将数组转换为R中的矩阵 [英] Converting array to matrix in R

问题描述

我有一个数组，在名为"comp"的项(是，否)上包括两个熟练程度变量(theta0，theta1).这需要转换为一个矩阵.有什么方法可以转换像底部那样的矩阵?

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?

我的数组如下:

>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"

很抱歉，我无法制作出您可以玩的东西.我正在寻找类似的东西:

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     ..         ..

关于...

推荐答案

您可以通过将矩阵的第3个边距展平来轻松获得值的列:

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]]))

剩下的您需要添加假名"列:

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

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

现在，您可以通过以下方式将它们合并到数据框中(肯定需要一个数据框而不是矩阵，因为z1的列是数字，而z2的列是字符).

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)

---

可复制的示例

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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