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基于50英里半径的位置分组

  •  1
  • user3570187  · 技术社区  · 6 年前

    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)
    

    我想创建一个组集群id,它基本上是去组半径在50英里以内的位置。让我知道我怎样才能做到这一点?非常感谢。下面是预期输出。

     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
    
    2 回复  |  直到 6 年前
        1
  •  1
  •   Wimpel    6 年前
    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
    
        2
  •  0
  •   Birger    6 年前

    可以使用位置之间的距离创建二维矩阵。这个 geosphere 有一个功能,为你做的重任。

    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).