在社交网络分析中查找间接节点（在 R 中，使用 dplyr）

Question

我想在 R 中使用 dplyr 解决以下问题。这个问题已在此处使用 data.table 得到解答：但因为我的代码的其余部分使用 dplyr我需要适应它。

我有关于在指定医院一起工作的医生组的信息。一名医生可以同时在多家医院工作。我想编写一个代码，输出在给定医院工作的给定医生的所有间接同事的信息。例如，如果我在某家医院与另一位也在另一家医院工作的医生一起工作，我想知道与我的同事一起在另一家医院工作的医生是谁。

考虑三个医院（1、2、3）和五个医生（A、B、C、D、E）的简单示例。 A、B、C医师在医院合作1.A、B、D医师在医院合作2.B、E医师在医院合作3.

对于在给定医院工作的每位医生，我希望通过他们的每位直接同事了解他们的间接同事的信息。例如，医生 A 通过医院 1 的医生 B 有一个间接同事：这是医院 3 的医生 E。另一方面，医生 B 在医院 1 没有通过医生 A 的间接同事。医生 C 有两个间接同事通过医院1的医生B：他们是医院2的医生D和医院3的医生E。依此类推..

以下是描述各医院医师网络的对象：

edges <- tibble(hosp  = c("1", "1", "1", "1", "1", "1", "2", "2", "2", "2", "2", "2", "3", "3"), 
             from = c("A", "A", "B", "B", "C", "C", "A", "A", "B", "B", "D", "D", "B", "E"), 
             to   = c("C", "B", "C", "A", "B", "A", "D", "B", "A", "D", "A", "B", "E", "B")) %>% arrange(hosp, from, to)

我想要一个产生以下输出的代码：

output <- tibble(hosp     = c("1", "1", "1", "1", "1", "1", "1", "2", "2", "2", "2", "2", "2", "2", "3", "3", "3", "3", "3"), 
             from     = c("A", "A", "B", "B", "C", "C", "C", "A", "A", "B", "B", "D", "D", "D", "B", "E", "E", "E", "E"), 
             to       = c("C", "B", "C", "A", "B", "A", "B", "D", "B", "A", "D", "A", "B", "B", "E", "B", "B", "B", "B"),
             hosp_ind = c("" , "3", "" , "" , "2", "2", "3", "" , "3", "" , "" , "1", "1", "3", "" , "1", "1", "2", "2"),
             to_ind   = c("" , "E", "" , "" , "D", "D", "E", "" , "E", "" , "" , "C", "C", "E", "" , "A", "C", "A", "D")) %>% arrange(hosp, from, to)

Answer 1

因为看起来你只想要网络中的第一层间接连接，所以没有图数据结构很简单。

get_indirects <- function(hosp_from) {
    x=hosp_from$from[1]
    hosp=hosp_from$hosp[1]
    directs <- edges %>% 
        filter(from==x) %>% 
        pull(to)
    indirects <- edges %>%
        filter(from %in% directs & !(to %in% append(directs,x))) %>% 
        rename(to = from, hosp_ind = hosp, to_ind = to) %>% 
        select(to, hosp_ind, to_ind) %>% 
        mutate(hosp=hosp,from=x,.before=to)
}

split_edges <- edges %>% 
    group_by(hosp,from) %>% 
    group_split()

indirect_df <- lapply(split_edges, get_indirects) %>% bind_rows()

direct_df <- anti_join(edges, indirect_df[,c("from","to")], by = c("from","to"))

output <- bind_rows(indirect_df,direct_df) %>% 
    replace_na(list(hosp_ind="",to_ind="")) %>% 
    arrange(hosp,from,to)

这给出了与示例的预期输出相同的输出。

Answer 2

其实你可以通过以下方式将data.table翻译成dplyr

g <- simplify(graph_from_data_frame(edges, directed = FALSE))
edges %>%
  rowwise() %>%
  do(cbind(., {
    to_ind <- setdiff(
      do.call(
        setdiff,
        Map(names, ego(g, 2, c(.$to, .$from), mindist = 2))
      ), .$from
    )
    if (!length(to_ind)) {
      hosp_ind <- to_ind <- NA_character_
    } else {
      hosp_ind <- lapply(to_ind, function(v) names(neighbors(g, v)))
    }
    data.frame(
      hosp_ind = unlist(hosp_ind),
      to_ind = rep(to_ind, lengths(hosp_ind))
    )
  }))

这给了你

# A tibble: 19 x 5
   hosp  from  to    hosp_ind to_ind
   <chr> <chr> <chr> <chr>    <chr>
 1 1     A     B     3        E
 2 1     A     C     NA       NA
 3 1     B     A     NA       NA
 4 1     B     C     NA       NA
 5 1     C     A     2        D
 6 1     C     B     2        D
 7 1     C     B     3        E
 8 2     A     B     3        E
 9 2     A     D     NA       NA
10 2     B     A     NA       NA
11 2     B     D     NA       NA
12 2     D     A     1        C
13 2     D     B     1        C
14 2     D     B     3        E
15 3     B     E     NA       NA
16 3     E     B     1        A
17 3     E     B     2        A
18 3     E     B     1        C
19 3     E     B     2        D

在社交网络分析中查找间接节点（在 R 中，使用 dplyr）

Finding indirect nodes in social network analysis (in R, using dplyr)

r

igraph

nodes

dplyr

在社交网络分析中查找间接节点（在 R 中，*使用 dplyr*）

Finding indirect nodes in social network analysis (in R, *using dplyr*)

r

igraph

nodes

dplyr

在社交网络分析中查找间接节点（在 R 中，使用 dplyr）

Finding indirect nodes in social network analysis (in R, using dplyr)