迭代查找另一个数据集中点的 x 距离的所有点

Question

我有两组数据点（df1 和 df2）。这个想法是找到（并计算）df2 中的所有点，这些点在 df1 中每个点的 x 半径距离内。让 x 从 500 m 开始，以 500 m 的增量增加到 20000 m。这个想法是使用循环或 purr 更有效地迭代和完成任务。

我试过这样的事情：

library(sf)
library(spData)
library(tidyverse)

# data set we can use for the example 

df1 <- cycle_hire %>% 
# transform for the buffer function
st_transform(., crs = st_crs(27700)) %>% 
select(id_1 = "id")

df2 <- cycle_hire_osm %>% 
# transform for the buffer function
st_transform(., crs = st_crs(27700)) %>% 
select(id_2 = "osm_id")

# create first 500 m buffer for df1

df1_500 <- st_buffer(df1, dist = 500)

# get points that fall within the 500 m buffer

within_500 <- st_join(df1_500, df2, join = st_intersects) %>% 
# for each point in df 1, count how many points from df2 are within 500 m 
  group_by(id_1) %>% 
  count() %>% 
  ungroup() %>% 
# in my real data some rows return NA next to the id_1
# but this is still being counted as 1. 
# I can deal with this later but I realize that in this toy example that this is a mistake. 
# remove geometry because I want to join everything later
  as_tibble() %>% 
  select(-geometry) %>% 
  # identify distance (will join later)
  mutate(within = "500")

# now x increases by 500 m and we create a 1000 m buffer

df1_1000 <- st_buffer(df1, dist = 1000)

# get points that fall within the 1000 m buffer

within_1000 <- st_join(df1_1000, df2, join = st_intersects) %>% 
  # for each point in df 1, count how many points from df2 are within 1000 m 
  group_by(id_1) %>% 
  count() %>% 
  ungroup() %>% 
  # remove geometry
  as_tibble() %>% 
  select(-geometry) %>% 
  # identify distance
  mutate(within = "1000")

# and repeat...

# eventually I bind into one tibble

bind_all <- bind_rows(within_500, within_1000)

我们重复这个直到我们达到 20000。

我将如何迭代它以便它可以扩展到 20000 米或更多？我还希望在起始距离和增量方面具有一定的灵活性 - 即，可能从 200 米开始并增加我的 200 米增量。或者从 200 米开始，增加 200 直到我们达到 20000，然后我们开始增加 1000 直到 40000。显然我的代码效率很低，我不应该手动执行此操作。

任何帮助将不胜感激

Answer 1

500 米以内的查找：

df_example <- df1 %>% 
  mutate(within_500 = lengths(st_is_within_distance(x = .,
                                                    y = df2,
                                                    dist = 500)))
df_example

Simple feature collection with 742 features and 2 fields
geometry type:  POINT
dimension:      XY
bbox:           xmin: 522502 ymin: 174408 xmax: 538733.2 ymax: 184421
projected CRS:  OSGB 1936 / British National Grid
First 10 features:
   id_1                  geometry within_500
1     1 POINT (531203.5 182832.1)         12
2     2 POINT (525208.1 179391.9)          9
3     3 POINT (532985.8 182001.6)          8
4     4   POINT (530437.8 182912)          8
5     5     POINT (528051 178742)          2
6     6 POINT (528858.4 181542.9)         11
7     7   POINT (527159 183300.8)          3
8     8 POINT (527032.7 182634.6)          5
9     9   POINT (532205 180434.6)          7
10   10 POINT (532464.9 180284.3)          5

为了找到所有这些（为简洁起见，我做了 5000 米）：

intervals <- seq(500, 5000, 500)

df_final <- map_dfr(intervals,
                ~ df1 %>% 
                  mutate(within = .x,
                         n = lengths(st_is_within_distance(x = .,
                                                               y = df2,
                                                               dist = .x))) %>% 
                  st_drop_geometry()
                )

head(df_final)

  id_1 within  n
1    1    500 12
2    2    500  9
3    3    500  8
4    4    500  8
5    5    500  2
6    6    500 11


tail(df_final)

     id_1 within   n
7415  772   5000 205
7416  773   5000 400
7417  774   5000 113
7418  775   5000 130
7419  776   5000 114
7420  777   5000 120

迭代查找另一个数据集中点的 x 距离的所有点

iterate finding all points with x distance of points in another dataset

r

dplyr

sf

tibble