Question

我有一个数组，包括一个名为＆＃34; comp＆＃34;的项目上的两个熟练变量（theta0，theta1）（是，否）。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吗？

我的数组看起来像这样：

>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"

我道歉，我无法制作出你可以玩的东西。我正在寻找类似的东西：

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     ..         ..

...问候

Answer 1

通过在第3个边距上展平矩阵，您可以轻松获取值列：

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]]))

剩下的就是附加＆＃34; dimnames＆＃34;列：

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

现在你可以将它们合并到一个数据框中（你肯定需要一个数据框而不是一个矩阵，因为z1的列是数字而z2的列是字符）来自：

data.frame(z2, z1)

可重复的示例

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

Answer 2

使用reshape2的替代方案是

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

library(reshape2)
df <- dcast(melt(x), Var1+Var2~Var3) 
df

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

在R中将数组转换为矩阵

2 个答案: