How to store values matrix in cell

Question

Tipu Sultan 2019-5-13

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/461764-how-to-store-values-matrix-in-cell

评论： Tipu Sultan 2019-5-13

采纳的回答： Alex Mcaulley

在 MATLAB Online 中打开

Hello,

I have an array as follows:

t= [ 1 2 3 ];

A matrix as follows:

 M = [t^2 t 1 0 0 0;
         0 0 0 t^2 t 1];

Now I want to store the process the M for 3 values of t i.e in this case [1 2 3].How do I do it.

like if t=1

then M=[1^2 1 1 0 0 0;0 0 0 1^2 1 1];

if t=2

then M=[2^2 2 1 0 0 0;0 0 0 2^2 2 1]; and so on.

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Answer 1

Alex Mcaulley 2019-5-13

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/461764-how-to-store-values-matrix-in-cell#answer_374727

编辑：Alex Mcaulley 2019-5-13

在 MATLAB Online 中打开

Try this:

t = [1 2 3];
M = arrayfun(@(t) [t^2 t 1 0 0 0; 0 0 0 t^2 t 1],t,'UniformOutput',false);

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Alex Mcaulley 2019-5-13

在 MATLAB Online 中打开

You can use it in your scrip as:

S = [100 0 0 0 0 0;
    0 100 0 0 0 0;
    0 0 100 0 0 0;
    0 0 0 100 0 0;
    0 0 0 0 100 0;
    0 0 0 0 0 100];
prev_S = S;
Big_lambda = eye(2);
a=0;
b=0;
c=0;
p=0;
q=0;
s=0;
est_vec=[ a ; b; c; p; q; s];
theta = [ 45 46 48] ;
t= [ 1 2 3 ];
r= [200 210 220];
%Wanted = num2cell(M,1)
%[r,b,distance,angle]=deal(Wanted{:})
x = [ r; theta ; t];
%dif_x=zeros(2,3);
M = arrayfun(@(t) [t^2 t 1 0 0 0; 0 0 0 t^2 t 1],t,'UniformOutput',false);
time=4;
for i=1:3
     dif_x = [(-cos(theta(i))) (r(i).*sin(theta(i))) (2*a.*t(i)+b);(-sin(theta(i))) (-r(i).*cos(theta(i))) (2*p.*t(i)+q)]
     
     W = dif_x(i)*Big_lambda(i)*dif_x(i)'
     
       pred_x = x+randn(3)
     
     y = M{i} * est_vec + dif_x * (x(:,i)-pred_x(:,i))
     
     K = prev_S*M{i}'/((W + M{i}*prev_S*M{i}'))
     
     %S =prev_S;
          
     est_vec_new = est_vec +  K*(y-M{i}*est_vec)
     
     cond = abs(est_vec_new - est_vec)
     
     if  cond < 0.003
         break 
     end
     
     est_vec = est_vec_new
     a=est_vec(1,1)
     b=est_vec(2,1)
     c=est_vec(3,1)
     p=est_vec(4,1)
     q=est_vec(5,1)
     s=est_vec(6,1)
     
     S_new = (eye(6) - K*M{i})*S
     
     S = S_new
     
     r_cosTheta = a*time.^2+b*time+c % To calculate x co-ordinate for t>=3
     r_sineTheta = p*time.^2+q*time+s % To calculate y co-ordinate for t>=3
     
     %figure,plot(t,meas_equa1)
     %new_t=[t,time];
%figure,plot(r_cosTheta,r_sineTheta)
end
%end

Tipu Sultan 2019-5-13

Thanks it is working!

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Answer 2

KSSV 2019-5-13

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/461764-how-to-store-values-matrix-in-cell#answer_374726

在 MATLAB Online 中打开

M = @(t) [t^2 t 1 0 0 0;
         0 0 0 t^2 t 1];
     
t = [1 2 3] ; 
iwant = zeros(2,6,length(t)) ; 
for i = 1:length(t)
    iwant(:,:,i) = M(t(i)) ; 
end

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Tipu Sultan 2019-5-13

在 MATLAB Online 中打开

S = [100 0 0 0 0 0;
    0 100 0 0 0 0;
    0 0 100 0 0 0;
    0 0 0 100 0 0;
    0 0 0 0 100 0;
    0 0 0 0 0 100];
prev_S = S;
Big_lambda = eye(2);
a=0;
b=0;
c=0;
p=0;
q=0;
s=0;
est_vec=[ a ; b; c; p; q; s];
theta = [ 45 46 48] ;
t= [ 1 2 3 ];
r= [200 210 220];
%Wanted = num2cell(M,1)
%[r,b,distance,angle]=deal(Wanted{:})
x = [ r; theta ; t];
%dif_x=zeros(2,3);
M = @(t) [t^2 t 1 0 0 0;
         0 0 0 t^2 t 1];
     
t = [1 2 3] ; 
iwant = zeros(2,6,length(t)) ; 
for i = 1:length(t)
    iwant(:,:,i) = M(t(i)) ; 
end
time=4;
for i=1:3
     dif_x = [(-cos(theta(i))) (r(i).*sin(theta(i))) (2*a.*t(i)+b);(-sin(theta(i))) (-r(i).*cos(theta(i))) (2*p.*t(i)+q)]
     
     W = dif_x(i)*Big_lambda(i)*dif_x(i)'
     
       pred_x = x+randn(3)
     
     y = M * est_vec + dif_x * (x(:,i)-pred_x(:,i))
     
     K = prev_S*M'/((W + M*prev_S*M'))
     
     %S =prev_S;
          
     est_vec_new = est_vec +  K*(y-M*est_vec)
     
     cond = abs(est_vec_new - est_vec)
     
     if  cond < 0.003
         break 
     end
     
     est_vec = est_vec_new
     a=est_vec(1,1)
     b=est_vec(2,1)
     c=est_vec(3,1)
     p=est_vec(4,1)
     q=est_vec(5,1)
     s=est_vec(6,1)
     
     S_new = (eye(6) - K*M)*S
     
     S = S_new
     
     r_cosTheta = a*time.^2+b*time+c % To calculate x co-ordinate for t>=3
     r_sineTheta = p*time.^2+q*time+s % To calculate y co-ordinate for t>=3
     
     %figure,plot(t,meas_equa1)
     %new_t=[t,time];
%figure,plot(r_cosTheta,r_sineTheta)
end
%end

This is my whole code now when I am checking in workspace it says M as function handle.

error as follows:

Undefined operator '*' for input arguments of type 'function_handle'.

KSSV 2019-5-13

在 MATLAB Online 中打开

clc; clear all ;
S = [100 0 0 0 0 0;
    0 100 0 0 0 0;
    0 0 100 0 0 0;
    0 0 0 100 0 0;
    0 0 0 0 100 0;
    0 0 0 0 0 100];
prev_S = S;
Big_lambda = eye(2);
a=0;
b=0;
c=0;
p=0;
q=0;
s=0;
est_vec=[ a ; b; c; p; q; s];
theta = [ 45 46 48] ;
t= [ 1 2 3 ];
r= [200 210 220];
%Wanted = num2cell(M,1)
%[r,b,distance,angle]=deal(Wanted{:})
x = [ r; theta ; t];
%dif_x=zeros(2,3);
M = @(t) [t^2 t 1 0 0 0;
    0 0 0 t^2 t 1];
t = [1 2 3] ;
iwant = zeros(2,6,length(t)) ;
for i = 1:length(t)
    iwant(:,:,i) = M(t(i)) ;
end
time=4;
for i=1:3
    M = iwant(:,:,i) ; 
    dif_x = [(-cos(theta(i))) (r(i).*sin(theta(i))) (2*a.*t(i)+b);(-sin(theta(i))) (-r(i).*cos(theta(i))) (2*p.*t(i)+q)]
    
    W = dif_x(i)*Big_lambda(i)*dif_x(i)'
    
    pred_x = x+randn(3)
    
    y = M * est_vec + dif_x * (x(:,i)-pred_x(:,i))
    
    K = prev_S*M'/((W + M*prev_S*M'))
    
    %S =prev_S;
    
    est_vec_new = est_vec +  K*(y-M*est_vec)
    
    cond = abs(est_vec_new - est_vec)
    
    if  cond < 0.003
        break
    end
    
    est_vec = est_vec_new
    a=est_vec(1,1)
    b=est_vec(2,1)
    c=est_vec(3,1)
    p=est_vec(4,1)
    q=est_vec(5,1)
    s=est_vec(6,1)
    
    S_new = (eye(6) - K*M)*S
    
    S = S_new
    
    r_cosTheta = a*time.^2+b*time+c % To calculate x co-ordinate for t>=3
    r_sineTheta = p*time.^2+q*time+s % To calculate y co-ordinate for t>=3
    
    %figure,plot(t,meas_equa1)
    %new_t=[t,time];
    %figure,plot(r_cosTheta,r_sineTheta)
end
%end