使用tidygraph实现igraph中的回路自动化
你好,希望一切顺利。 我对上一个问题进行了修改,希望它能让问题更清楚 我创建了一个使用tidygraph实现igraph中的回路自动化,r,loops,dplyr,igraph,R,Loops,Dplyr,Igraph,你好,希望一切顺利。 我对上一个问题进行了修改,希望它能让问题更清楚 我创建了一个igraph对象,希望多次运行相同的分析,并在每次迭代中提取一些信息 我不能共享整个数据,所以我只共享一小部分。 df_edge如下所示: library(dplyr) job_1 <-c(1,2,6,6,5,6,7,8,6,8,8,6,6,8) job_2 <- c(2,4,5,8,3,1,4,6,1,7,3,2,4,5) weight <- c(1,1,1,2,1,1,2,1,1,1,2,1,
igraph
对象,希望多次运行相同的分析,并在每次迭代中提取一些信息
我不能共享整个数据,所以我只共享一小部分。
df_edge
如下所示:
library(dplyr)
job_1 <-c(1,2,6,6,5,6,7,8,6,8,8,6,6,8)
job_2 <- c(2,4,5,8,3,1,4,6,1,7,3,2,4,5)
weight <- c(1,1,1,2,1,1,2,1,1,1,2,1,1,1)
df_edge <- tibble(job_1,job_2,weight)
df_edge %>% glimpse()
Rows: 14
Columns: 3
$ job_1 <dbl> 1, 2, 6, 6, 5, 6, 7, 8, 6, 8, 8, 6, 6, 8
$ job_2 <dbl> 2, 4, 5, 8, 3, 1, 4, 6, 1, 7, 3, 2, 4, 5
$ weight <dbl> 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1
job_id <- c(1,2,3,4,5,6,7,8)
job_type <- c(1,2,0,0,3,1,1,1)
df_node <- tibble(job_id,job_type)
df_node %>% glimpse()
Rows: 8
Columns: 2
$ job_id <dbl> 1, 2, 3, 4, 5, 6, 7, 8
$ job_type <dbl> 1, 2, 0, 0, 3, 1, 1, 1
创建igraph
对象:
library(igraph)
library(tidygraph)
tp_network_subset <- graph.data.frame(df_edge,vertices = df_node,directed = F)
我正在手动执行以下操作:
### finding a job_id that belongs to job_type==1 category
df_node %>% filter(job_type==1) %>%
select(job_id)
A tibble: 4 x 1
job_id
<dbl>
1 1
2 6
3 7
4 8
# for instance, I picked one of them and it is job_id = 6
如果job_rate>0.5
,我希望保留子图的job_rate
和job_type=4
类别的行(对应节点)。在本例中,作业率为0.6
,因此我保留以下内容
df_final <- as_tbl_graph(node_test[[1]]) %>%
activate(nodes) %>%
filter(!node_is_isolated()) %>%
as_tibble() %>% filter(job_type==0)
# A tibble: 1 x 2
name job_type
<chr> <dbl>
1 4 0
因此,我需要执行此过程并为所有
job\u type==1
节点创建子图。如果图形长度不为零且其作业率>0.5
,则提取子图形中的所有对应节点,以及作业率
和收藏夹结果中显示的其他列。这对您有用吗
dflst <- split(df_node, job_type)
tpe <- as.numeric(names(dflst))
out <- tibble()
for (i in seq_along(dflst)) {
df <- dflst[[i]]
node_test_lst <- make_ego_graph(tp_network_subset, order = 1, nodes = df$job_id)
origin_id <- df$job_id
jtpe <- tpe[i]
for (j in seq_along(node_test_lst)) {
node_test <- node_test_lst[[j]]
df_test <- as_tbl_graph(node_test) %>%
activate(nodes) %>%
filter(!node_is_isolated()) %>%
as_tibble()
if (nrow(df_test %>% filter(job_type == 0)) > 0 & any(df_test$job_type %in% 1:3)) {
job_rate <- with(df_test, sum(job_type == jtpe) / sum(job_type %in% 1:3))
if (job_rate > 0.5) {
df_final <- df_test %>%
filter(job_type == 0) %>%
mutate(
subgraph_origin_id = origin_id[j],
job_rate = job_rate,
subgraph_size = nrow(df_test)
) %>%
cbind(
setNames(
as.list(table(factor(df_test$job_type, levels = 0:3))),
sprintf("no_(job_type==%s)_in_subgrapgh", 0:3)
)
)
out <- out %>% rbind(df_final)
}
}
}
}
@菲尔,谢谢你的编辑。你有什么办法帮忙吗?非常感谢!如果您能创建一个最小的可复制示例(示例数据),我可以尝试帮助您。@Brigadeiro,谢谢您的反馈。我只是提供了一些数据,希望能有所帮助。非常感谢!请(1)加载运行代码所需的软件包,(2)在说明手动解决问题的方法之前,请清楚说明您试图解决的问题。@ThomaslsCoding,感谢您提供答案。因此,我在list2DF(as.list(表(factor(df\u test$job\u type,levels=0:3))中得到了
错误:找不到函数“list2DF”
。我已经安装了library(base)
和library(base)
,但仍然收到相同的错误。@Alex对不起,我的错。现在我已经修好了。请重试。@Alex您是在您的帖子中的数据上尝试了我的代码还是我们的真实数据?我没有看到你的帖子数据有任何错误。@Alex不用担心。我认为这取决于你如何定义你的job\u rate
。从您的代码来看,似乎是作业类型=0
与作业类型=1,2或3的比率。在这种情况下,如果有许多行的值0
,但1、2或3的值很少,那么肯定会给出大于1
的速率。我不认为这是一个编码问题。它由您如何定义job\u rate
@Alex决定,您应该使用node\u test\lst
df_test %>% glimpse()
Rows: 6
Columns: 2
$ name <chr> "1", "2", "4", "5", "6", "8"
$ job_type <dbl> 1, 2, 0, 3, 1, 1
## subgraph size is 6 which will be an outcome of interest
### if the graph is zero length , I should stop here and pick another job_id that belongs to job_type==1 category
### calculating the measure of interest in respect to job_type==1 category
df_test %>%
summarise(job_rate= (nrow(df_test %>% filter(job_type==1)))/(nrow(df_test %>%
filter(job_type %in% c(1,2,3)))))
# 0.6
df_final <- as_tbl_graph(node_test[[1]]) %>%
activate(nodes) %>%
filter(!node_is_isolated()) %>%
as_tibble() %>% filter(job_type==0)
# A tibble: 1 x 2
name job_type
<chr> <dbl>
1 4 0
name job_type subgraph_origin_id job_rate subgraph_size no_(job_type==0)_in_subgrapgh no_(job_type==1)_in_subgrapgh no_(job_type==2)_in_subgrapgh no_(job_type==3)_in_subgrapgh
<chr> <dbl>
1 4 0 6 0.6 6
dflst <- split(df_node, job_type)
tpe <- as.numeric(names(dflst))
out <- tibble()
for (i in seq_along(dflst)) {
df <- dflst[[i]]
node_test_lst <- make_ego_graph(tp_network_subset, order = 1, nodes = df$job_id)
origin_id <- df$job_id
jtpe <- tpe[i]
for (j in seq_along(node_test_lst)) {
node_test <- node_test_lst[[j]]
df_test <- as_tbl_graph(node_test) %>%
activate(nodes) %>%
filter(!node_is_isolated()) %>%
as_tibble()
if (nrow(df_test %>% filter(job_type == 0)) > 0 & any(df_test$job_type %in% 1:3)) {
job_rate <- with(df_test, sum(job_type == jtpe) / sum(job_type %in% 1:3))
if (job_rate > 0.5) {
df_final <- df_test %>%
filter(job_type == 0) %>%
mutate(
subgraph_origin_id = origin_id[j],
job_rate = job_rate,
subgraph_size = nrow(df_test)
) %>%
cbind(
setNames(
as.list(table(factor(df_test$job_type, levels = 0:3))),
sprintf("no_(job_type==%s)_in_subgrapgh", 0:3)
)
)
out <- out %>% rbind(df_final)
}
}
}
}
> out
name job_type subgraph_origin_id job_rate subgraph_size
1 4 0 6 0.60 6
2 4 0 7 1.00 3
3 3 0 8 0.75 5
no_(job_type==0)_in_subgrapgh no_(job_type==1)_in_subgrapgh
1 1 3
2 1 2
3 1 3
no_(job_type==2)_in_subgrapgh no_(job_type==3)_in_subgrapgh
1 1 1
2 0 0
3 0 1