r语言 - 如何通过对节点进行分组来降低邻接矩阵的分辨率

Question

>我有一个邻接矩阵，表示物种之间的喂食链接(列吃行(

mat1<-matrix(data=c(0,0,0,1,0,0,1,0,0,1,0,0,0,0,0,0), 
nrow=4, 
ncol=4, 
byrow = TRUE,
dimnames = list(c("a","b","c","d"), 
c("a","b","c","d")))

我想使用显示每个物种所属哪个家族的数据帧将此矩阵的分辨率降低到家庭级别，

df <- data.frame(Species = c("a","b","c","d"), Family = c("E","E","F","F"))

以便生成的矩阵将给出家庭之间的喂养链接的数量

mat2<-matrix(data=c(0,2,1,0), 
nrow=2, 
ncol=2, 
byrow = TRUE,
dimnames = list(c("E","F"), 
c("E","F")))

谢谢你的时间

Answer 1

我相信有一种更优雅的方法，但这里有一种data.table的方法。如果您的邻接矩阵非常大，这可能比tidyr的透视方法更快。

首先，我们将两个对象转换为data.tables。然后我们将Family连接到邻接矩阵上。然后我们按组对每列求Family。最后，我们转置并再次做同样的事情。

library(data.table)
setDT(df)
dt <- as.data.table(cbind(Species = rownames(mat1),as.data.frame(mat1)))
a <- df[dt,on = "Species"][,-"Species"][,lapply(.SD, sum), by = Family]
b <- data.table::transpose(a, keep.names = "Family", make.names = 1)
setnames(b,"Family","Species")
c <- df[b,on = "Species"][,-"Species"][,lapply(.SD,sum), by = Family]
data.table::transpose(c, keep.names = "Family", make.names = 1)
Family E F
1:      E 0 2
2:      F 1 0

Answer 2

因为这是我知道如何做的唯一方法，所以这里有一个使用整洁的解决方案。
它将矩阵变成长形式的 tibble，按族聚合，然后再次使其变宽。

library(tidyverse)
# create a tibble that looks like the desired end-result matrix
df2 <- mat1 %>% 
as_tibble(rownames = "Species_from") %>% # make a tibble
pivot_longer(cols = -Species_from,
names_to = "Species_to") %>% # turn into long form
left_join(df, by = c("Species_from" = "Species")) %>% # add Family_from and Family_to
left_join(df, by = c("Species_to" = "Species"), suffix = c("_from", "_to")) %>% 
group_by(Family_from, Family_to) %>% # aggregate Family_from and Family_to
summarise(value = sum(value)) %>% # ... by taking their sum
pivot_wider(names_from = Family_to,
values_from = value) # turn back into wide form
# turn into a matrix
mat2 <- as.matrix(df2[, c("E", "F")])
rownames(mat2) <- df2$Family_from
mat2
#   E F
# E 0 2
# F 1 0