R等价于Matlab中的排列数组维数permute(A，dimorder)

Question

我正在寻找一个等价物从Matlab的permute(A,dimorder)，以便将一些Matlab代码转换为R。循环需要如下所示的行：

x = permute(a{i}(b(i,ii),:,:,:,:,:),[2 3 4 5 6 1])

单元阵列结构例如a{1}(1,:,:,:,:,:)导致选择单元阵列a{}内的第一行矩阵。permute()中的[2 3 4 5 6 1]指的是dimorder。可以在这里找到matlab函数permute()的文档，包括示例输出：https://de.mathworks.com/help/matlab/ref/permute.html

在R中有几个函数以这样或那样的方式引用排列，但它们似乎都不是我想要的，尽管我可能弄错了什么。

Answer 1

我相信我已经成功地在R中复制了MATLAB脚本。我不认为你真的需要permute的等价物。在MATLAB脚本中，permute似乎只是简单地删除了多余的维度。R默认情况下这样做，除非您在子集数组时指定drop = FALSE，例如，

lnA[[tau, modal]] <- a[[modal]][outcomes[modal, tau],,,drop = FALSE]

如果我在最终的for循环之前将lnA = cell(T, NumModalities);添加到MATLAB脚本中，然后将循环的内部修改为

lnA{tau, modal} = permute(a{modal}(outcomes(modal,tau),:,:,:,:,:),[2 3 4 5 6 1]);

然后，对于MATLAB和R实现，我在lnA中得到了相同的矩阵数组。

在R中，我使用一个列表数组作为MATLAB 2+维单元格数组的等价物：

lnA1 = cell(T, 1); # MATLAB
lnA1 <- vector("list", Time) # R    
lnA2 = cell(T, NumModalities); # MATLAB
lnA2 <- array(vector("list", Time*NumModalities), c(Time, NumModalities)) # R
lnA2 <- matrix(vector("list", Time*NumModalities), Time) # R
lnA3 = cell(T, NumModalities, 2); # MATLAB
lnA3 <- array(vector("list", Time*NumModalities*2), c(Time, NumModalities, 2)) # R

下面是实现：

nat_log <- function (x) { # necessary as log(0) not defined...
  x <- log(x + exp(-16))
}

# Set up a list for D and A
D <- list(c(1, 0),       # (left better, right better)
          c(1, 0, 0, 0)) #(start, hint, choose-left, choose-right)
A <- c(rep(list(array(0, c(3, 2, 4))), 2), list(array(0, c(4, 2, 4))))

Ns <- lengths(D) # number of states in each state factor (2 and 4)
A[[1]][,,1:Ns[2]] <- matrix(c(1,1,  # No Hint
                              0,0,  # Machine-Left Hint
                              0,0), # Machine-Right Hint
                      ncol = 2, nrow = 3, byrow = TRUE)

pHA <- 1
A[[1]][,,2] <- matrix(c(0,       0,       # No Hint
                        pHA,     1 - pHA, # Machine-Left Hint
                        1 - pHA, pHA),    # Machine-Right Hint
                      nrow = 3, ncol = 2, byrow = TRUE)

A[[2]][,,1:2] <- matrix(c(1, 1,   # Null
                          0, 0,   # Loss
                          0, 0),  # Win
                        ncol = 2, nrow = 3, byrow = TRUE)

pWin <- 0.8
A[[2]][,,3] <- matrix(c(0,        0,         # Null        
                        1 - pWin, pWin,      # Loss
                        pWin,     1 - pWin), # Win
                      ncol = 2, nrow = 3, byrow = TRUE)

A[[2]][,,4] <- matrix(c(0,        0,        # Null        
                        pWin,     1 - pWin, # Loss
                        1 - pWin, pWin),    # Win
                      ncol = 2, nrow = 3, byrow = TRUE)

for (i in 1:Ns[2]) {
  A[[3]][i,,i] <- c(1,1)
}

# Set up a list of matrices:
a <- lapply(1:3, function(i) A[[i]]*200)
a[[1]][,,2] <- matrix(c(0,    0,     # No Hint
                        0.25, 0.25,  # Machine-Left Hint
                        0.25, 0.25), # Machine-Right Hint
                      nrow = 3, ncol = 2, byrow = TRUE)

outcomes <- matrix(c(1, 2, 1,
                     1, 1, 2,
                     1, 2, 4),
                   ncol = 3, nrow = 3, byrow = TRUE)

NumModalities <- length(a)       # number of outcome factors
Time <- 3L
lnA <- array(vector("list", Time*NumModalities), c(Time, NumModalities))

for (tau in 1:Time){
  for (modal in 1:NumModalities){
    lnA[[tau, modal]] <- a[[modal]][outcomes[modal, tau],,]
  }
}