2

如果我有一个 tidygraph 对象列表，我想根据某些条件删除列表元素。例如，如果我有一个整洁的图形对象列表，如下所示：

nodes <- data.frame(name = c("Hadley", "David", "Romain", "Julia"))
edges <- data.frame(from = c(1, 1, 1, 2, 3, 3, 4, 4, 4),
                    to = c(2, 3, 4, 1, 1, 2, 1, 2, 3))
nodes_1 <- data.frame(name = c(NA))
edges_1 <- c()

tg <- tbl_graph(nodes = nodes, edges = edges)
tg_1 <- tbl_graph(nodes = nodes_1, edges = edges_1)

myList <- list(tg, tg_1)

查看 的输出，myList我们可以看到其中一个 tidygraph 对象没有与之关联的边，并且names列是NA.

> myList
[[1]]
# A tbl_graph: 4 nodes and 9 edges
#
# A directed simple graph with 1 component
#
# Node Data: 4 × 1 (active)
  name  
  <chr> 
1 Hadley
2 David 
3 Romain
4 Julia 
#
# Edge Data: 9 × 2
   from    to
  <int> <int>
1     1     2
2     1     3
3     1     4
# … with 6 more rows

[[2]]
# A tbl_graph: 1 nodes and 0 edges
#
# A rooted tree
#
# Node Data: 1 × 1 (active)
  name 
  <lgl>
1 NA   
#
# Edge Data: 0 × 2
# … with 2 variables: from <int>, to <int>

我想知道是否有一种方法可以遍历列表并检查是否有任何 tidygraph 对象没有边缘数据或者NA（如上面的示例）并从列表中删除该对象。

我知道我可以使用手动删除列表myList <- myList[-2]。但是当我的列表中有大量 tidygraph 对象时，我正在寻找更通用的解决方案？

4

2 回答 2

1

我们可以做

library(purrr)
library(tibble)
library(dplyr)
keep(myList, ~ .x %>% 
              as_tibble %>%
              pull(name) %>%
              is.na %>% 
              `!` %>% 
               any)

-输出

[[1]]
# A tbl_graph: 4 nodes and 9 edges
#
# A directed simple graph with 1 component
#
# Node Data: 4 x 1 (active)
  name  
  <chr> 
1 Hadley
2 David 
3 Romain
4 Julia 
#
# Edge Data: 9 x 2
   from    to
  <int> <int>
1     1     2
2     1     3
3     1     4
# … with 6 more rows
于 2021-09-09T01:31:57.947 回答
1

您可以过滤使用igraph::gsize()返回边缘计数的列表：

library(tidygraph)
library(igraph)

Filter(function(x) gsize(x) > 0, myList)  # Filter(gsize, myList) also works but potentially less safe.

# Or

purrr::discard(myList, ~ gsize(.x) == 0)

[[1]]
# A tbl_graph: 4 nodes and 9 edges
#
# A directed simple graph with 1 component
#
# Node Data: 4 x 1 (active)
  name  
  <chr> 
1 Hadley
2 David 
3 Romain
4 Julia 
#
# Edge Data: 9 x 2
   from    to
  <int> <int>
1     1     2
2     1     3
3     1     4
# ... with 6 more rows
于 2021-09-09T01:37:22.267 回答