您可以使用merge并指定要使用row.names进行合并。

combined_adj <- merge(x = adj1,
      y = adj2,
      by = c('row.names', 
             intersect(colnames(adj1), 
                       colnames(adj2))
             ), 
      all = TRUE
)
combined_adj
  Row.names g1 g2 g3 g5 g6 g4
1      John  0  0  0  0  0  0
2      Mary  0 NA NA NA NA  0
3     Paolo  0 NA NA NA NA  0
4     Sarah  0  0  0  0  0 NA
5       Tim  0  0  0  0  0  0

这会将其转换为data.frame，因此如果需要，您需要将其转换回矩阵。

row.names(combined_adj) <- combined_adj[,1]
combined_adj <- combined_adj[,-1]

编辑：合并多个矩阵

我们使用Reduce将其应用于所有矩阵。我们首先转换为data.frame，然后创建一个带有row_names的列来简化操作。

# create sample data
adj1 <- matrix(
  0, 3, 5,
  dimnames = list(c("Tim", "John", "Sarah"), 
                  c("g1", "g2", "g3", "g5", "g6"))
)

adj2 <- matrix(
  0, 4, 2, 
  dimnames = list(c("Tim", "Mary", "John", "Paolo"),
                  c("g1", "g4"))
)

adj3 <- matrix(
  0, 3, 3, 
  dimnames = list(c("Tim2", "Mary2", "John"), c("g1", "g4", 'g7'))
)

# create a list 
list_matrices <- list(adj1, adj2, adj3)

# convert to dataframes and create a column with row.names
list_matrices <- lapply(list_matrices, function(mat){
  mat <- as.data.frame(mat)
  mat$row_names <- row.names(mat)
  mat
})

# successively combine them, merge 1..2 and then merge result with 3 and so on
res <- Reduce(function(mat1, mat2) merge(mat1, mat2, all = TRUE), x = list_matrices)

res
  g1 row_names g4 g2 g3 g5 g6 g7
1  0      John  0  0  0  0  0  0
2  0      Mary  0 NA NA NA NA NA
3  0     Mary2  0 NA NA NA NA  0
4  0     Paolo  0 NA NA NA NA NA
5  0     Sarah NA  0  0  0  0 NA
6  0       Tim  0  0  0  0  0 NA
7  0      Tim2  0 NA NA NA NA  0

这可能是一个解决方案。但是，我假设这些单元格中确实存在的信息对于相同的行名称和列名组合总是相同的。除此之外，它还依赖于dplyr：

require(tidyverse)

list_adj <- list(
  adj1, adj2
)

df.adj <- NULL

for (adj in list_adj) {
  df.adj.temp <- adj %>% as_tibble(rownames = "row_names")
  
  if (is.null(df.adj)) {
    df.adj <- df.adj.temp
  } else {
    c.colnames.join.by <- c(intersect(colnames(df.adj), colnames(df.adj.temp)))
    
    df.adj <- df.adj %>% 
      full_join(df.adj.temp, by = c.colnames.join.by) %>% 
      mutate(across(.cols = - row_names, .fns = \(x) replace_na(x, 10)))
  }
}

df.adj %>% 
  arrange(row_names) %>% # ordering rows
  select(all_of(sort(colnames(df.adj)))) %>% # ordering columns
  column_to_rownames(var = "row_names") %>% 
  as.matrix()

# output
      g1 g2 g3 g5 g6 g4
John   0  0  0  0  0  0
Mary   0 10 10 10 10  0
Paolo  0 10 10 10 10  0
Sarah  0  0  0  0  0 10
Tim    0  0  0  0  0  0

下面是一个带有reshape的基本R选项

df <- unique(
    rbind(
        as.data.frame(as.table(adj1)),
        as.data.frame(as.table(adj2))
    )
)

reshape(
    df,
    direction = "wide",
    idvar = "Var1",
    timevar = "Var2"
)

它给出了

Var1 Freq.g1 Freq.g2 Freq.g3 Freq.g5 Freq.g6 Freq.g4
1    Tim       0       0       0       0       0       0
2   John       0       0       0       0       0       0
3  Sarah       0       0       0       0       0      NA
17  Mary       0      NA      NA      NA      NA       0
19 Paolo       0      NA      NA      NA      NA       0

或者，我们使用xtabs

mat <- `class<-`(xtabs(Freq ~ ., df) * NA, "matrix")
mat[as.matrix(df[-3])] <- df$Freq

它给出了

> mat
       Var2
Var1    g1 g2 g3 g5 g6 g4
  Tim    0  0  0  0  0  0
  John   0  0  0  0  0  0
  Sarah  0  0  0  0  0 NA
  Mary   0 NA NA NA NA  0
  Paolo  0 NA NA NA NA  0
attr(,"call")
xtabs(formula = Freq ~ ., data = df)