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我正在处理的数据是一个矩阵，其中矩阵中的每一列都是一个包含 2 个元素的列表。我要做的是计算有多少列是相同的。

我正在从tidygraph对象中提取列表矩阵。我下面的例子应该更好地解释我的问题。首先，我创建一些数据，将它们转换为tidygraph对象并将它们全部放入一个列表中，如下所示：

library(ggraph)
library(tidygraph)

# create some nodes and edges data
nodes  <- data.frame(name = c("x4", NA, NA))
nodes1 <- data.frame(name = c("x4", "x2", NA, NA, "x1", NA, NA))
nodes2 <- data.frame(name = c("x1", NA, NA))
nodes3 <- data.frame(name = c("x6", NA, NA))
nodes4 <- data.frame(name = c("x10", "x3", NA, NA, NA))
nodes5 <- data.frame(name = c("x1", "x2", NA, NA, "x7", NA, NA))

edges  <- data.frame(from = c(1,1), to = c(2,3))
edges1 <- data.frame(from = c(1, 2, 2, 1, 5, 5), to = c(2, 3, 4, 5, 6, 7))
edges2 <- data.frame(from = c(1,1), to = c(2,3))
edges3 <- data.frame(from = c(1,1), to = c(2,3))
edges4 <- data.frame(from = c(1,2,2,1), to = c(2,3,4,5))
edges5 <- data.frame(from = c(1, 2, 2, 1, 5, 5), to = c(2, 3, 4, 5, 6, 7))


# create the tbl_graphs
tg   <- tbl_graph(nodes = nodes,  edges = edges)
tg_1 <- tbl_graph(nodes = nodes1, edges = edges1)
tg_2 <- tbl_graph(nodes = nodes2, edges = edges2)
tg_3 <- tbl_graph(nodes = nodes3, edges = edges3)
tg_4 <- tbl_graph(nodes = nodes4, edges = edges4)
tg_5 <- tbl_graph(nodes = nodes5, edges = edges5)


# put into list
myList <- list(tg, tg_1, tg_2, tg_3, tg_4, tg_5)

为清楚起见，查看以下元素之一myList：

myList[1]
[[1]]
# A tbl_graph: 3 nodes and 2 edges
#
# A rooted tree
#
# Node Data: 3 × 1 (active)
  name 
  <chr>
1 x4   
2 NA   
3 NA   
#
# Edge Data: 2 × 2
   from    to
  <int> <int>
1     1     2
2     1     3

本质上，我想做的是遍历列表的每个元素，查看边缘数据，看看有多少是相同的。我确信有多种方法可以做到这一点......但我尝试通过使用一个tidygraph函数来提取边缘数据并返回列表矩阵来做到这一点：

# extract just the edges data
resEdges <- sapply(myList, function(x) {
  nodes <- tidygraph::activate(x, edges) %>% 
    tibble::as_tibble()
})

同样，为了清楚起见，查看第 1 列resEdges如下所示：

> resEdges[,1]
$from
[1] 1 1

$to
[1] 2 3

所以，我想做的是遍历resEdges's 列并计算相同列的频率。

在我的示例中，只有 3 个唯一列。所以，我想要的输出看起来像这样：

> edgeFreq
# A tibble: 3 × 3
  from             to             frequency
  1 1              2 3            3
  1 2 2 1 5 5      2 3 4 5 6 7    2
  1 2 2 1          2 3 4 5        1
4

1 回答 1

1 
myList %>%
  map_chr(~as_tibble(activate(.x, edges))%>%
        map_chr(str_c, collapse = " ") %>%
        toString())%>%
  table()%>%
  as_tibble() %>%
  setNames(c("data", "frequency")) %>%
  separate(data, c("From", "to"), ", ")
  
# A tibble: 3 x 3
  From        to          frequency
  <chr>       <chr>           <int>
1 1 1         2 3                 3
2 1 2 2 1 5 5 2 3 4 5 6 7         2
3 1 2 2 1     2 3 4 5             1
于 2021-11-01T19:54:45.730 回答