R-成对距离计算嵌套数据框

Question

我正在寻找一种以成对方式计算点之间的分离距离，并将每个点的结果存储在随附的嵌套数据框架中。

例如，我有此数据框架（来自地图包），其中包含有关美国城市（包括其物理位置）的信息。我丢弃了其余信息，并将坐标嵌套在嵌套的数据框架中。我打算使用geosphere软件包中的distHaversine()来计算这些距离。

library(tidyverse)
df <- maps::us.cities %>% 
  slice(1:20) %>% 
  group_by(name) %>% 
  nest(long, lat, .key = coords)
                   name            coords
                  <chr>           <list>
 1           Abilene TX <tibble [1 x 2]>
 2             Akron OH <tibble [1 x 2]>
 3           Alameda CA <tibble [1 x 2]>
 4            Albany GA <tibble [1 x 2]>
 5            Albany NY <tibble [1 x 2]>
 ...(With 15 more rows)

我研究了使用函数的地图家族，加上突变的功能，但我遇到了艰难的时期。所需结果的形式如下：

                   name            coords            sep_dist
                  <chr>           <list>            <list>
 1           Abilene TX <tibble [1 x 2]> <tibble [19 x 2]>
 2             Akron OH <tibble [1 x 2]> <tibble [19 x 2]>
 3           Alameda CA <tibble [1 x 2]> <tibble [19 x 2]>
 4            Albany GA <tibble [1 x 2]> <tibble [19 x 2]>
 5            Albany NY <tibble [1 x 2]> <tibble [19 x 2]>
 ...(With 15 more rows)

用sep_dist tibbles看着这样的东西：

               location  distance
                  <chr>     <dbl> 
 1             Akron OH      1003
 2           Alameda CA       428
 3            Albany GA      3218
 4            Albany NY      3627
 5            Albany OR        97
 ...(With 14 more rows)                       -distances completely made up

位置是要与名称进行比较的点（在这种情况下为Abilene）。

Answer 1

geosphere提供了将全能距离与distm

进行比较的能力

可重复的数据

set.seed(1)
df <- data.frame(name=letters[1:4],
                 lon=runif(4)*10,
                 lat=runif(4)*10)

distm

library(geosphere)
ans <- as.data.frame(distm(df[,2:3], df[,2:3], fun=distHaversine))
         # a        b        c        d
# 1      0.0 784506.1 894320.6 877440.5
# 2 784506.1      0.0 226504.3 647666.7
# 3 894320.6 226504.3      0.0 486290.8
# 4 877440.5 647666.7 486290.8      0.0

整理所需格式

colnames(ans) <- df$name
library(dplyr)
library(tidyr)
desired <- ans %>%
             gather(pos1, distance) %>%
             mutate(pos2 = rep(df$name, nrow(df))) %>%
             filter(pos1!=pos2) %>%
             select(pos1, pos2, distance)
   # pos1 pos2 distance
# 1     a    b 784506.1
# 2     a    c 894320.6
# 3     a    d 877440.5
# 4     b    a 784506.1
# 5     b    c 226504.3
# 6     b    d 647666.7
# 7     c    a 894320.6
# 8     c    b 226504.3
# 9     c    d 486290.8
# 10    d    a 877440.5
# 11    d    b 647666.7
# 12    d    c 486290.8

Answer 2

我们可以通过位置名称和坐标的所有组合扩展一个"网格"，但以相同的位置名称删除组合。之后，使用map2_dbl应用distHaversine功能。

library(tidyverse)
library(geosphere)
df2 <- df %>%
  # Create the grid
  mutate(name1 = name) %>%
  select(starts_with("name")) %>%
  complete(name, name1) %>%
  filter(name != name1) %>%
  left_join(df, by = "name") %>%
  left_join(df, by = c("name1" = "name")) %>%
  # Grid completed. Calcualte the distance by distHaversine
  mutate(distance = map2_dbl(coords.x, coords.y, distHaversine))
df2
# A tibble: 380 x 5
         name          name1         coords.x         coords.y  distance
        <chr>          <chr>           <list>           <list>     <dbl>
 1 Abilene TX       Akron OH <tibble [1 x 2]> <tibble [1 x 2]> 1881904.4
 2 Abilene TX     Alameda CA <tibble [1 x 2]> <tibble [1 x 2]> 2128576.9
 3 Abilene TX      Albany GA <tibble [1 x 2]> <tibble [1 x 2]> 1470577.2
 4 Abilene TX      Albany NY <tibble [1 x 2]> <tibble [1 x 2]> 2542025.1
 5 Abilene TX      Albany OR <tibble [1 x 2]> <tibble [1 x 2]> 2429367.3
 6 Abilene TX Albuquerque NM <tibble [1 x 2]> <tibble [1 x 2]>  702287.5
 7 Abilene TX  Alexandria LA <tibble [1 x 2]> <tibble [1 x 2]>  700093.2
 8 Abilene TX  Alexandria VA <tibble [1 x 2]> <tibble [1 x 2]> 2161594.6
 9 Abilene TX    Alhambra CA <tibble [1 x 2]> <tibble [1 x 2]> 1718967.5
10 Abilene TX Aliso Viejo CA <tibble [1 x 2]> <tibble [1 x 2]> 1681868.8
# ... with 370 more rows

要创建最终输出，我们可以基于名称和nest group_by和所有其他所需的列。

df3 <- df2 %>%
  select(-starts_with("coord")) %>%
  group_by(name) %>%
  nest()
df3
# A tibble: 20 x 2
                   name              data
                  <chr>            <list>
 1           Abilene TX <tibble [19 x 2]>
 2             Akron OH <tibble [19 x 2]>
 3           Alameda CA <tibble [19 x 2]>
 4            Albany GA <tibble [19 x 2]>
 5            Albany NY <tibble [19 x 2]>
 6            Albany OR <tibble [19 x 2]>
 7       Albuquerque NM <tibble [19 x 2]>
 8        Alexandria LA <tibble [19 x 2]>
 9        Alexandria VA <tibble [19 x 2]>
10          Alhambra CA <tibble [19 x 2]>
11       Aliso Viejo CA <tibble [19 x 2]>
12             Allen TX <tibble [19 x 2]>
13         Allentown PA <tibble [19 x 2]>
14             Aloha OR <tibble [19 x 2]>
15          Altadena CA <tibble [19 x 2]>
16 Altamonte Springs FL <tibble [19 x 2]>
17           Altoona PA <tibble [19 x 2]>
18          Amarillo TX <tibble [19 x 2]>
19              Ames IA <tibble [19 x 2]>
20           Anaheim CA <tibble [19 x 2]>

现在data中的每个数据框架看起来像这样。

df3$data[[1]]
# A tibble: 19 x 2
                  name1  distance
                  <chr>     <dbl>
 1             Akron OH 1881904.4
 2           Alameda CA 2128576.9
 3            Albany GA 1470577.2
 4            Albany NY 2542025.1
 5            Albany OR 2429367.3
 6       Albuquerque NM  702287.5
 7        Alexandria LA  700093.2
 8        Alexandria VA 2161594.6
 9          Alhambra CA 1718967.5
10       Aliso Viejo CA 1681868.8
11             Allen TX  296560.4
12         Allentown PA 2342363.5
13             Aloha OR 2457938.8
14          Altadena CA 1719207.6
15 Altamonte Springs FL 1805480.9
16           Altoona PA 2102993.0
17          Amarillo TX  361520.0
18              Ames IA 1194234.7
19           Anaheim CA 1694698.9