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"
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"