R-在给定半径内查找最近邻点和邻数，坐标为lat long
我试图弄清楚在我的数据集中某些点是多么孤立。我使用两种方法来确定隔离度，最近邻居的距离和给定半径内相邻站点的数量。我所有的坐标都是经纬度 以下是我的数据：R-在给定半径内查找最近邻点和邻数，坐标为lat long,r,distance,latitude-longitude,R,Distance,Latitude Longitude,我试图弄清楚在我的数据集中某些点是多么孤立。我使用两种方法来确定隔离度，最近邻居的距离和给定半径内相邻站点的数量。我所有的坐标都是经纬度 以下是我的数据： 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
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
谢谢。最好的选择是使用库
sprgeoslibrary(sp)
library(rgeos)
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"
这将提供更好的结果，如果点分散在地球上，坐标以度为单位，@Zbynek提出的解决方案非常好，但是如果你想寻找两个邻居之间的距离，或者像我一样以km为单位，我会提出这个解决方案
earth.dist<-function(lat1,long1,lat2,long2){
rad <- pi/180
a1 <- lat1 * rad
a2 <- long1 * rad
b1 <- lat2 * rad
b2 <- long2 * rad
dlat <- b1-a1
dlon<- b2-a2
a <- (sin(dlat/2))^2 +cos(a1)*cos(b1)*(sin(dlon/2))^2
c <- 2*atan2(sqrt(a),sqrt(1-a))
R <- 6378.145
dist <- R *c
return(dist)
}
Dist <- matrix(0,ncol=length(mydata),nrow=length(mydata.sp))
for (i in 1:length(mydata)){
for(j in 1:length(mydata.sp)){
Dist[i,j] <- earth.dist(mydata$lat[i],mydata$long[i],mydata.sp$lat[j],mydata.sp$long[j])
}}
DDD <- matrix(0, ncol=5,nrow=ncol(Dist)) ### RECTIFY the nb of col by the number of variable you want
for(i in 1:ncol(Dist)){
sub<- sort(Dist[,i])[2]
DDD[i,1] <- names(sub)
DDD[i,2] <- sub
DDD[i,3] <- rownames(Dist)[i]
sub_neig_atr <- Coord[Coord$ID==names(sub),]
DDD[i,4] <- sub_neig_atr$area
DDD[i,5] <- sub_neig_atr$canopy
### Your can add any variable you want here
}
DDD <- as.data.frame(DDD)
names(DDD)<-c("neigboor_ID","distance","pond","n.area","n.canopy")
data <- merge(mydata,DDD, by="pond")
earth.dist我使用spacerisk
包在下面添加了一个解决方案。这个包中的关键功能是用C++编写的，因此非常快。
首先，加载数据：
df <- data.frame(pond = c("A10", "AA006", "Blacksmith", "Borrow.Pit.1",
"Borrow.Pit.2", "Borrow.Pit.3", "Boulder"),
lat = c(41.95928, 41.96431, 41.95508, 41.95601,
41.95571, 41.95546, 41.918223),
long = c(-72.14605, -72.121, -72.123803, -72.15419,
-72.15413, -72.15375, -72.14978),
area = c(1500, 250, 361, 0, 0, 0, 1392),
canopy = c(66, 0, 77, 0, 0, 0, 98),
avg.depth = c(60.61538462, 57.77777778, 71.3125, 41.44444444,
37.7, 29.22222222, 43.53333333))
现在计算目标池塘1500m范围内的池塘数量。函数spatialrisk:：concentration
对从中心点开始的半径内的观察次数求和。从池塘数量中减去1，以排除池塘本身
df$npond <- 1
radius1500 <- spatialrisk::concentration(df, df, npond, lon_sub = long,
lon_full = long, radius = 1500,
display_progress = FALSE)$concentration - 1
我在下面添加了一个使用较新的sf
软件包的替代解决方案，供感兴趣的人使用，现在就来此页面（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)
要在计算距离后获得最近的邻居，可以使用sort（）
和partial=2
参数。根据数据量的不同，这可能比在上一个解决方案中使用order
要快得多。包Rfast
可能更快，但我避免在这里包含额外的包。有关各种解决方案的讨论和基准测试，请参阅此相关帖子：在Rfast中，有一个名为“dista”的函数，只计算欧几里德距离或曼哈顿距离（目前）。它提供了计算k-最小距离的选项。或者，它可以返回距离最小的观测值的索引。余弦距离基本上与欧氏距离相同（我认为排除常数2） 另一个答案，虽然速度可能较慢，但对dplyr上瘾者可能有直观的吸引力
您可以创建一个由每个可能的纬度/纬度组合组成的巨型网格，然后使用geosphere找到距离最小的网格
例如，您有两个具有不同点的数据集要比较，但是您可以通过复制第一个数据集来轻松地进行调整
library(tidyverse)
library(geosphere)
library(data.table)
#This function creates a big dataframe with every possible combination
expand.grid.df <- function(...) Reduce(function(...) merge(..., by=NULL), list(...))
shortest_distance <- expand.grid.df(df1,df2) %>%
mutate(distance = distHaversine(p1 = cbind(lon_2,lat_2),
p2 = cbind(lon,lat))) %>%
group_by(ACCIDENT_NO) %>%
slice(which.min(distance))
库（tidyverse）
图书馆（地球圈）
库（数据表）
#此函数使用各种可能的组合创建一个大数据帧
展开.grid.df%
组别(意外数字)%>%
切片（哪个.min（距离））
非常感谢。你建议的图书馆正是我需要的！这是一些非常有用和可读的代码，谢谢！然而，我无法将其调整到我略有不同的用例：我需要在两个不同的数据集之间找到最近的点（我有一个tweet数据集，我需要每个tweet最近的城市）。我应该更改什么？使用以下函数：排序（x[x>0]，递减=F）[1]
@ike<代码>排序x[x>0]geospheredistmdf <- data.frame(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(1500, 250, 361, 0, 0, 0, 1392),
canopy = c(66, 0, 77, 0, 0, 0, 98),
avg.depth = c(60.61538462, 57.77777778, 71.3125, 41.44444444,
37.7, 29.22222222, 43.53333333))
ans1 <- purrr::map2_dfr(df$long, df$lat,
~spatialrisk::points_in_circle(df, .x, .y,
lon = long,
radius = 100000)[2,])
colnames(ans1) <- c("neighbor", "n.lat", "n.long", "n.area",
"n.canopy", "n.avg.depth", "distance_m")
neighbor n.lat n.long n.area n.canopy n.avg.depth distance_m
1 Borrow.Pit.1 41.95601 -72.15419 0 0 41.44444 765.87823
2 Blacksmith 41.95508 -72.12380 361 77 71.31250 1053.35200
3 AA006 41.96431 -72.12100 250 0 57.77778 1053.35200
4 Borrow.Pit.2 41.95571 -72.15413 0 0 37.70000 33.76321
5 Borrow.Pit.1 41.95601 -72.15419 0 0 41.44444 33.76321
6 Borrow.Pit.2 41.95571 -72.15413 0 0 37.70000 42.00128
7 Borrow.Pit.3 41.95546 -72.15375 0 0 29.22222 4158.21978
df$npond <- 1
radius1500 <- spatialrisk::concentration(df, df, npond, lon_sub = long,
lon_full = long, radius = 1500,
display_progress = FALSE)$concentration - 1
cbind(df, ans1, radius1500)
pond lat long area canopy avg.depth neighbor n.lat n.long n.area n.canopy n.avg.depth distance_m radius1500
1 A10 41.95928 -72.14605 1500 66 60.61538 Borrow.Pit.1 41.95601 -72.15419 0 0 41.44444 765.87823 3
2 AA006 41.96431 -72.12100 250 0 57.77778 Blacksmith 41.95508 -72.12380 361 77 71.31250 1053.35200 1
3 Blacksmith 41.95508 -72.12380 361 77 71.31250 AA006 41.96431 -72.12100 250 0 57.77778 1053.35200 1
4 Borrow.Pit.1 41.95601 -72.15419 0 0 41.44444 Borrow.Pit.2 41.95571 -72.15413 0 0 37.70000 33.76321 3
5 Borrow.Pit.2 41.95571 -72.15413 0 0 37.70000 Borrow.Pit.1 41.95601 -72.15419 0 0 41.44444 33.76321 3
6 Borrow.Pit.3 41.95546 -72.15375 0 0 29.22222 Borrow.Pit.2 41.95571 -72.15413 0 0 37.70000 42.00128 3
7 Boulder 41.91822 -72.14978 1392 98 43.53333 Borrow.Pit.3 41.95546 -72.15375 0 0 29.22222 4158.21978 0
# 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)
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
library(tidyverse)
library(geosphere)
library(data.table)
#This function creates a big dataframe with every possible combination
expand.grid.df <- function(...) Reduce(function(...) merge(..., by=NULL), list(...))
shortest_distance <- expand.grid.df(df1,df2) %>%
mutate(distance = distHaversine(p1 = cbind(lon_2,lat_2),
p2 = cbind(lon,lat))) %>%
group_by(ACCIDENT_NO) %>%
slice(which.min(distance))