如何删除R中的节点子节点
这与我几个小时前刚问的问题有关,这是linkv 现在我需要删除数据集中的所有子节点。如果我选择了如何删除R中的节点子节点,r,R,这与我几个小时前刚问的问题有关,这是linkv 现在我需要删除数据集中的所有子节点。如果我选择了dt$source==“learn1”,则“learn1”下的任何节点都将消失,因此输出将如下c(“learn1”、“learn2”、“learn3”、“learn4”、“learn5”) 我使用@MrFlick和@iancampbell建议的igraph,但我找不到解决方案 dt <- data.frame(source = c("learn","learn1&
dt$source==“learn1”
,则“learn1”下的任何节点都将消失,因此输出将如下c(“learn1”、“learn2”、“learn3”、“learn4”、“learn5”)
我使用@MrFlick和@iancampbell建议的igraph
,但我找不到解决方案
dt <- data.frame(source = c("learn","learn1", "disc","learn2","learn3","disc1","lb","learn4", "learn5"),
new = c("learn","learn","disc","learn1","learn1","disc","lb","learn2","learn4"))
> dt
source new
1 learn learn
2 learn1 learn
3 disc disc
4 learn2 learn1
5 learn3 learn1
6 disc1 disc
7 lb lb
8 learn4 learn2
9 learn5 learn4
dt
来源新
1学习
2学1学
3光盘
4学2学1
5学3学1
6碟1碟
7磅
8学习4学习2
9学5学4
我认为在本例中,您正在查找子组件
以查找所有可访问的节点
library(igraph)
gg <- graph_from_data_frame(dt)
names(subcomponent(gg, "learn1", mode = "in"))
# [1] "learn1" "learn2" "learn3" "learn4" "learn5"
库(igraph)
gg我认为在本例中,您正在寻找子组件
,以查找所有可访问的节点
library(igraph)
gg <- graph_from_data_frame(dt)
names(subcomponent(gg, "learn1", mode = "in"))
# [1] "learn1" "learn2" "learn3" "learn4" "learn5"
库(igraph)
gg这里是另一个选项(但我认为通过subcomponent
by的方法更简单)
g这里是另一种选择(但我认为,采用subcomponent
by的方法更简单)
g谢谢,我也使用子组件,但我错过了mode=“in”谢谢,我也使用子组件,但我错过了mode=“in”