您可以在此处使用full_join()：

full_join(
  d %>% filter(is.na(b_name)) %>% select(-b_name),
  d %>% filter(is.na(a_name)) %>% select(b_name, min.distance),
) %>% relocate(b_name, .before=latitude)

获得相同结果的另一种方法是将filter()步骤合并到对split()的单个调用中

d=split(d,is.na(d$a_name))
full_join(select(d[[1]], -b_name), select(d[[2]], c(b_name, min.distance)))

输出：

a_name   b_name latitude longitude min.distance
1   jongone   jongro  38.1241  126.1215     95.62616
2   jongtwo dongmoon  37.1161  127.1345     12.53152
3 jongthree  kangseo  37.1562  126.9762     73.95093
4  jongfour     <NA>  37.1551  127.6262     17.12411
5  jongfive   jamsil  36.1515  127.9627    175.15416
6      <NA> kangbook       NA        NA    128.17918
7      <NA>   nolboo       NA        NA     85.15162
8      <NA>    ilsan       NA        NA     16.11515

输入：

d = structure(list(a_name = c(NA, NA, NA, NA, NA, NA, NA, "jongone", 
"jongtwo", "jongthree", "jongfour", "jongfive"), b_name = c("kangseo", 
"kangbook", "jongro", "dongmoon", "jamsil", "nolboo", "ilsan", 
NA, NA, NA, NA, NA), latitude = c(37.1562, 38.1255, 37.8226, 
37.1161, 36.1215, 37.3213, 37.1111, 38.1241, 37.1161, 37.1562, 
37.1551, 36.1515), longitude = c(126.9762, 126.1952, 127.1724, 
127.1345, 127.9722, 127.3211, 127.1331, 126.1215, 127.1345, 126.9762, 
127.6262, 127.9627), min.distance = c(73.950928, 128.179185, 
95.626161, 12.531519, 175.154161, 85.151616, 16.115151, 95.626161, 
12.531519, 73.950928, 17.124115, 175.154161)), class = "data.frame", row.names = c(NA, 
-12L))

我们可以按"min.distance"分组，并将NA元素修改为非NA值，从而获得distinct行，而无需进行任何连接

library(dplyr)
df1 %>% 
   group_by(min.distance) %>%
   mutate(across(c(a_name, b_name), ~ (.x[!is.na(na_if(.x, ""))][1]))) %>% 
   ungroup %>%
   distinct(min.distance, .keep_all = TRUE)

• 输出

# A tibble: 8 × 5
  a_name    b_name   latitude longitude min.distance
  <chr>     <chr>       <dbl>     <dbl>        <dbl>
1 jongthree kangseo      37.2      127.         74.0
2 <NA>      kangbook     38.1      126.        128. 
3 jongone   jongro       37.8      127.         95.6
4 jongtwo   dongmoon     37.1      127.         12.5
5 jongfive  jamsil       36.1      128.        175. 
6 <NA>      nolboo       37.3      127.         85.2
7 <NA>      ilsan        37.1      127.         16.1
8 jongfour  <NA>         37.2      128.         17.1

数据

df1 <- structure(list(a_name = c(NA, NA, NA, NA, NA, NA, NA, "jongone", 
"jongtwo", "jongthree", "jongfour", "jongfive"), b_name = c("kangseo", 
"kangbook", "jongro", "dongmoon", "jamsil", "nolboo", "ilsan", 
NA, NA, NA, NA, NA), latitude = c(37.1562, 38.1255, 37.8226, 
37.1161, 36.1215, 37.3213, 37.1111, 38.1241, 37.1161, 37.1562, 
37.1551, 36.1515), longitude = c(126.9762, 126.1952, 127.1724, 
127.1345, 127.9722, 127.3211, 127.1331, 126.1215, 127.1345, 126.9762, 
127.6262, 127.9627), min.distance = c(73.950928, 128.179185, 
95.626161, 12.531519, 175.154161, 85.151616, 16.115151, 95.626161, 
12.531519, 73.950928, 17.124115, 175.154161)), class = "data.frame", 
row.names = c(NA, 
-12L))

如果不想依赖于min.distance值的精确匹配，您可以尝试真正的空间匹配（并对距离差异进行一定容差的过滤），如下所示：

library(sf) ## library for spatial processing

根据a_name的is.na()状态，将df拆分为两个 Dataframe 的列表，并将列表成员命名为A和B：

df_list <- setNames(
  df1 |>  split(is.na(df1$a_name)),
  c('A', 'B')
)

一次性将两个 Dataframe 转换为空间 Dataframe （包含geometry column）（通过Map列表）：

df_list <- 
  df_list |> 
  map( ~ st_as_sf(.x, coords = c('latitude', 'longitude'))
)

为A的每一行找到B的最近特征（行），并将该行列绑定到A：

df_list$A |>
  bind_cols(
    df_list$B |>
    slice(st_nearest_feature(df_list$A, df_list$B))
  ) |>
## select (only) and rename desired columns):
select('a_name' = 1, 'b_name' = 6,
       'a_min_dist' = 3, 'b_min_dist' = 7
       ) |>
mutate(mutual_distance = abs(a_min_dist - b_min_dist))

输出

##      a_name   b_name a_min_dist b_min_dist mutual_distance
## 1   jongone kangbook   95.62616  128.17918        32.55302
## 2   jongtwo dongmoon   12.53152   12.53152         0.00000
## 3 jongthree  kangseo   73.95093   73.95093         0.00000
## 4  jongfour   nolboo   17.12411   85.15162        68.02750
## 5  jongfive   jamsil  175.15416  175.15416         0.00000

现在您可以按可接受的mutual_distance进行过滤。