根据50英里半径分组位置

问题描述 投票:1回答:2

嗨,我有一个数据集,我试图获得基于50英里半径的组群ID。这是数据集的结构

g_lat<- c(45.52306, 40.26719, 34.05223, 37.38605, 37.77493)
g_long<- c(-122.67648,-86.13490, -118.24368, -122.08385, -122.41942)
df<- data.frame(g_lat, g_long)

我想创建一个组群集ID,它基本上将分组在50英里范围内的位置。让我知道如何实现这一目标?非常感谢。以下是预期产量。

 g_lat      g_long      clusterid
45.52306   -122.67648    1 
40.26719    -86.13490    2
34.05223    -118.24368   3
37.38605    -122.08385   4
37.77493    -122.41942   4
r dplyr sp geosphere
2个回答
1
投票
g_lat<- c(45.52306, 40.26719, 34.05223, 37.38605, 37.77493)
g_long<- c(-122.67648,-86.13490, -118.24368, -122.08385, -122.41942)
df<- data.frame(point = c(1:5), longitude = g_long, latitude = g_lat)

library(sf)
my.sf.point <- st_as_sf(x = df, 
                        coords = c("longitude", "latitude"),
                        crs = "+proj=longlat +datum=WGS84")

#distance matrix in feet
st_distance(my.sf.point)

#which poiint are within 50 miles (~80467.2 meters)
l <- st_is_within_distance(my.sf.point, dist = 80467.2 )

l
# Sparse geometry binary predicate list of length 5, where the predicate was `is_within_distance'
# 1: 1
# 2: 2
# 3: 3
# 4: 4, 5
# 5: 4, 5

df$within_50 <- rowSums(as.matrix(l))-1

df
#   point longitude latitude within_50
# 1     1 -122.6765 45.52306         0
# 2     2  -86.1349 40.26719         0
# 3     3 -118.2437 34.05223         0
# 4     4 -122.0838 37.38605         1
# 5     5 -122.4194 37.77493         1


m <- as.matrix(l)
colnames(m) <- c(1:nrow(df))
rownames(m) <- c(1:nroe(df))
df$points_within_50 <- apply( m, 1, function(u) paste( names(which(u)), collapse="," ) )
df$clusterid <- dplyr::group_indices(df, df$points_within_50) 

#   point longitude latitude within_50 points_within_50 clusterid
# 1     1 -122.6765 45.52306         0                1         1
# 2     2  -86.1349 40.26719         0                2         2
# 3     3 -118.2437 34.05223         0                3         3
# 4     4 -122.0838 37.38605         1              4,5         4
# 5     5 -122.4194 37.77493         1              4,5         4

0
投票

您可以创建具有位置之间距离的2d矩阵。 geosphere有一个功能,可以为您提供繁重的工作。

library(geosphere)
library(magrittr)

g_lat <- c(45.52306, 40.26719, 34.05223, 37.38605, 37.77493)
g_long <- c(-122.67648,-86.13490, -118.24368, -122.08385, -122.41942)
m <- cbind(g_long, g_lat) 

(matrix <- distm(m) / 1609.34)
#>           [,1]     [,2]      [,3]      [,4]      [,5]
#> [1,]    0.0000 1872.882  825.4595  562.3847  534.8927
#> [2,] 1872.8818    0.000 1812.5862 1936.5786 1946.4373
#> [3,]  825.4595 1812.586    0.0000  315.2862  347.3751
#> [4,]  562.3847 1936.579  315.2862    0.0000   32.5345
#> [5,]  534.8927 1946.437  347.3751   32.5345    0.0000
matrix < 50 
#>       [,1]  [,2]  [,3]  [,4]  [,5]
#> [1,]  TRUE FALSE FALSE FALSE FALSE
#> [2,] FALSE  TRUE FALSE FALSE FALSE
#> [3,] FALSE FALSE  TRUE FALSE FALSE
#> [4,] FALSE FALSE FALSE  TRUE  TRUE
#> [5,] FALSE FALSE FALSE  TRUE  TRUE
colSums(matrix < 50)
#> [1] 1 1 1 2 2

Created on 2018-09-16 by the [reprex package](http://reprex.tidyverse.org) (v0.2.0).
© www.soinside.com 2019 - 2024. All rights reserved.