Converting a vector into a matrix (in R)

Question

I want to create a "n x 8" Matrix from a "n x 1" vector

-- Question: Why do I want to do this?

-- Answer: In order to matrix multiply this against an "8 x 8" markov chain probability transition matrix, and return an "n x 8" Matrix of the predicted states

-- Solution: I have solved this in Attempt 3 below - but want to know if there is a better way to resolve this (rather than using two transpose functions)?

R code

Create a dummy "n x 1" vector: (here we use n = 2)

> temp_vector <- c("state 4", "state 7")
> temp_vector
[1] "state 4" "state 7"

Expected Output:

NA NA NA TRUE NA NA NA NA
NA NA NA NA NA NA TRUE NA

Attempt 1: Convert to matrix:

> temp_matrix <- matrix(temp_vector, 
                ncol = 8, # there are 8 states
                nrow = length(temp_vector) # there are 10 rows in the vector
                )
> temp_matrix
     [,1]      [,2]      [,3]      [,4]      [,5]      [,6]      [,7]      [,8]     
[1,] "state 4" "state 4" "state 4" "state 4" "state 4" "state 4" "state 4" "state 4"
[2,] "state 7" "state 7" "state 7" "state 7" "state 7" "state 7" "state 7" "state 7"

Attempt 1 FAIL: This is not ideal, I want a matrix with ONE entry per row, not EIGHT.

Attempt 2: Compare the stateSpace above with the matrix, to give a matrix made up of TRUE/FALSE:

> stateSpace <- c("state 1", "state 2", "state 3", "state 4", "state 5", "state 6", "state 7", "state 8")

> temp_matrix == stateSpace
     state 1 state 2 state 3 state 4 state 5 state 6 state 7 state 8
[1,]   FALSE   FALSE   FALSE   FALSE   FALSE   FALSE   FALSE   FALSE
[2,]   FALSE   FALSE   FALSE   FALSE   FALSE   FALSE   FALSE   FALSE

Attempt 2 FAIL: expected each row to have one TRUE and the rest FALSE

Reason: (I THINK) matrices are compared column-wise.

Looking into Attempt 2 further, on an element by element level this works:

> temp_matrix[1,1] == colnames(temp_matrix)[1]
state 1 
  FALSE 
> temp_matrix[1,2] == colnames(temp_matrix)[2]
state 2 
  FALSE 
> temp_matrix[1,3] == colnames(temp_matrix)[3]
state 3 
  FALSE 
> temp_matrix[1,4] == colnames(temp_matrix)[4]
state 4 
   TRUE

Looking into Attempt 2 further, on a row by row level this works:

> temp_matrix[1,] == colnames(temp_matrix)[]
state 1 state 2 state 3 state 4 state 5 state 6 state 7 state 8 
  FALSE   FALSE   FALSE    TRUE   FALSE   FALSE   FALSE   FALSE 

> temp_matrix[2,] == colnames(temp_matrix)[]
state 1 state 2 state 3 state 4 state 5 state 6 state 7 state 8 
  FALSE   FALSE   FALSE   FALSE   FALSE   FALSE    TRUE   FALSE

Attempt 3: after noting the above learnings of column wise comparison in R

> t(stateSpace == t(temp_matrix))
     state 1 state 2 state 3 state 4 state 5 state 6 state 7 state 8
[1,]    TRUE   FALSE   FALSE   FALSE   FALSE   FALSE   FALSE   FALSE
[2,]   FALSE    TRUE   FALSE   FALSE   FALSE   FALSE   FALSE   FALSE

Attempt 3 SUCCESS: created this stackoverflow post to see if there is a better way to resolve this (rather than using two transpose functions)

Other options: dcast, reshape, spread; sadly did NOT work either.

I tried reshape():

reshape(temp_vector, direction = "wide")
> Error in data[, timevar] : incorrect number of dimensions

I tried spread():

library(tidyr)
spread(temp_vector, key = numbers, value = value)
> Error in UseMethod("spread_") : 
  no applicable method for 'spread_' applied to an object of class "factor"

This is my first ever stackoverflow post - so please let me know if anything is incorrect above. — Kunal, May 25 '18 at 14:04
Please show your desired output. Also, you example data is too big. I would create a data of 10-20 rows. See [here](https://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example) for some examples. — David Arenburg, May 29 '18 at 08:44
David - thank you - removed all the excess info - and made the question more specific. — Kunal, May 29 '18 at 09:59
Most important is expected output... and also try to remove even more excess info... temp_vector can be made shorter I suspect. — s_baldur, May 29 '18 at 10:40
I think the issue is that matrices are compared column-wise, and I need to make the comparison row-wise — Kunal, May 29 '18 at 12:44
my current solution to this question: t(colnames(temp_matrix) == t(temp_matrix)) — Kunal, May 29 '18 at 12:50

score 0 · Accepted Answer · answered Jun 04 '18 at 06:30

Try this:

> v <- c("state 4", "state 7")
> states <- c("state 1", "state 2", "state 3", "state 4",
+             "state 5", "state 6", "state 7", "state 8")
> m <- matrix(states, byrow = TRUE, nrow = 2, ncol = 8)
> m
#      [,1]      [,2]      [,3]      [,4]      [,5]      [,6]      [,7]      # [,8]     
# [1,] "state 1" "state 2" "state 3" "state 4" "state 5" "state 6" "state 7" "state 8"
# [2,] "state 1" "state 2" "state 3" "state 4" "state 5" "state 6" "state 7" "state 8"
> v == m
#       [,1]  [,2]  [,3]  [,4]  [,5]  [,6]  [,7]  [,8]
# [1,] FALSE FALSE FALSE  TRUE FALSE FALSE FALSE FALSE
# [2,] FALSE FALSE FALSE FALSE FALSE FALSE  TRUE FALSE

In R, a matrix is basically a vector under the hood. When m is created above, the matrix function "recycles" its argument spaces because it needs to create a matrix with 16 elements. In other words, the following two function calls produce the same result:

> matrix(states, byrow = TRUE, nrow = 2, ncol = 8)
> matrix(rep(states, 2), byrow = TRUE, nrow = 2, ncol = 8)

Similarly, when v and m are compared for equality, v is recycled 8 times to produce a vector of length 16. In other words, the following two equality comparisons produce the same results:

> v == m
> rep(v, 8) == m

You can think of the above two comparisons as happening between two vectors, where the matrix m is converted back into a vector by stacking the columns. You can use as.vector to see the vector that m corresponds to:

> as.vector(m)
#  [1] "state 1" "state 1" "state 2" "state 2" "state 3" "state 3" "state 4" "state 4" "state 5"
# [10] "state 5" "state 6" "state 6" "state 7" "state 7" "state 8" "state 8"

Converting a vector into a matrix (in R)

1 Answers1