs = structure(list(
input = "EPSG:4326", wkt = "GEOGCS[\"WGS 84\",\n DATUM[\"WGS_1984\",\n
SPHEROID[\"WGS 84\",6378137,298.257223563,\n AUTHORITY[\"EPSG\",\"7030\"]],\n
AUTHORITY[\"EPSG\",\"6326\"]],\n PRIMEM[\"Greenwich\",0,\n
AUTHORITY[\"EPSG\",\"8901\"]],\n UNIT[\"degree\",0.0174532925199433,\n
AUTHORITY[\"EPSG\",\"9122\"]],\n AUTHORITY[\"EPSG\",\"4326\"]]"), class = "crs"), class =
c("sfc_LINESTRING",
"sfc"), bbox = structure(c(xmin = 21, ymin = 10.5, xmax = 42.5,
ymax = 15), class = "bbox"))
为了让这个答案更简单，而不是使用您数据中的许多行，我只使用从某些点创建的几个线串来回答这个问题。
关于问题：“对于共享相同对象 B 坐标的线串，我想找到对应的对象 A 坐标和对象 C 坐标之间的距离。”
创建点和线：
# Create the lines
x1 <- c(2,3,4) # str line 1
y1 <- c (3,3,3)
y2 <- c (4,4,4)
x2 <- c(x1 - 5) # str line 2
x3 <- c(x1-2) # str line 3
ya <- c(3,4,5) # diagonal line
l1 <- matrix(data = c(x1, y1), nrow = 3, ncol = 2)
l2 <- matrix(data = c(x2, y1), nrow = 3, ncol = 2)
l3 <- matrix(data = c(x3, y1), nrow = 3, ncol = 2)
l4 <- matrix(data = c(x1, ya), nrow = 3, ncol = 2)
创建简单要素线串 (sfc) 对象：
l1 <- st_linestring(l1, dim = "XY")
l2 <- st_linestring(l2, dim = "XY")
l3 <- st_linestring(l3, dim = "XY")
l4 <- st_linestring(l4, dim = "XY")
g1 <- c(l1, l3) # new var with multiple linestrings
g2 <- c(l1, l3, l4) # new var with multiple linestrings
绘制新变量：
par(mfrow=c(1,2)) # set the plotting area into a 1*2 array
plot(g1, axes = T) # plot horizontal line
plot(g2, axes = T) # plot horizontal + diag line
确定线串的长度：
st_length(g1)
st_length(g2)
4
6.82842712474619
所以这个答案表明当两个或多个线串被分配给一个变量时，当这些线串共享一个公共坐标时，结果长度与连接字符串的长度相同。这条线是直的还是不直的并不重要。变量 g2 与三个线串共享一个公共点。可以证明返回的值 6.82。只需使用勾股定理找到斜边的长度，并将该值添加到其他两个水平线段的长度：( 2 + 2 + 2.828)
我有两个 LINESTRING 类型的 shapefile。
一个是对象A和对象B之间的9行列表。
另一个是对象B和对象C之间的18行列表
对于共享相同对象 B 坐标的线串，我想找到相应的对象 A 坐标和对象 C 坐标之间的距离。
例如；
对象 B 到对象 C 线串： 线绳 (21 15, 40.5 10.5) 线串(23 15,40.5 10.5)
对象 B 到对象 A 线串： 线串(53 11, 40.5 10.5)
那我想计算坐标(21, 15)和(53,11)之间的距离，计算(23, 15)和(53,11)之间的距离
要重现我的 LINESTRING shapefile，您可以运行以下代码：
对象 B 到对象 A
ObjectBtoA <-structure(list(structure(c(53, 40.5, 11, 10.5), .Dim = c(2L,
2L), class = c("XY", "LINESTRING", "sfg")), structure(c(53, 41.5,
11, 10.5), .Dim = c(2L, 2L), class = c("XY", "LINESTRING", "sfg"
)), structure(c(53, 42.5, 11, 10.5), .Dim = c(2L, 2L), class = c("XY",
"LINESTRING", "sfg")), structure(c(53, 40.5, 11.5, 11.5), .Dim = c(2L,
2L), class = c("XY", "LINESTRING", "sfg")), structure(c(53, 41.5,
11.5, 11.5), .Dim = c(2L, 2L), class = c("XY", "LINESTRING",
"sfg")), structure(c(53, 42.5, 11.5, 11.5), .Dim = c(2L, 2L), class = c("XY",
"LINESTRING", "sfg")), structure(c(53, 40.5, 12.5, 12.5), .Dim = c(2L,
2L), class = c("XY", "LINESTRING", "sfg")), structure(c(53, 41.5,
12.5, 12.5), .Dim = c(2L, 2L), class = c("XY", "LINESTRING",
"sfg")), structure(c(53, 42.5, 12.5, 12.5), .Dim = c(2L, 2L), class = c("XY",
"LINESTRING", "sfg"))), n_empty = 0L, precision = 0, crs = structure(list(
input = "EPSG:4326", wkt = "GEOGCS[\"WGS 84\",\n DATUM[\"WGS_1984\",\n
SPHEROID[\"WGS 84\",6378137,298.257223563,\n AUTHORITY[\"EPSG\",\"7030\"]],\n
AUTHORITY[\"EPSG\",\"6326\"]],\n PRIMEM[\"Greenwich\",0,\n
AUTHORITY[\"EPSG\",\"8901\"]],\n UNIT[\"degree\",0.0174532925199433,\n
AUTHORITY[\"EPSG\",\"9122\"]],\n AUTHORITY[\"EPSG\",\"4326\"]]"), class = "crs"), class =
c("sfc_LINESTRING",
"sfc"), bbox = structure(c(xmin = 40.5, ymin = 10.5, xmax = 53,
ymax = 12.5), class = "bbox"))
对象 B 到对象 C：
structure(list(structure(c(21, 40.5, 15, 10.5), .Dim = c(2L,
2L), class = c("XY", "LINESTRING", "sfg")), structure(c(21, 41.5,
15, 10.5), .Dim = c(2L, 2L), class = c("XY", "LINESTRING", "sfg"
)), structure(c(21, 42.5, 15, 10.5), .Dim = c(2L, 2L), class = c("XY",
"LINESTRING", "sfg")), structure(c(21, 40.5, 15, 11.5), .Dim = c(2L,
2L), class = c("XY", "LINESTRING", "sfg")), structure(c(21, 41.5,
15, 11.5), .Dim = c(2L, 2L), class = c("XY", "LINESTRING", "sfg"
)), structure(c(21, 42.5, 15, 11.5), .Dim = c(2L, 2L), class = c("XY", "LINESTRING",
"sfg")), structure(c(21, 40.5, 15, 12.5), .Dim = c(2L,
2L), class = c("XY", "LINESTRING", "sfg")), structure(c(21, 41.5,
15, 12.5), .Dim = c(2L, 2L), class = c("XY", "LINESTRING", "sfg"
)), structure(c(21, 42.5, 15, 12.5), .Dim = c(2L, 2L), class = c("XY",
"LINESTRING", "sfg")), structure(c(23, 40.5, 15, 10.5), .Dim = c(2L,
2L), class = c("XY", "LINESTRING", "sfg")), structure(c(23, 41.5,
15, 10.5), .Dim = c(2L, 2L), class = c("XY", "LINESTRING", "sfg"
)), structure(c(23, 42.5, 15, 10.5), .Dim = c(2L, 2L), class = c("XY",
"LINESTRING", "sfg")), structure(c(23, 40.5, 15, 11.5), .Dim = c(2L,
2L), class = c("XY", "LINESTRING", "sfg")), structure(c(23, 41.5,
15, 11.5), .Dim = c(2L, 2L), class = c("XY", "LINESTRING", "sfg"
)), structure(c(23, 42.5, 15, 11.5), .Dim = c(2L, 2L), class = c("XY",
"LINESTRING", "sfg")), structure(c(23, 40.5, 15, 12.5), .Dim = c(2L,
2L), class = c("XY", "LINESTRING", "sfg")), structure(c(23, 41.5,
15, 12.5), .Dim = c(2L, 2L), class = c("XY", "LINESTRING", "sfg"
)), structure(c(23, 42.5, 15, 12.5), .Dim = c(2L, 2L), class = c("XY",
"LINESTRING", "sfg"))), n_empty = 0L, precision = 0, crs = structure(list(
input = "EPSG:4326", wkt = "GEOGCS[\"WGS 84\",\n DATUM[\"WGS_1984\",\n
SPHEROID[\"WGS 84\",6378137,298.257223563,\n AUTHORITY[\"EPSG\",\"7030\"]],\n
AUTHORITY[\"EPSG\",\"6326\"]],\n PRIMEM[\"Greenwich\",0,\n
AUTHORITY[\"EPSG\",\"8901\"]],\n UNIT[\"degree\",0.0174532925199433,\n
AUTHORITY[\"EPSG\",\"9122\"]],\n AUTHORITY[\"EPSG\",\"4326\"]]"), class = "crs"), class =
c("sfc_LINESTRING",
"sfc"), bbox = structure(c(xmin = 21, ymin = 10.5, xmax = 42.5,
ymax = 15), class = "bbox"))
为了让这个答案更简单，而不是使用您数据中的许多行，我只使用从某些点创建的几个线串来回答这个问题。
关于问题：“对于共享相同对象 B 坐标的线串，我想找到对应的对象 A 坐标和对象 C 坐标之间的距离。”
创建点和线：
# Create the lines
x1 <- c(2,3,4) # str line 1
y1 <- c (3,3,3)
y2 <- c (4,4,4)
x2 <- c(x1 - 5) # str line 2
x3 <- c(x1-2) # str line 3
ya <- c(3,4,5) # diagonal line
l1 <- matrix(data = c(x1, y1), nrow = 3, ncol = 2)
l2 <- matrix(data = c(x2, y1), nrow = 3, ncol = 2)
l3 <- matrix(data = c(x3, y1), nrow = 3, ncol = 2)
l4 <- matrix(data = c(x1, ya), nrow = 3, ncol = 2)
创建简单要素线串 (sfc) 对象：
l1 <- st_linestring(l1, dim = "XY")
l2 <- st_linestring(l2, dim = "XY")
l3 <- st_linestring(l3, dim = "XY")
l4 <- st_linestring(l4, dim = "XY")
g1 <- c(l1, l3) # new var with multiple linestrings
g2 <- c(l1, l3, l4) # new var with multiple linestrings
绘制新变量：
par(mfrow=c(1,2)) # set the plotting area into a 1*2 array
plot(g1, axes = T) # plot horizontal line
plot(g2, axes = T) # plot horizontal + diag line
确定线串的长度：
st_length(g1)
st_length(g2)
4
6.82842712474619
所以这个答案表明当两个或多个线串被分配给一个变量时，当这些线串共享一个公共坐标时，结果长度与连接字符串的长度相同。这条线是直的还是不直的并不重要。变量 g2 与三个线串共享一个公共点。可以证明返回的值 6.82。只需使用勾股定理找到斜边的长度，并将该值添加到其他两个水平线段的长度：( 2 + 2 + 2.828)