一种igraph方法，因此使用Joseph的答案中的df：

library(igraph)

g <- graph_from_data_frame(df)

E(g)$weight = 1/degree(g, mode="out")[df$Time1] # get counts

as_adj(g, attr = "weight", sparse=FALSE) # output weighted adjacency matrix

    A B   C
A 0.5 0 0.5
B 1.0 0 0.0
C 0.0 0 0.0

这是另一个base R解决方案。

df <- data.frame(Time1 = c("A","B","A"), Time2 = c("A","A","C"), stringsAsFactors = FALSE)

myStates <- sort(unique(c(df$Time1, df$Time2)))
lenSt <- length(myStates)

currState <- match(df$Time1, myStates)
nextState <- match(df$Time2, myStates)
transMat <- matrix(0L, lenSt, lenSt)

transMat[cbind(currState, nextState)] <- 1L
transMat <- transMat/rowSums(transMat)
transMat[is.na(transMat)] <- 0

transMat
     [,1] [,2] [,3]
[1,]  0.5    0  0.5
[2,]  1.0    0  0.0
[3,]  0.0    0  0.0

当然有更好的方法。这是我在一个无聊的星期五下午用循环涂鸦。

lvls <- sort(unique(unlist(df[,-1])))

dat <- matrix(0, nrow= length(lvls), ncol= length(lvls))

colnames(dat) <- lvls
rownames(dat) <- lvls

concat <- paste0(df[,2], df[,3])

for (i in 1:length(lvls)) {
  for (j in 1:length(lvls)) {
    dat[i,j] <- paste0(rownames(dat)[i], colnames(dat)[j])
  }
}

dat <- matrix(sapply(dat, function(x) length(grep(x, concat))), 
       nrow= length(lvls), ncol= length(lvls))

colnames(dat) <- lvls
rownames(dat) <- lvls

dat

##   A B C
## A 1 0 1
## B 1 0 0
## C 0 0 0

dat <- dat / rowSums(dat)
dat[is.na(dat)] <- 0

dat

##    A B   C
##A 0.5 0 0.5
##B 1.0 0 0.0
##C 0.0 0 0.0