R中的空间最近邻居分配

Question

我正在进行一项研究，该研究试图根据特定人员的地址将颗粒物暴露分配给特定人员。 我有两个具有经度和纬度坐标的数据集。 如果是个人，则为一个，如果是下午，则为一个。 我想根据最接近的区块为每个主题分配一个pm曝光区块。

library(sp)
library(raster)
library(tidyverse)

#subject level data
subjectID<-c("A1","A2","A3","A4")

subjects<-data.frame(tribble(
~lon,~lat,
-70.9821391,    42.3769511,
-61.8668537,    45.5267133,
-70.9344039,    41.6220337,
-70.7283830,    41.7123494
))

row.names(subjects)<-subjectID

#PM Block Locations 
blockID<-c("B1","B2","B3","B4","B5")

blocks<-data.frame(tribble(
~lon,~lat,
-70.9824591,    42.3769451,
-61.8664537,    45.5267453,
-70.9344539,    41.6220457,
-70.7284530,    41.7123454,
-70.7284430,    41.7193454
))

row.names(blocks)<-blockID

#Creating distance matrix
dis_matrix<-pointDistance(blocks,subjects,lonlat = TRUE)

###The above code doesnt preserve the row names. Is there a way to to do 
that?

###I'm unsure about the below code
colnames(dis_matrix)<-row.names(subjects)
row.names(dis_matrix)<-row.names(blocks)

dis_data<-data.frame(dis_matrix)

###Finding nearst neighbor and coercing to usable format 
getname <-function(x) {
row.names(dis_data[which.min(x),])
}

nn<-data.frame(lapply(dis_data,getname)) %>% 
gather(key=subject,value=neighbor)

这段代码为我提供了有意义的输出，但是我不确定其有效性和效率。 任何有关如何改进和修复此代码的建议都将受到赞赏。 我还收到错误消息：

Warning message:
attributes are not identical across measure variables;
they will be dropped 

我无法确定其来源。

谢谢参观！

Answer 1

这是一些示例数据，如何使用pointDistance ：

library(raster)

#subject level data
subjectID <- c("A1","A2","A3","A4")
subxy <- matrix(c(-65, 42, -60, 4.5, -70, 20, -75, 41 ), ncol=2, byrow=TRUE)
#PM Block Locations 
blockID <- c("B1","B2","B3","B4","B5")
blockxy <- matrix(c(-68, 22, -61, 25, -70, 31, -65, 11,-63, 21), ncol=2, byrow=TRUE)

# distance of all subxy to all blockxy points
d <- pointDistance(subxy, blockxy, lonlat=TRUE)

# get the blockxy record nearest to each subxy record
r <- apply(d, 1, which.min)
r
#[1] 3 4 1 3

因此，对是：

p <- data.frame(subject=subjectID, block=blockID[r])
p

#  subject block
#1      A1    B3
#2      A2    B4
#3      A3    B1
#4      A4    B3

说明它有效：

plot(rbind(blockxy, subxy), ylim=c(0,45), xlab='longitude', ylab='latitude')
points(blockxy, col="red", pch=20, cex=2)
points(subxy, col="blue", pch=20, cex=2)
text(subxy, subjectID, pos=1)
text(blockxy, blockID, pos=1)
for (i in 1:nrow(subxy)) {
    arrows(subxy[i,1], subxy[i,2], blockxy[r[i],1], blockxy[r[i],2])
}

Answer 2

如果您的数据集很大，则可能要使用此答案中 @ user3507085解释的非常高效的nabor包。 由于问题是关闭的，所以我已经复制粘贴了下面的答案，因此它在此线程中“保持活跃”。 我不知道这是否被认为是不好的做法，如果愿意，我很乐意删除/编辑（请注意knn给出的距离不是地理距离，但是我想可以通过简单的转换将其转换为球形距离，包括反正弦）：

lonlat2xyz=function (lon, lat, r) 
{
lon = lon * pi/180
lat = lat * pi/180
if (missing(r)) 
    r <- 6378.1
x <- r * cos(lat) * cos(lon)
y <- r * cos(lat) * sin(lon)
z <- r * sin(lat)
return(cbind(x, y, z))
}

lon1=runif(100,-180,180);lon2=runif(100,-180,180);lat1=runif(100,-90,90);lat2=runif(100,-90,90)

xyz1=lonlat2xyz(lon1,lat1)
xyz2=lonlat2xyz(lon2,lat2)

library(nabor)

out=knn(data=xyz1,query = xyz2,k=20)

library(maps)

map()
points(lon1,lat1,pch=16,col="black")
points(lon2[1],lat2[1],pch=16,col="red")
points(lon1[out$nn.idx[1,]],lat1[out$nn.idx[1,]],pch=16,col="blue")