use*_*942 25 r distance latitude-longitude
我试图弄清楚我的数据集中有多少孤立的某些点.我使用两种方法来确定隔离,最近邻居的距离和给定半径内的相邻站点的数量.我所有的坐标都是纬度和经度
这就是我的数据:
pond lat long area canopy avg.depth neighbor n.lat n.long n.distance n.area n.canopy n.depth n.avg.depth radius1500
A10 41.95928 -72.14605 1500 66 60.61538462
AA006 41.96431 -72.121 250 0 57.77777778
Blacksmith 41.95508 -72.123803 361 77 71.3125
Borrow.Pit.1 41.95601 -72.15419 0 0 41.44444444
Borrow.Pit.2 41.95571 -72.15413 0 0 37.7
Borrow.Pit.3 41.95546 -72.15375 0 0 29.22222222
Boulder 41.918223 -72.14978 1392 98 43.53333333
我想把最近的邻近池塘的名称放在列邻居,它的纬度和长度在n.lat和n.long,两个池塘之间的距离n.distance,以及区域,冠层和avg.depth in每个适当的列.
其次,我想把目标池塘1500米范围内的池塘数量调到半径1500.
有谁知道一个功能或包,可以帮助我计算我想要的距离/数字？如果这是一个问题,输入我需要的其他数据并不困难,但最近邻居的名字和距离加上1500米以内的池塘数量是我真正需要帮助的.
谢谢.
Zby*_*nek 36
最好的选择是使用库sp和rgeos,从而使您能够构建空间类和执行地理处理.
library(sp)
library(rgeos)
读取数据并将其转换为空间对象:
mydata <- read.delim('d:/temp/testfile.txt', header=T)
sp.mydata <- mydata
coordinates(sp.mydata) <- ~long+lat
class(sp.mydata)
[1] "SpatialPointsDataFrame"
attr(,"package")
[1] "sp"
现在计算点之间的成对距离
d <- gDistance(sp.mydata, byid=T)
找到第二个最短距离(最近距离指向自身,因此使用第二个最短距离)
min.d <- apply(d, 1, function(x) order(x, decreasing=F)[2])
使用所需变量构造新数据框
newdata <- cbind(mydata, mydata[min.d,], apply(d, 1, function(x) sort(x, decreasing=F)[2]))
colnames(newdata) <- c(colnames(mydata), 'neighbor', 'n.lat', 'n.long', 'n.area', 'n.canopy', 'n.avg.depth', 'distance')
newdata
pond lat long area canopy avg.depth neighbor n.lat n.long n.area n.canopy n.avg.depth
6 A10 41.95928 -72.14605 1500 66 60.61538 Borrow.Pit.3 41.95546 -72.15375 0 0 29.22222
3 AA006 41.96431 -72.12100 250 0 57.77778 Blacksmith 41.95508 -72.12380 361 77 71.31250
2 Blacksmith 41.95508 -72.12380 361 77 71.31250 AA006 41.96431 -72.12100 250 0 57.77778
5 Borrow.Pit.1 41.95601 -72.15419 0 0 41.44444 Borrow.Pit.2 41.95571 -72.15413 0 0 37.70000
4 Borrow.Pit.2 41.95571 -72.15413 0 0 37.70000 Borrow.Pit.1 41.95601 -72.15419 0 0 41.44444
5.1 Borrow.Pit.3 41.95546 -72.15375 0 0 29.22222 Borrow.Pit.2 41.95571 -72.15413 0 0 37.70000
6.1 Boulder 41.91822 -72.14978 1392 98 43.53333 Borrow.Pit.3 41.95546 -72.15375 0 0 29.22222
distance
6 0.0085954872
3 0.0096462277
2 0.0096462277
5 0.0003059412
4 0.0003059412
5.1 0.0004548626
6.1 0.0374480316
编辑:如果坐标以度为单位并且您想要以公里为单位计算距离,请使用包geosphere
library(geosphere)
d <- distm(sp.mydata)
# rest is the same
如果点分散在地球上并且坐标以度为单位,则应该提供更好的结果
我在下面添加了一个使用较新sf软件包的替代解决方案，供那些感兴趣并现在访问此页面的人使用（就像我所做的那样）。
首先，加载数据并创建sf对象。
# Using sf
mydata <- structure(
list(pond = c("A10", "AA006", "Blacksmith", "Borrow.Pit.1",
"Borrow.Pit.2", "Borrow.Pit.3", "Boulder"),
lat = c(41.95928, 41.96431, 41.95508, 41.95601, 41.95571, 41.95546,
41.918223),
long = c(-72.14605, -72.121, -72.123803, -72.15419, -72.15413,
-72.15375, -72.14978),
area = c(1500L, 250L, 361L, 0L, 0L, 0L, 1392L),
canopy = c(66L, 0L, 77L, 0L, 0L, 0L, 98L),
avg.depth = c(60.61538462, 57.77777778, 71.3125, 41.44444444,
37.7, 29.22222222, 43.53333333)),
class = "data.frame", row.names = c(NA, -7L))
library(sf)
data_sf <- st_as_sf(mydata, coords = c("long", "lat"),
# Change to your CRS
crs = "+proj=longlat +ellps=WGS84 +datum=WGS84 +no_defs")
st_is_longlat(data_sf)
sf::st_distance 使用纬度/经度数据时，使用大圆距离计算以米为单位的距离矩阵。
dist.mat <- st_distance(data_sf) # Great Circle distance since in lat/lon
# Number within 1.5km: Subtract 1 to exclude the point itself
num.1500 <- apply(dist.mat, 1, function(x) {
sum(x < 1500) - 1
})
# Calculate nearest distance
nn.dist <- apply(dist.mat, 1, function(x) {
return(sort(x, partial = 2)[2])
})
# Get index for nearest distance
nn.index <- apply(dist.mat, 1, function(x) { order(x, decreasing=F)[2] })
n.data <- mydata
colnames(n.data)[1] <- "neighbor"
colnames(n.data)[2:ncol(n.data)] <-
paste0("n.", colnames(n.data)[2:ncol(n.data)])
mydata2 <- data.frame(mydata,
n.data[nn.index, ],
n.distance = nn.dist,
radius1500 = num.1500)
rownames(mydata2) <- seq(nrow(mydata2))
mydata2
pond lat long area canopy avg.depth neighbor n.lat n.long n.area n.canopy
1 A10 41.95928 -72.14605 1500 66 60.61538 Borrow.Pit.1 41.95601 -72.15419 0 0
2 AA006 41.96431 -72.12100 250 0 57.77778 Blacksmith 41.95508 -72.12380 361 77
3 Blacksmith 41.95508 -72.12380 361 77 71.31250 AA006 41.96431 -72.12100 250 0
4 Borrow.Pit.1 41.95601 -72.15419 0 0 41.44444 Borrow.Pit.2 41.95571 -72.15413 0 0
5 Borrow.Pit.2 41.95571 -72.15413 0 0 37.70000 Borrow.Pit.1 41.95601 -72.15419 0 0
6 Borrow.Pit.3 41.95546 -72.15375 0 0 29.22222 Borrow.Pit.2 41.95571 -72.15413 0 0
7 Boulder 41.91822 -72.14978 1392 98 43.53333 Borrow.Pit.3 41.95546 -72.15375 0 0
n.avg.depth n.distance radius1500
1 41.44444 766.38426 3
2 71.31250 1051.20527 1
3 57.77778 1051.20527 1
4 37.70000 33.69099 3
5 41.44444 33.69099 3
6 37.70000 41.99576 3
7 29.22222 4149.07406 0
为了计算距离之后获得的最近的邻居，你可以使用sort()与partial = 2论证。根据数据量，这可能比order在之前的解决方案中使用要快得多。该软件包Rfast可能会更快，但我避免在此处包含其他软件包。有关各种解决方案的讨论和基准测试，请参阅此相关帖子：https : //stackoverflow.com/a/53144760/12265198