CODEWITHSUNDEEP
rmatrix
Mayou
Mayou

Reputation: 8848

Transition matrix

Consider the following dataframe:

 df = data.frame(cusip = paste("A", 1:10, sep = ""), xt = c(1,2,3,2,3,5,2,4,5,1), xt1 = c(1,4,2,1,1,4,2,2,2,5))

The data is divided in five states, which are quantiles in reality: 1,2,3,4,5. The first column of the dataframe represents the state at time t, and the second column is the state at time t+1.

I would like to compute a sort of a transition matrix for the five states. The meaning of the matrix would be as follows:

I am really not sure how to do this in an efficient way. I have the feeling the answer is trivial, but I just can't get my head around it.

Could anyone please help?

Upvotes: 5

Views: 683

Answers (2)

Sandipan Dey
Sandipan Dey

Reputation: 23129

If you want to have all the states (1..5) in the column of the transition matrix, you can try this:

levs <- sort(union(df$xt, df$xt1)) 
tbl <- table(factor(df$xt, levs), factor(df$xt1, levs))
tbl / rowSums(tbl)  

         1         2         3         4         5
  1 0.5000000 0.0000000 0.0000000 0.0000000 0.5000000
  2 0.3333333 0.3333333 0.0000000 0.3333333 0.0000000
  3 0.5000000 0.5000000 0.0000000 0.0000000 0.0000000
  4 0.0000000 1.0000000 0.0000000 0.0000000 0.0000000
  5 0.0000000 0.5000000 0.0000000 0.5000000 0.0000000

Upvotes: 0

Josh O&#39;Brien
Josh O&#39;Brien

Reputation: 162431

res <- with(df, table(xt, xt1)) ## table() to form transition matrix
res/rowSums(res)                ## /rowSums() to normalize by row
#    xt1
# xt          1         2         4         5
#   1 0.5000000 0.0000000 0.0000000 0.5000000
#   2 0.3333333 0.3333333 0.3333333 0.0000000
#   3 0.5000000 0.5000000 0.0000000 0.0000000
#   4 0.0000000 1.0000000 0.0000000 0.0000000
#   5 0.0000000 0.5000000 0.5000000 0.0000000

## As an alternative to  2nd line above, use sweep(), which won't rely on 
## implicit recycling of vector returned by rowSums(res)
sweep(res, MARGIN = 1, STATS = rowSums(res), FUN = `/`)

Upvotes: 5

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