【发布时间】:2018-01-03 14:38:13
【问题描述】:
【问题讨论】:
标签: r reshape gephi network-analysis
【问题讨论】:
标签: r reshape gephi network-analysis
我认为这段代码应该可以完成这项工作。这不是最好的最优雅的方式,但它有效:)
# Data
x <-
structure(
list(
Persons = c(1L, 1L, 1L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L),
Organizations = c("A", "B", "E", "F", "A", "E", "C", "D", "C", "A", "E")
),
.Names = c("Persons", "Organizations"),
class = "data.frame",
row.names = c(NA, -11L)
)
# This will merge n:n
edgelist <- merge(x, x, by = "Organizations")[,2:3]
# We don't want autolinks
edgelist <- subset(edgelist, Persons.x != Persons.y)
# Removing those that are repeated
edgelist <- unique(edgelist)
edgelist
#> Persons.x Persons.y
#> 2 1 3
#> 3 1 2
#> 4 3 1
#> 6 3 2
#> 7 2 1
#> 8 2 3
嗨
由reprex package (v0.1.1.9000) 于 2018 年 1 月 3 日创建。
【讨论】:
以x开头:
structure(list(Persons = c(1L, 1L, 1L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L), Organizations = c("A", "B", "E", "F", "A", "E", "C", "D", "C", "A", "E")), .Names = c("Persons", "Organizations"), class = "data.frame", row.names = c(NA,-11L))
创建一个具有不同名称的新data.frame。只需将Organizations 转换为因子,然后使用数值:
> y=data.frame(Source=x$Persons, Target=as.numeric(as.factor(x$Organizations)))
> y
Source Target
1 1 1
2 1 2
3 1 5
4 2 6
5 2 1
6 2 5
7 2 3
8 3 4
9 3 3
10 3 1
11 3 5
不管怎样,我很确定 gephi 可以处理字符串。
【讨论】: