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我有可以用两列表示的共现数据。每列中的条目来自同一组可能性。最终，我的目标是绘制一个有向网络，但首先我想将表格分成倒数（即 X->Y 和 Y->X）和仅在一个方向上发生的那些（即只有 Y->Z ）。这是一个例子：

library(tidyverse)

# Example data
from <-  c("A", "B", "F", "Q", "T", "S", "D", "E", "A", "T", "F")
to <- c("E", "D", "Q", "S", "F", "T", "B", "A", "D", "A", "E")
df <- data_frame(from, to)
df
# A tibble: 11 x 2
   from  to   
   <chr> <chr>
 1 A     E    
 2 B     D    
 3 F     Q    
 4 Q     S    
 5 T     F    
 6 S     T    
 7 D     B    
 8 E     A    
 9 A     D    
10 T     A    
11 F     E

这是我想要的输出：

# Desired output 1 - reciprocal co-occurrences
df %>% 
  slice(c(1,2)) %>% 
  rename(item1 = from, item2 = to)

# A tibble: 2 x 2
  item1 item2
  <chr> <chr>
1 A     E    
2 B     D

# Desired output 2 - single occurrences
df %>% 
  slice(c(3,4,6,6,9,10,11))

# A tibble: 7 x 2
  from  to   
  <chr> <chr>
1 F     Q    
2 Q     S    
3 S     T    
4 S     T    
5 A     D    
6 T     A    
7 F     E

如果共现是互惠的，则条目的顺序无关紧要，我只需要它们的名称共现不需要知道方向。

这感觉像是一个图问题，所以我试了一下，但我不熟悉使用这种类型的数据，而且大多数教程似乎都涵盖了无向图。查看tidygraph我理解的igraph包使用我尝试过的包：

library(tidygraph)

df %>% 
  as_tbl_graph(directed = TRUE) %>%
  activate(edges) %>% 
  mutate(recip_occur = edge_is_mutual()) %>% 
  as_tibble() %>%
  filter(recip_occur == TRUE) 
# A tibble: 4 x 3
   from    to recip_occur
  <int> <int> <lgl>      
1     1     8 TRUE       
2     2     7 TRUE       
3     7     2 TRUE       
4     8     1 TRUE

然而，这使边缘与节点分离并重复相互的共现。有没有人有这种数据的经验？

4

1 回答 1

1

尝试这个：

数据：

from <-  c("A", "B", "F", "Q", "T", "S", "D", "E", "A", "T", "F")
to <- c("E", "D", "Q", "S", "F", "T", "B", "A", "D", "A", "E")
df <- data_frame(from, to)

代码：

recursive_IND <-
1:nrow(df) %>% 
sapply(function(x){
    if(any((as.character(df[x,]) == t(df[,c(2,1)])) %>% {.[1,] & .[2,]}))
        return(T) else return(F)
    })

df[recursive_IND,][!(df[recursive_IND,] %>% apply(1,sort) %>% t %>% duplicated(.)),]
df[!recursive_IND,]

结果：

# A tibble: 2 x 2
#  from  to   
#  <chr> <chr>
#1 A     E    
#2 B     D    

# A tibble: 7 x 2
#  from  to   
#  <chr> <chr>
#1 F     Q    
#2 Q     S    
#3 T     F    
#4 S     T    
#5 A     D    
#6 T     A    
#7 F     E
于 2018-07-25T08:19:51.833 回答