在TidyGraph中计算Everett-Valente经纪评分
在TidyGraph中计算Everett-Valente经纪评分
提问于 2018-10-03 04:36:42
我想计算我的定向网络中每个节点的Everett-Valente Brokerage得分(Everett和Valente 2016)。这个分数是基于中间的中心性。本质上，这控制了网络的大小。代理控制信息/资源流的能力受到网络大小和/或绑定冗余的限制。对于无向图，计算Everett - Valente Brokerage评分如下：
- 计算节点之间的中心性。
- 将每个节点的计算中间度中心度加倍，并将(n - 1)添加到每个非挂起项中。
- 将每个非零分数除以节点的程度。
我计划使用if_else语句来处理非悬挂式和零分。
g <- g %>%
activate(nodes) %>%
mutate(betweenness = centrality_betweenness(),
ev_brokerage = if_else(..if_else(..)..))我不知道如何实现ev_brokerage (条件语句)。为了将其适用于直接案件，Everett和Valente (2016)规定了以下规则：
-EV经纪业务：
- 计算节点间的中心性为v。
- 如果节点之间的中心性=0，则添加j，其中j=可以到达v的顶点数。
- 将每个非零和除以v的内次。
为out-EV经纪：
- 计算节点间的中心性为v。
- 如果节点之间的中心度=0，则添加k，其中k=v可以到达的顶点数。
- 将每个非零和除以v的出度。
EV的经纪业务v= in-EV和out-EV的平均值。
如果有人能帮我处理变体()的陈述，我将不胜感激。我想知道如何在有向情况下求出j和k，并在无向情况下计算出非挂起的节点。
回答 2
Stack Overflow用户
回答已采纳
发布于 2018-10-03 21:36:58
如果你只是把它变成一个独立的函数，计算出一个into对象的分数，那么这个推理(和概括)就会简单得多。然后，它可以被调整成对版面图友好的东西。
suppressPackageStartupMessages(library(tidygraph))
if_else <- dplyr::if_else
case_when <- dplyr::case_when
map2_dbl <- purrr::map2_dbl使用无向图非常简单，因为您不需要嵌套任何控制流。
create_notable("Zachary") %>%
mutate(pendant = centrality_degree() == 1, # is a node a pendant?
btwn = centrality_betweenness()) %>% # raw betweenness
mutate(ev_step1 = if_else(pendant, # if it's a pendant...
btwn * 2, # double betweenness...
btwn * 2 + (graph_order() - 1)), # else double it AND subtract n (nodes) - 1
ev_brok = if_else(ev_step1 == 0, # if it's 0...
ev_step1, # leave it as is...
ev_step1 / centrality_degree()) # else divide it by raw degree
) %>%
select(ev_brok, btwn, pendant)
#> # A tbl_graph: 34 nodes and 78 edges
#> #
#> # An undirected simple graph with 1 component
#> #
#> # Node Data: 34 x 3 (active)
#> ev_brok btwn pendant
#> <dbl> <dbl> <lgl>
#> 1 30.9 231. FALSE
#> 2 10.00 28.5 FALSE
#> 3 18.5 75.9 FALSE
#> 4 7.60 6.29 FALSE
#> 5 11.2 0.333 FALSE
#> 6 16.2 15.8 FALSE
#> # ... with 28 more rows
#> #
#> # Edge Data: 78 x 2
#> from to
#> <int> <int>
#> 1 1 2
#> 2 1 3
#> 3 1 4
#> # ... with 75 more rows这里有一个有向图..。
(g <- matrix(c(1, 2,
1, 3,
3, 4,
4, 1,
2, 5,
5, 6, # 6 is pendant with in-tie
7, 2, # 7 is pendant with out-ie
4, 8, # 8 is pendant with in-tie
9, 10,
10, 11,
11, 12, # 12 is a pendant with in-tie
11, 13,
9, 13),
ncol = 2, byrow = TRUE) %>%
igraph::graph_from_edgelist()) %>% plot()
与其在彼此内部嵌套ifelse()，不如用dplyr::case_when()包装它们(但它仍然应该放在一个可以测试和验证的适当函数中)。
(
res <- g %>%
as_tbl_graph() %>%
mutate(btwn = centrality_betweenness(),
in_reach = local_size(order = graph_order(), mode = "in") - 1, # reach being max. ego graph order - 1 for ego
out_reach = local_size(order = graph_order(), mode = "out") - 1,
in_deg = centrality_degree(mode = "in"),
out_deg = centrality_degree(mode = "out")) %>%
mutate(ev_in = case_when(
btwn == 0 ~ if_else(btwn + in_reach == 0, # if btwn is 0 and if btwn + in_reach is 0
btwn + in_reach, # then btwn + in_reach (0)
(btwn + in_reach) / in_deg), # else add btwn and in_reach, then divide by in_deg
btwn != 0 ~ btwn / in_deg
)) %>%
mutate(ev_out = case_when(
btwn == 0 ~ if_else(btwn + out_reach == 0,
btwn + out_reach,
(btwn + out_reach) / out_deg),
btwn != 0 ~ btwn / out_deg
)) %>%
mutate(ev_brok = map2_dbl(ev_in, ev_out, ~ mean(c(.x, .y)))) %>%
select(ev_brok, starts_with("ev_"), btwn, everything())
)
#> # A tbl_graph: 13 nodes and 13 edges
#> #
#> # A directed simple graph with 2 components
#> #
#> # Node Data: 13 x 8 (active)
#> ev_brok ev_in ev_out btwn in_reach out_reach in_deg out_deg
#> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
#> 1 5.25 7 3.5 7 2 6 1 2
#> 2 6 4 8 8 4 2 2 1
#> 3 2 2 2 2 2 6 1 1
#> 4 4.5 6 3 6 2 6 1 2
#> 5 5 5 5 5 5 1 1 1
#> 6 3 6 0 0 6 0 1 0
#> # ... with 7 more rows
#> #
#> # Edge Data: 13 x 2
#> from to
#> <int> <int>
#> 1 1 2
#> 2 1 3
#> 3 3 4
#> # ... with 10 more rows这是检查数学的完整表格：
res %>% as_tibble()
#> # A tibble: 13 x 8
#> ev_brok ev_in ev_out btwn in_reach out_reach in_deg out_deg
#> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
#> 1 5.25 7 3.5 7 2 6 1 2
#> 2 6 4 8 8 4 2 2 1
#> 3 2 2 2 2 2 6 1 1
#> 4 4.5 6 3 6 2 6 1 2
#> 5 5 5 5 5 5 1 1 1
#> 6 3 6 0 0 6 0 1 0
#> 7 1.5 0 3 0 0 3 0 1
#> 8 1.5 3 0 0 3 0 1 0
#> 9 1 0 2 0 0 4 0 2
#> 10 2 2 2 2 1 3 1 1
#> 11 2.25 3 1.5 3 2 2 1 2
#> 12 1.5 3 0 0 3 0 1 0
#> 13 0.75 1.5 0 0 3 0 2 0Stack Overflow用户
发布于 2018-10-09 03:28:23
在对照Everett和Valente (2016)中的露营网示例之后，可以计算出定向网络的EV经纪评分：
g <- g %>%
activate(nodes) %>%
# compute in-degree, out-degree, and betweenness centrality
mutate(betweenness = centrality_betweenness(),
in_degree = centrality_degree(mode = "in"),
out_degree = centrality_degree(mode = "out"),
in_reach = local_size(order = graph_order(), mode = "in") - 1,
out_reach = local_size(order = graph_order(), mode = "out") - 1) %>%
# compute everett-valente brokerage score
mutate(ev_in = if_else(betweenness != 0, betweenness + in_reach, betweenness),
ev_in = if_else(ev_in != 0, ev_in / in_degree, ev_in),
ev_out = if_else(betweenness != 0, betweenness + out_reach, betweenness),
ev_out = if_else(ev_out != 0, ev_out / out_degree, ev_out),
ev_brokerage = (ev_in + ev_out) / 2) 使用Granovetter (1973)在Everett和Valente (2016)中提出的假设无向网络，可以计算EV经纪业务分数如下：
edgelist <- data.frame(from = c(1,1,1,2,2,2,3,3,3,3,4,4,4,4,5,5,5,6,6,6,7,7,8,8,8,8,9,
9,10,10,10,11,11,11,11,11,12,12,12,13,13,13,13,14,14,
14,14,15,15,15,16,16,17,17,17,18,18,18,18,19,19,20,20,
20,20,20,21,21,22,22,22,23,23,23,24,24,24,25,25,25,25),
to = c(2,3,24,1,3,4,1,2,4,5,2,3,5,6,3,4,6,5,5,7,6,8,9,10,11,
14,8,10,9,8,11,10,8,12,14,13,11,14,13,11,12,14,15,8,11,
12,13,13,16,17,15,17,15,16,18,17,19,20,21,18,20,19,18,
21,25,22,18,20,20,25,23,24,25,22,1,25,23,24,23,22,20))
g <- igraph::graph_from_edgelist(as.matrix(edgelist), directed = F) %>% simplify()
g <- as_tbl_graph(g) %>%
activate(nodes) %>%
# compute brokerage
mutate(betweenness = centrality_betweenness(),
degree = centrality_degree(),
ev_condition = if_else(betweenness != 0, betweenness * 2 + graph_order() - 1, betweenness),
ev_brokerage = if_else(ev_condition != 0, ev_condition / degree, ev_condition))
data <- g %>% as.tibble()根据Everett和Valente (2016)，我还没有将EV经纪业务的得分正常化。
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/52619762
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