q=zeros(size(R)); % initialize Q as zero,q的行数和列数等于矩阵R的。 q1=ones(size(R))*inf; % initialize previous Q as big number count=0; % counter for episode=0:50000 % random initial state y=randperm(size(R,1));%产生1到6的随机数%a=size(R,1)把矩阵R的行数返回给a,b=size...
function [X,FVAL,EXITFLAG,OUTPUT,LAMBDA,GRAD,HESSIAN] = fmincon(FUN,X,A,B,Aeq,Beq,LB,UB,NONLCON,options,varargin) /*fmincon可以在多元函数中找到最小值 FMINCON attempts to solve problems of the form: min F(X) subject to: A*X <= B, Aeq*X = Beq (linear constraints)线性约束 X C(X...
% Copy each example into a patch on the display array curr_ex = 1; for j = 1:display_rows for i = 1:display_cols if curr_ex > m, break; end % Copy the patch % Get the max value of the patch max_val = max(abs(X(curr_ex, :))); display_array(pad + (j - 1) * (ex...
Another way to create a matrix is to use a function, such as ones, zeros, or rand. For example, create a 5-by-1 column vector of zeros. 创建矩阵的另一种方法是使用函数,ones(),zeros(),rand()。例如,创建一个五行一列的零向量。 z = zeros(5,1) z = 0 0 0 0 0 } Matrix and A...
datain = vertcat(datain,zeros(N-size1,1)); else datain = horzcat(datain,zeros(1,N-size2)); end end % Transform datain into frequency domain via FFT and shift output (A) % so that zero-frequency amplitude is in the middle of the array ...
2、The expression,A() = [] delete rows or columns of A(删除A的行或列) 答案: 1)第一步:A([4 6])=[76 0] 2)第二步:A([4 5 7 8])=zeros 3)第三步:A(3,:) = [] 九、Colon Operator(冒号操作,即等差级数) 1、Want to create a long array:A=[1 2 3 ... 100] ...
I'm trying to solve fzero for each value of x in an array, and I get theis error: Operands to the || and && operators must be convertible to logical scalar values. Error in fzero (line 444) while fb ~= 0 && a ~= b
E=zeros(10,5,3)表示生成十行五列且为三位 的0矩阵命令行窗口中E=(:,:,1)即表示一维矩阵 E=(:,:,1)=rand(10,5) rand生成均匀分布的伪函数,分布在(0~1) 之间 E=(:,:,1)=randn(10,5) randn生成标准正态分布的伪随机数(均值为 0,方差为1) ...
function K = karray (k,N) K = @(x) [zeros(N,N)]; for i=1:1:N for j=1:1:N K(i,j) = (k(i)*gamma(i)+k(i+1)*gamma(i+1))*x(i)^2+k(i)+k(i+1); K(i,j+1) = -3*k(i+1)*gamma(i+1)*x(i)^2-k(i+1); ...
to be zeros and make it 65 untis long. then set the first value of the matrix to be -pi/2. then increase the interval or the point specified in the matrix and then just add the previous value in the matrix plus pi/2*1/64. i am trying to figure out if there is ...