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r - 基于与 sf 的距离对选择点

问题描述

我想计算点位置之间的欧几里得距离 ( df.dist) 并在这对距离大于 20000 米 ( dplyr::filter(as.numeric(df.dist$dist) < 20000)) 时删除一些位置，但不起作用。在我的例子中：

# Package
library(sf)
#> Linking to GEOS 3.9.0, GDAL 3.2.1, PROJ 7.2.1
library(dplyr)

# Create data
set.seed(1)
df <- data.frame(
  x  = rnorm(10),
  y = rnorm(10)
)
df <- st_as_sf(df,coords = c("x","y"),remove = F, crs = 4326)
df.laea = st_transform(
  df,
  crs = "+proj=laea +x_0=4600000 +y_0=4600000 +lon_0=0.13 +lat_0=0.24 +datum=WGS84 +units=m"
)

# Calculate matrix distance and selection pair of distance < 20000 meters
df.dist <- df %>%
  mutate(
    dist = st_distance(geometry)
  )
distance.target <- df.dist  %>% 
        dplyr::filter(as.numeric(df.dist$dist) < 20000)

#Erro: Problem with `filter()` input `..1`.
#i Input `..1` is `as.numeric(df.dist$dist) < 20000`.
#x Input `..1` must be of size 10 or 1, not size 100.
#Run `rlang::last_error()` to see where the error occurred.

拜托，有什么想法可以解决吗？

提前致谢！

标签： rsf

解决方案

st_distance在您的情况下返回矩阵 10x10 这是 100 个值，因此错误Input ..1 must be of size 10 or 1, not size 100.。输出df.dist非常具有误导性：

为了df.dist

Simple feature collection with 10 features and 3 fields
geometry type:  POINT
dimension:      XY
bbox:           xmin: -0.8356286 ymin: -2.2147 xmax: 1.595281 ymax: 1.511781
geographic CRS: WGS 84
            x           y                       geometry          dist
1  -0.6264538  1.51178117    POINT (-0.6264538 1.511781)      0.00 [m]
2   0.1836433  0.38984324    POINT (0.1836433 0.3898432) 153363.06 [m]
3  -0.8356286 -0.62124058  POINT (-0.8356286 -0.6212406) 237004.19 [m]
4   1.5952808 -2.21469989       POINT (1.595281 -2.2147) 480555.51 [m]
5   0.3295078  1.12493092     POINT (0.3295078 1.124931) 114666.44 [m]
6  -0.8204684 -0.04493361 POINT (-0.8204684 -0.04493361) 173482.34 [m]
7   0.4874291 -0.01619026  POINT (0.4874291 -0.01619026) 209564.83 [m]
8   0.7383247  0.94383621    POINT (0.7383247 0.9438362) 164361.86 [m]
9   0.5757814  0.82122120    POINT (0.5757814 0.8212212) 154058.72 [m]
10 -0.3053884  0.59390132   POINT (-0.3053884 0.5939013) 107601.29 [m]

因为df.dist$dist是矩阵：

Units: [m]
          [,1]      [,2]     [,3]     [,4]      [,5]      [,6]      [,7]      [,8]      [,9]     [,10]
 [1,]      0.0 153363.06 237004.2 480555.5 114666.44 173482.34 209564.83 164361.86 154058.72 107601.29
 [2,] 153363.1      0.00 159289.4 328063.3  82887.65 121676.58  56207.83  86974.75  64657.24  58927.70
 [3,] 237004.2 159289.36      0.0 322838.3 232597.70  63747.10 161762.69 246263.77 223885.09 146756.58
 [4,] 480555.5 328063.26 322838.3      0.0 395238.69 360338.42 272578.66 362043.43 354354.31 375761.40
 [5,] 114666.4  82887.65 232597.7 395238.7      0.00 181986.32 127397.41  49713.21  43349.25  91879.46
 [6,] 173482.3 121676.58  63747.1 360338.4 181986.32      0.00 145629.14 205089.33 182564.02  90980.32
 [7,] 209564.8  56207.83 161762.7 272578.7 127397.41 145629.14      0.00 109766.72  93117.07 111084.52
 [8,] 164361.9  86974.75 246263.8 362043.4  49713.21 205089.33 109766.72      0.00  22608.55 122449.40
 [9,] 154058.7  64657.24 223885.1 354354.3  43349.25 182564.02  93117.07  22608.55      0.00 101253.41
[10,] 107601.3  58927.70 146756.6 375761.4  91879.46  90980.32 111084.52 122449.40 101253.41      0.00

你可能想要这样的东西：

m <- df$geometry %>% st_distance()
indices <- which(m < units::set_units(20000, "m"), arr.ind = TRUE)

df.dist <- as.data.frame(indices)
df.dist$dist <- as.numeric(m[indices])

   row col dist
1    1   1    0
2    2   2    0
3    3   3    0
4    4   4    0
5    5   5    0
6    6   6    0
7    7   7    0
8    8   8    0
9    9   9    0
10  10  10    0

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