amisos55 amisos55 - 3 days ago 6
R Question

Converting array to matrix in R

I have an array, including two proficiency variables (theta0, theta1) over an item (Yes, No) called "comp". This needs to be converted to one matrix. Is there any way that I could convert a matrix like the one at the bottom?

My array looks like this:

>priCPT.i6

, , comp = Yes

theta1
theta0 Low Med High
Low 0.8377206 0.6760511 0.4576021
Med 0.6760511 0.4576021 0.2543239
High 0.4576021 0.2543239 0.1211734

, , comp = No

theta1
theta0 Low Med High
Low 0.1622794 0.3239489 0.5423979
Med 0.3239489 0.5423979 0.7456761
High 0.5423979 0.7456761 0.8788266

attr(,"class")
[1] "CPA" "array"


I apologize, I could not produce something that you could play with. I am looking for something like:

theta0 theta1 Yes No
Low Low 0.8377206 0.1622794
Low Med .. ..
Low High .. ..
Med Low .. ..
Med Med .. ..
Med High .. ..
High Low .. ..
High Med .. ..
High High .. ..


Regards...

Answer

You can easily get columns of values by flattening the matrix on the 3rd margin:

z1 <- apply(priCPT.i6, 3L, c)
## we can also simply use `matrix`; but remember to set `dimnames`
## otherwise we lose dimnames
## z1 <- matrix(priCPT.i6, ncol = 2L,
##              dimnames = list(NULL, dimnames(priCPT.i6)[[3]]))

What you need for the rest is to append the "dimnames" columns:

z2 <- expand.grid(dimnames(priCPT.i6)[1:2])

Now you can merge them into a data frame (you definitely need a data frame than a matrix, because columns of z1 are numeric while columns of z2 are character) via:

data.frame(z2, z1)

Reproducible example

x <- array(1:18, dim = c(3L, 3L, 2L), dimnames = list(
           c("Low", "Medium", "High"), c("Low", "Medium", "High"), c("Yes", "No")))

#, , Yes
#
#       Low Medium High
#Low      1      4    7
#Medium   2      5    8
#High     3      6    9
#
#, , No
#
#       Low Medium High
#Low     10     13   16
#Medium  11     14   17
#High    12     15   18

z1 <- apply(x, 3L, c)
## z1 <- matrix(x, ncol = 2L, dimnames = list(NULL, dimnames(x)[[3]]))
z2 <- expand.grid(dimnames(x)[1:2])
data.frame(z2, z1)

#    Var1   Var2 Yes No
#1    Low    Low   1 10
#2 Medium    Low   2 11
#3   High    Low   3 12
#4    Low Medium   4 13
#5 Medium Medium   5 14
#6   High Medium   6 15
#7    Low   High   7 16
#8 Medium   High   8 17
#9   High   High   9 18
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