在TidyGraph中计算Everett-Valente经纪人得分
我想为我的定向网络中的每个节点计算Everett-Valente经纪评分（Everett和Valente 2016）。该分数基于中间性中心。本质上，这控制了网络大小。代理控制信息/资源流的能力受网络规模和/或联系冗余的影响。对于无向图，可以这样计算Everett-Valente经纪人得分：
- 计算节点之间的中心性。
- 将每个节点的计算中间性中心度加倍，并将（n-1）添加到每个非下垂条目。
- 将每个非零得分除以节点的度数。
我计划使用if_else语句来处理非下垂和零分，例如
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）。为了将此扩展到定向案例，Everett和Valente（2016）提供了以下规则：
对于电动汽车经纪人：
- 计算v的节点中间性。
- 如果节点中间性中心度= 0，则添加j，其中j =可以达到v的顶点数。
- 将每个非零和除以v的度数。
对于电动汽车经纪人：
- 计算v的节点中间性。
- 如果节点中间性中心度= 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。
v的EV经纪人=内向和外向的平均值。
如果有人可以通过mutate（）语句帮助我，我将不胜感激。我想知道如何在有向情况下计算j和k，并在无向情况下找出非悬挂节点。
2 个答案:
答案 0 :(得分：1)
如果您将其简化为一个独立函数来计算igraph对象的分数，那么（将其推广）这个理由将容易得多。然后，可以将其调整为对整笔记录友好的内容。
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
这是一个有向图示例...
(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()
您可以将ifelse()包装在dplyr::case_when()中，而不是将(
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
彼此嵌套（但它仍应位于可以测试和验证的适当函数中）。
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 0
这是检查数学的完整表格：
import UIKit
import Firebase
import FirebaseUI
@UIApplicationMain
class AppDelegate: UIResponder, UIApplicationDelegate,FUIAuthDelegate {
var window: UIWindow?
func application(_ application: UIApplication, didFinishLaunchingWithOptions launchOptions: [UIApplication.LaunchOptionsKey: Any]?) -> Bool {
// Override point for customization after application launch.
FirebaseApp.configure()
let authUI = FUIAuth.defaultAuthUI()
// You need to adopt a FUIAuthDelegate protocol to receive callback
authUI!.delegate = self
let providers: [FUIAuthProvider] = [
FUIGoogleAuth(),
FUIPhoneAuth(authUI:FUIAuth.defaultAuthUI()!),
]
authUI!.providers = providers
return true
}
func application(_ app: UIApplication, open url: URL,
options: [UIApplication.OpenURLOptionsKey : Any]) -> Bool {
let sourceApplication = options[UIApplication.OpenURLOptionsKey.sourceApplication] as! String?
if FUIAuth.defaultAuthUI()?.handleOpen(url, sourceApplication: sourceApplication) ?? false {
return true
}
// other URL handling goes here.
return false
}
}
答案 1 :(得分：0)
在核实Everett和Valente（2016）中提供的Campnet示例后，定向网络的EV经纪人得分可以计算如下：
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Everett和Valente（2016）中提出的Granovetter（1973）的假设无向网络，可以计算出EV经纪人得分：
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），我尚未对电动汽车经纪人评分进行标准化。
- 我写了这段代码，但我无法理解我的错误
- 我无法从一个代码实例的列表中删除 None 值，但我可以在另一个实例中。为什么它适用于一个细分市场而不适用于另一个细分市场？
- 是否有可能使 loadstring 不可能等于打印？卢阿
- java中的random.expovariate()
- Appscript 通过会议在 Google 日历中发送电子邮件和创建活动
- 为什么我的 Onclick 箭头功能在 React 中不起作用？
- 在此代码中是否有使用“this”的替代方法？
- 在 SQL Server 和 PostgreSQL 上查询，我如何从第一个表获得第二个表的可视化
- 每千个数字得到
- 更新了城市边界 KML 文件的来源？