使用 R 创建关系矩阵

Question

我的数据框包含参与其中的不同个人的项目，以及项目执行的年份。

我如何为每一年创建一个 nxn 关系矩阵（n 是个人的数量）来计算个人之间的合作数量。

考虑以下重现所需结构的示例：

# Example dataframe
set.seed(1)
tp=cbind(paste(rep("project",10),1:10,sep=""),sample(2005:2010,10,replace=T))
tp=tp[sample(1:10,50,T),]
id=sample(paste(rep("id",10),1:10,sep=""),50,T)
df=as.data.frame(cbind(tp,id));rm(tp,id)
names(df)=c("project","year","id")
df=df[order(df$project,df$id),]

df[1:10,]
# project  year id
# project1 2006 id1
# project1 2006 id3
# project1 2006 id5
# project1 2006 id5
# project4 2006 id3
# project4 2006 id4
# project5 2006 id3
# project5 2006 id4
# project6 2008 id2
# project6 2008 id3

例如，2006 年的关系矩阵如下所示

    id1 id2 id3 id4 id5
id1  0   0   1   0   1
id2  0   0   0   0   0
id3  1   0   0   2   1
id4  0   0   2   0   0
id5  1   0   1   0   0

# link between 1 and 3, 1 and 5, 3 and 5 on project 1
# links between 3 and 4 on project 4 and project 5
# the matrix is symmetric
# the diagonal is O because an individual cannot collaborate with himself

Answer 1

我稍微修改了您的采样代码，使项目维度与 id 维度不同，因为我正在研究矩阵的维度以确保我得到正确的 n x n 矩阵。这是有效的代码：

set.seed(1)
tp=cbind(paste(rep("project",5),1:5,sep=""),sample(2008:2010,5,replace=T))
tp=tp[sample(1:5,20,T),]
id=sample(paste(rep("id",10),1:10,sep=""),20,T)
df=as.data.frame(cbind(tp,id));rm(tp,id)
names(df)=c("project","year","id")
df=df[order(df$project,df$id),]

spl=split(df,df$year)
net=lapply(spl,function(x){
  m = table(x$id, x$project)
  res = tcrossprod(m)  ## equivalently: res = m %*% t(m)
  diag(res) <- 0
  res <- ifelse(res > 0, 1, 0)
  res
})
net

拆分数据：

$`2008`
    project year  id
5  project1 2008 id4
7  project1 2008 id6
19 project1 2008 id6
2  project5 2008 id1
13 project5 2008 id2
1  project5 2008 id4
16 project5 2008 id9

$`2009`
    project year  id
9  project2 2009 id2
6  project2 2009 id5
20 project2 2009 id6
17 project2 2009 id7
14 project2 2009 id8
11 project3 2009 id7

$`2010`
    project year  id
3  project4 2010 id4
8  project4 2010 id5
15 project4 2010 id5
12 project4 2010 id8
18 project4 2010 id8
4  project4 2010 id9
10 project4 2010 id9

每年项目的邻接矩阵：

$`2008`

      id1 id2 id4 id5 id6 id7 id8 id9
  id1   0   1   1   0   0   0   0   1
  id2   1   0   1   0   0   0   0   1
  id4   1   1   0   0   1   0   0   1
  id5   0   0   0   0   0   0   0   0
  id6   0   0   1   0   0   0   0   0
  id7   0   0   0   0   0   0   0   0
  id8   0   0   0   0   0   0   0   0
  id9   1   1   1   0   0   0   0   0

$`2009`

      id1 id2 id4 id5 id6 id7 id8 id9
  id1   0   0   0   0   0   0   0   0
  id2   0   0   0   1   1   1   1   0
  id4   0   0   0   0   0   0   0   0
  id5   0   1   0   0   1   1   1   0
  id6   0   1   0   1   0   1   1   0
  id7   0   1   0   1   1   0   1   0
  id8   0   1   0   1   1   1   0   0
  id9   0   0   0   0   0   0   0   0

$`2010`

      id1 id2 id4 id5 id6 id7 id8 id9
  id1   0   0   0   0   0   0   0   0
  id2   0   0   0   0   0   0   0   0
  id4   0   0   0   1   0   0   1   1
  id5   0   0   1   0   0   0   1   1
  id6   0   0   0   0   0   0   0   0
  id7   0   0   0   0   0   0   0   0
  id8   0   0   1   1   0   0   0   1
  id9   0   0   1   1   0   0   1   0

Answer 2

您也可以将 dplyr 与 tidyr 一起使用：

library(dplyr)
library(tidyr)

df %>%
  unique %>%
  mutate(val = 1) %>%
  spread(id, val) %>%
  select(-project) %>%
  group_by(year) %>%
  do({
    mat <- select(., -year) %>% as.matrix
    mat[is.na(mat)] <- 0
    cp <- crossprod(mat)
    diag(cp) <- 0
    cp %>% as.data.frame %>%
      add_rownames(var = 'id')
  }) %>%
  ungroup

使用 R 创建关系矩阵

Creating relational matrices with R

r

relational

matrix