我正在從 Matlab 中尋找等效的 permute(A,dimorder)，以便將一些 Matlab 代碼轉換為 R。一個回圈需要一條看起來像這樣的行：
x = permute(a{i}(b(i,ii),:,:,:,:,:),[2 3 4 5 6 1])
單元陣列結構例如a{1}(1,:,:,:,:,:)導致選擇單元陣列a{}內的第一行矩陣。permute() 中的 [2 3 4 5 6 1] 指的是二階。可以在此處找到包括示例輸出在內的 matlab 函式 permute() 的檔案：https :
//de.mathworks.com/help/matlab/ref/permute.html
R 中有幾個函式以某種方式指代排列，但似乎沒有一個是我正在尋找的，盡管我可能弄錯了。
這將是一些示例 Matlabcode。我把它從另一個腳本放在一起，這是一個嵌套回圈。所以下面的代碼可能沒有太大意義，但它提供了結果。相關行（回圈）在下部：
D{1} = [1 0]'; % {'left better','right better'}
D{2} = [1 0 0 0]'; % {'start','hint','choose-left','choose-right'}
Ns = [length(D{1}) length(D{2})]; % number of states in each state factor (2 and 4)
for i = 1:Ns(2)
A{1}(:,:,i) = [1 1; % No Hint
0 0; % Machine-Left Hint
0 0];% Machine-Right Hint
end
pHA = 1; % By default we set this to 1, but try changing its value to
% see how it affects model behavior
A{1}(:,:,2) = [0 0; % No Hint
pHA 1-pHA; % Machine-Left Hint
1-pHA pHA]; % Machine-Right Hint
#
for i = 1:2
A{2}(:,:,i) = [1 1; % Null
0 0; % Loss
0 0]; % Win
end
pWin = .8; % By default we set this to 1, but try changing its value to
% see how it affects model behavior
A{2}(:,:,3) = [0 0; % Null
1-pWin pWin; % Loss
pWin 1-pWin]; % Win
A{2}(:,:,4) = [0 0; % Null
pWin 1-pWin; % Loss
1-pWin pWin]; % Win
for i = 1:Ns(2)
A{3}(i,:,i) = [1 1];
end
a{1} = A{1}*200;
a{2} = A{2}*200;
a{3} = A{3}*200;
a{1}(:,:,2) = [0 0; % No Hint
.25 .25; % Machine-Left Hint
.25 .25]; % Machine-Right Hint
outcomes= [1,2,1;
1,1,2;
1,2,4]
NumModalities = numel(a);
T = 3
%RELEVANT PART:
for tau = 1:T
for modal = 1:NumModalities
lnA = permute(a{modal}(outcomes(modal,tau),:,:,:,:,:),[2 3 4 5 6 1])
%lnA = permute(nat_log(a{modal}(outcomes(modal,tau),:,:,:,:,:)),[2 3 4 5 6 1])
% with nat_log
end
end
% natural log that replaces zero values with very small values for numerical reasons.
function y = nat_log(x)
y = log(x exp(-16));
end
我在 R 中的嘗試如下所示：
nat_log = function (x) { # necessary as log(0) not defined...
x = log(x exp(-16))
}
# Setup a list for D
D = vector("list", 2*1)
dim(D) = matrix(c(2,1))
D[[1,1]] = t(matrix(c(1, 0)))
D[[2,1]] = t(matrix(c(1, 0, 0, 0)))
# Setup a list:
A = vector("list", 3*4)
dim(A) = matrix(c(3,4))
Ns = t(matrix(c(ncol(D[[1]]), ncol(D[[2]]))))
Ns
### A[1,1:4]
for (i in 1:Ns[,2]){
# [[i,,]] for all behavior states
A[[1,i]] <- matrix(c(1,1, # No Hint
0,0, # Machine-Left Hint
0,0), # Machine-Right Hint
ncol = 2, nrow = 3, byrow = TRUE)
}
### A[1,2]
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)
for (i in 1:2){
A[[2,i]] = matrix(c( 1, 1, # Null
0, 0, # Loss
0, 0), # Win
ncol = 2, nrow = 3, byrow = TRUE)
}
pWin = .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]){
# x will be added to a 2*4 Matrix
x = matrix(c(1,1), ncol = 2, nrow = 1, byrow=TRUE)
A3 = matrix(c(0,0, 0,0 ,0,0, 0,0), # c('start','hint','choose-left','choose-right')
ncol = 2, nrow = 4, byrow=TRUE)
#y = A3[col = i]
A3[i,] = 1
A[[3,i]] <- A3
}
# See to understand the consequence of the loop
A[3,]
A
# Setup a list of lists:
a = vector("list", 3*4)
dim(a) = matrix(c(3,4))
a
for (i in 1:4){
apre = as.matrix(A)
a[[1,i]] = apre[[1,i]]*200
a[[2,i]] = apre[[2,i]]*200
a[[3,i]] = apre[[3,i]]*200
}
a[[1,2]] = matrix(c(0, 0, # No Hint
.25, .25, # Machine-Left Hint
.25, .25), # Machine-Right Hint
nrow = 3, ncol = 2, byrow = TRUE)
a
NumModalities = nrow(a) # number of outcome factors
Time = 3
outcomes = matrix(c(1,2,1,
1,1,2,
1,2,4), ncol = 3, nrow = 3, byrow = TRUE)
# RELEVANT PART:
lnA = vector("list", 2*4)
dim(lnA) = matrix(c(2,4))
x = vector("list",2*4)
dim(x) = matrox(c(2,4))
for (tau in 1:Time){
for (modal in 1:NumModalities){
x = a[[modal]][outcomes[[modal,tau]]]
lnA = aperm(as.array(x),c(2, 3, 4, 5, 6, 1))
}
}
as.array(x)
is.array(x)
lnA
uj5u.com熱心網友回復：
我相信我在 R 中成功復制了 MATLAB 腳本。我認為您實際上不需要permute. 在 MATLAB 腳本中，permute似乎只是洗掉多余的維度。R 默認情況下會這樣做，除非您指定drop = FALSE何時對陣列進行子集化，例如，
lnA[[tau, modal]] <- a[[modal]][outcomes[modal, tau],,,drop = FALSE]
如果我lnA = cell(T, NumModalities);在最終for回圈之前添加到 MATLAB 腳本，然后將回圈內部修改為
lnA{tau, modal} = permute(a{modal}(outcomes(modal,tau),:,:,:,:,:),[2 3 4 5 6 1]);
然后我lnA為 MATLAB 和 R 實作獲得相同的矩陣陣列。
在 R 中，我使用串列陣列作為 MATLAB 2 維元胞陣列的等效項：
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
這是實作：
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],,]
}
}
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