我有2个变量......输入数量N和历史记录M的长度。这两个变量决定了矩阵V的大小，即n×m，即n行，m列。
我很难想出一个算法，它使我能够产生一定量的排列（或序列，你如何看待适合）。
如果有人能够用算法帮助我，我会非常高兴，如果可能的话，在Matlab中，但是伪算法也是非常好的。
我给你3个例子：
如果输入的数量是N = 1并且历史的长度是M = 2，那么我有（M + 1）^ N个不同的组合，在这种情况下是3.排列是：（如果您对matlab矩阵符号不熟悉,单独列;分隔行）。
V(1) = [1,0,0] V(2) = [0,1,0] V(3) = [0,0,1] 如果输入的数量是N = 2并且历史的长度是M = 2，则我有（M + 1）^ N个不同的组合，在这种情况下为9。排列如下：
V(1) = [1,0,0; 1,0,0] V(2) = [1,0,0; 0,1,0] V(3) = [1,0,0; 0,0,1] V(4) = [0,1,0; 1,0,0] V(5) = [0,1,0; 0,1,0] V(6) = [0,1,0; 0,0,1] V(7) = [0,0,1; 1,0,0] V(8) = [0,0,1; 0,1,0] V(9) = [0,0,1; 0,0,1] 如果输入的数量是N = 3并且历史的长度是M = 3，则我有（M + 1）^ N个不同的组合，在这种情况下是64。排列如下：
V(1) = [1,0,0,0; 1,0,0,0; 1,0,0,0] V(2) = [1,0,0,0; 1,0,0,0; 0,1,0,0] V(3) = [1,0,0,0; 1,0,0,0; 0,0,1,0] V(4) = [1,0,0,0; 1,0,0,0; 0,0,0,1] V(5) = [1,0,0,0; 0,1,0,0; 1,0,0,0] ... V(8) = [1,0,0,0; 0,1,0,0; 0,0,0,1] V(9) = [1,0,0,0; 0,0,1,0; 1,0,0,0] ... V(16) = [1,0,0,0; 0,0,0,1; 0,0,0,1] V(17) = [0,1,0,0; 1,0,0,0; 1,0,0,0] ... V(64) = [0,0,0,1; 0,0,0,1; 0,0,0,1]编辑：我只是找到一种方法来产生真正的大矩阵W，其中每一行代表V（i）
对于第一种情况：
W = eye(3)这里eye(k)创建一个大小为kxk的单位矩阵
对于第二种情况：
W = [kron(eye(3), ones(3,1)), ... kron(ones(3,1), eye(3))]这里kron是kronecker产品 ，并且ones(k,l) 创建了一个大小为kxl 的矩阵
对于第三种情况：
W = [kron(kron(eye(4), ones(4,1)), ones(4,1)), ... kron(kron(ones(4,1), eye(4)), ones(4,1)), ... kron(kron(ones(4,1), ones(4,1)), eye(4))]现在我们创建了矩阵W，其中每行代表矢量形式的V（i），V（i）还不是矩阵。
观察两件事情：
当输入N增加时，额外的列被添加额外的克罗内克积，单位矩阵沿着向量移动。 当历史M的长度增加时，身份矩阵向量增加，例如，眼睛（4） - >眼睛（5），（4,1） - >个（5,1）。I have 2 variables... the number of inputs N and the length of the history M. These two variables determine the size of the matrix V which is n x m, i.e., n rows, m columns.
I have difficulties to come up with a algorithm which enables me to generate a certain amount of permutations (or sequences, how you see fit).
I would be really glad if someone could help me with a algorithm, if possible in Matlab, but a pseudo-algorithm would also be very nice.
I give you 3 examples:
If the number of inputs is N = 1 and the length of the history is M = 2, I have (M+1)^N different combinations, in this case 3. The permutations are:(In case you are not familiar with matlab matrix notation, , seperates columns, ; seperates rows.)
V(1) = [1,0,0] V(2) = [0,1,0] V(3) = [0,0,1] If the number of inputs is N = 2 and the length of the history is M = 2, I have (M+1)^N different combinations, in this case 9.The permutations are:
V(1) = [1,0,0; 1,0,0] V(2) = [1,0,0; 0,1,0] V(3) = [1,0,0; 0,0,1] V(4) = [0,1,0; 1,0,0] V(5) = [0,1,0; 0,1,0] V(6) = [0,1,0; 0,0,1] V(7) = [0,0,1; 1,0,0] V(8) = [0,0,1; 0,1,0] V(9) = [0,0,1; 0,0,1] If the number of inputs is N = 3 and the length of the history is M = 3, I have (M+1)^N different combinations, in this case 64.The permutations are:
V(1) = [1,0,0,0; 1,0,0,0; 1,0,0,0] V(2) = [1,0,0,0; 1,0,0,0; 0,1,0,0] V(3) = [1,0,0,0; 1,0,0,0; 0,0,1,0] V(4) = [1,0,0,0; 1,0,0,0; 0,0,0,1] V(5) = [1,0,0,0; 0,1,0,0; 1,0,0,0] ... V(8) = [1,0,0,0; 0,1,0,0; 0,0,0,1] V(9) = [1,0,0,0; 0,0,1,0; 1,0,0,0] ... V(16) = [1,0,0,0; 0,0,0,1; 0,0,0,1] V(17) = [0,1,0,0; 1,0,0,0; 1,0,0,0] ... V(64) = [0,0,0,1; 0,0,0,1; 0,0,0,1]Edit: I just found a way to generate really large matrices W in which each row represents V(i)
For the first case:
W = eye(3)Herein eye(k) creates an identity matrix of size k x k
For the second case:
W = [kron(eye(3), ones(3,1)), ... kron(ones(3,1), eye(3))]Herein kron is the kronecker product, and ones(k,l) creates a matrix with ones of size k x l
For the third case:
W = [kron(kron(eye(4), ones(4,1)), ones(4,1)), ... kron(kron(ones(4,1), eye(4)), ones(4,1)), ... kron(kron(ones(4,1), ones(4,1)), eye(4))]Now we have created the matrices W in which each row represents V(i) in vector form, V(i) is not yet a matrix.
Observe two things:
When the input N is increased an extra column is added with an extra kronecker product and the identity matrix moves along the vector. When the length of the history M is increased the identity matrices vectors are increased, e.g., eye(4) -> eye(5), ones(4,1) -> ones(5,1).最满意答案
我想这可以满足你的所有要求。 即使顺序似乎对我来说是正确的：
M=3;N=3; mat1=eye(M+1); vectors=mat2cell(repmat(1:M+1,N,1),ones(N,1),[M+1]);超高效的笛卡尔产品，从这里采取：
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% n = numel(vectors); %// number of vectors combs = cell(1,n); %// pre-define to generate comma-separated list [combs{end:-1:1}] = ndgrid(vectors{end:-1:1}); %// the reverse order in these two %// comma-separated lists is needed to produce the rows of the result matrix in %// lexicographical order combs = cat(n+1, combs{:}); %// concat the n n-dim arrays along dimension n+1 combs = reshape(combs,[],n); %// reshape to obtain desired matrix %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% V=cell(size(combs,1),1); for i=1:size(combs,1) for j=1:size(combs,2) V{i,1}=[V{i,1};mat1(combs(i,j),:)]; end end输出：
M=2,N=2; V= [1,0,0;1,0,0] [1,0,0;0,1,0] [1,0,0;0,0,1] [0,1,0;1,0,0] [0,1,0;0,1,0] [0,1,0;0,0,1] [0,0,1;1,0,0] [0,0,1;0,1,0] [0,0,1;0,0,1] M=3;N=3; %order verified for the indices given in the question V(1) = [1,0,0,0; 1,0,0,0; 1,0,0,0] V(2) = [1,0,0,0; 1,0,0,0; 0,1,0,0] V(3) = [1,0,0,0; 1,0,0,0; 0,0,1,0] V(4) = [1,0,0,0; 1,0,0,0; 0,0,0,1] V(5) = [1,0,0,0; 0,1,0,0; 1,0,0,0] ... V(8) = [1,0,0,0; 0,1,0,0; 0,0,0,1] V(9) = [1,0,0,0; 0,0,1,0; 1,0,0,0] ... V(16) = [1,0,0,0; 0,0,0,1; 0,0,0,1] V(17) = [0,1,0,0; 1,0,0,0; 1,0,0,0] ... V(64) = [0,0,0,1; 0,0,0,1; 0,0,0,1]I guess this satisfies all your requirements. Even the order seems correct to me:
M=3;N=3; mat1=eye(M+1); vectors=mat2cell(repmat(1:M+1,N,1),ones(N,1),[M+1]);Super-efficient cartesian product, taken from here:
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% n = numel(vectors); %// number of vectors combs = cell(1,n); %// pre-define to generate comma-separated list [combs{end:-1:1}] = ndgrid(vectors{end:-1:1}); %// the reverse order in these two %// comma-separated lists is needed to produce the rows of the result matrix in %// lexicographical order combs = cat(n+1, combs{:}); %// concat the n n-dim arrays along dimension n+1 combs = reshape(combs,[],n); %// reshape to obtain desired matrix %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% V=cell(size(combs,1),1); for i=1:size(combs,1) for j=1:size(combs,2) V{i,1}=[V{i,1};mat1(combs(i,j),:)]; end endOutputs:
M=2,N=2; V= [1,0,0;1,0,0] [1,0,0;0,1,0] [1,0,0;0,0,1] [0,1,0;1,0,0] [0,1,0;0,1,0] [0,1,0;0,0,1] [0,0,1;1,0,0] [0,0,1;0,1,0] [0,0,1;0,0,1] M=3;N=3; %order verified for the indices given in the question V(1) = [1,0,0,0; 1,0,0,0; 1,0,0,0] V(2) = [1,0,0,0; 1,0,0,0; 0,1,0,0] V(3) = [1,0,0,0; 1,0,0,0; 0,0,1,0] V(4) = [1,0,0,0; 1,0,0,0; 0,0,0,1] V(5) = [1,0,0,0; 0,1,0,0; 1,0,0,0] ... V(8) = [1,0,0,0; 0,1,0,0; 0,0,0,1] V(9) = [1,0,0,0; 0,0,1,0; 1,0,0,0] ... V(16) = [1,0,0,0; 0,0,0,1; 0,0,0,1] V(17) = [0,1,0,0; 1,0,0,0; 1,0,0,0] ... V(64) = [0,0,0,1; 0,0,0,1; 0,0,0,1]更多推荐
发布评论