hi everyone , i need help to do this , thank u

clear all

clc

A=10*randn(1,8);

B=10*randn(1,8);

EA=A+2

EB=B+2

E=min(EA,EB)

% i need the value of A and B in a row vector that corrispond to the value of E

thank u

Answer 1

[Adam Danz]

The variable 'idx' tells you whether the minimum came from EA (idx=1) or EB (idx=2). Use that to pull out values from A and B as needed.

[E, idx] = min([EA;EB]);   %only works if EA and EB are same size

% If you want one row vector of A and one for B
A0 = A(idx==1); 
B0 = B(idx==2); 

% If you want one row vector that combines A and B
AB = nan(1, length(E));
AB(idx==1) = A(idx==1);
AB(idx==2) = B(idx==2);

[mina massoud]

thank u very much for ur responding

but how can i do it if i want the value of A and B in a row vector and not the index

thank u again

[mina massoud]

sorry i get the variabile AB and that what i want

thank u very much

u r the best

i accept ur answer

thak u

[Adam Danz]

Any time.

[mina massoud]

%sorry for asking u again , what if A and B is a matrix 2x8 and i need to get out the column %vector of AB that corrispond to the minimum value of E

A=10*randn(2,8);

B=10*randn(2,8);

MA=mean(A,1)

MB=mean(B,1)

E=min(MA,MB)

[E, idx] = min([MA;MB]); %only works if EA and EB are same size

% If you want one row vector that combines A and B

AB = nan(1, E);

AB(idx==1) = A(idx==1 );

AB(idx==2) = B(idx==2);

[Adam Danz]

The min value of E probably won't be represented at all in A or B since E will be from MA or MB which are means of the columns in A or B.

If I've misunderstood the question, maybe it would help to provide an example.

[mina massoud]

that really what i mean that E corrispond to the min value of MA and MB and what i need is the index of MA and MB to get out the corrispondent column vector of A and B

A >>>>> MA >>>>>> min E >>>> from this i need the corrispondat column vector of A

B>>>>>> MB >>>> min E >>>>> from this i need the corrispondat column vector of B

so at the end i need a matrix 2x8 that contain only the the mix of A and B hat corrispond to the minimum value of MA and MB

A=10*randn(2,8); % from this matrix in need only the column vector that corrispond to the min value of MA

B=10*randn(2,8); %from this matrix in need only the column vector that corrispond to the min value of MB

MA=mean(A,1)

MB=mean(B,1)

E=min(MA,MB)

thank u again for your willingness

[Adam Danz]

Is this what you're looking for?

[E, idx] = min([MA;MB]); 
AB = nan(1, length(E)); 
AB(idx==1) = A(idx==1 ); 
AB(idx==2) = B(idx==2);

[mina massoud]

% yes but it doesnt work on my code the last two line , can u fix it to me , thank u adam

clear all

clc

r0=20000; % raggio

fiA=pi/4; fiB=pi/4; fiC=pi/4; % angolo di elevazione

tetaA=(pi/4); tetaB=(pi/4)+(pi); tetaC=(pi/4)+(3*pi/2); % angolo azimutale

%% i range misurati del bersaglio uno e due

b1=[1200 100] ; % la posizione del primo bersaglio

R0TgT1=sqrt(b1(1,1)^2+b1(1,2)^2); % il modulo

%% posizione dei trasmettitori

TXA=[ r0*cos(fiA)*cos(tetaA) , r0*cos(fiA)*sin(tetaA) , r0*sin(fiA)];

TXB=[ r0*cos(fiB)*cos(tetaB) , r0*cos(fiB)*sin(tetaB) , r0*sin(fiB)];

TXC=[ r0*cos(fiC)*cos(tetaC) , r0*cos(fiC)*sin(tetaC) , r0*sin(fiC)];

%% la base line

rbA=sqrt(TXA(1,1)^2+TXA(1,2)^2+TXA(1,3)^2);

rbB=sqrt(TXB(1,1)^2+TXB(1,2)^2+TXB(1,3)^2);

rbC=sqrt(TXC(1,1)^2+TXC(1,2)^2+TXC(1,3)^2);

%% la distanza tra il satellite e il bersaglio uno

TXA_1=sqrt((TXA(1,1)-b1(1,1)).^2+(TXA(1,2)-b1(1,2)).^2+TXA(1,3).^2);

TXB_1=sqrt((TXB(1,1)-b1(1,1)).^2+(TXB(1,2)-b1(1,2)).^2+TXB(1,3).^2);

TXC_1=sqrt((TXC(1,1)-b1(1,1)).^2+(TXC(1,2)-b1(1,2)).^2+TXC(1,3).^2);

%% la distanza tra il satellite e il bersaglio due

%TXA_2=sqrt((TXA(1,1)-1000).^2+(TXA(1,2)-0).^2+TXA(1,3).^2);

%TXB_2=sqrt((TXB(1,1)-1000).^2+(TXB(1,2)-0).^2+TXB(1,3).^2);

%TXC_2=sqrt((TXC(1,1)-1000).^2+(TXC(1,2)-0).^2+TXC(1,3).^2);

%% range misurato del primo bersaglio

rA_1=TXA_1+R0TgT1-rbA;

rB_1=TXB_1+R0TgT1-rbB;

rC_1=TXC_1+R0TgT1-rbC;

%% range misurato del secondo bersaglio

%rA_2=TXA_2+R0TgT2-rbA;

%rB_2=TXB_2+R0TgT2-rbB;

%rC_2=TXB_2+R0TgT2-rbC;

%% adesso vogliamo calcolare la norma del errore che esce fuori dovuta al fatto che sto facendo delle combinazioni sbagliate

k=0;

intx=10; inty=10; D_x=3000; D_y=1000;

x=[10:intx:D_x]; y=[-D_y/2:inty:D_y/2-1];

var_A=15^2; var_B=15^2; var_C=15^2;

for i=1:length(x)

for j=1:length(y)

TgT=[x(i),y(j)];

k=k+1;

TGT(k,:)=TgT;

TXA_2=sqrt((TXA(1,1)-x(i)).^2+(TXA(1,2)-y(j)).^2+TXA(1,3).^2);

TxA_2(i,j)=TXA_2;

TXB_2=sqrt((TXB(1,1)-x(i)).^2+(TXB(1,2)-y(j)).^2+TXB(1,3).^2);

TxB_2(i,j)=TXB_2;

TXC_2=sqrt((TXC(1,1)-x(i)).^2+(TXC(1,2)-y(j)).^2+TXC(1,3).^2);

TxC_2(i,j)=TXC_2;

R0TgT2=sqrt(x(i)^2+y(j)^2);

rA_2=TXA_2+R0TgT2-rbA;

rB_2=TXB_2+R0TgT2-rbB;

rC_2=TXC_2+R0TgT2-rbC;

RA_2(i,j)=(rA_2); % una matrice 300x100

RB_2(i,j)=(rB_2);

RC_2(i,j)=(rC_2);

for n=1:8

Rg=[rA_1 rA_2 ;rB_1 rB_2 ;rC_1 rC_2 ] ;

[X,Y,Z] = ndgrid(1:2 );

Cm = [Z(:),Y(:),X(:)] ;

r = Rg(sub2ind(size(Rg),repmat(1:3,8,1),Cm ));

%sub2ind (MATLAB Functions) The sub2ind command determines the equivalent single index corresponding to a set

%of subscript values. IND = sub2ind(siz,I,J) returns the linear index equivalent to the row and column

%subscripts I and J for a matrix of size siz .

A= [ TXA; TXB ;TXC ] ;% è una matrice 3x2

A(:,3)=[] ;

%b) definizione del vettore costante K

K= 1/2* [ rbA^2-(r(n,1)+rbA)^2 ; rbB^2-(r(n,2)+rbB)^2 ; rbC^2-(r(n,3)+rbC)^2 ] ; % è una matrice 2x1

% definizione del vettore C

% il vettore C è la somma di due vettori costanti ri+Rbi

C = [ r(n,1) ; r(n,2) ;r(n,3) ] + [ rbA ; rbB; rbC ] ; % è una matrice 2x1

% definizione del vettore a , b

invATA_AT= inv(A'*A)*A'; % è una matrice 3*3

%adesso definisco 'a' e 'b'

a= invATA_AT*K;% è una matrice 3*1

b= invATA_AT*C; % è una matrice 3*1

aTa=a'*a; % 1x1

aTb=a'*b; % 1x1

bTb=b'*b; %1x1

Rt_num_n=-aTb + (sqrt((aTb)^2 -(bTb-1)*aTa));

Rt_denum_n= bTb-1;

R0_n(n)=Rt_num_n/Rt_denum_n;% la misura di R0 positivo 1x1

Rn(n)=norm(R0_n(n));

R_N{i,j}=Rn;

Rt_num_p=-aTb - (sqrt((aTb)^2 -(bTb-1)*aTa));

Rt_denum_p= bTb-1;

R0_p(n)=Rt_num_p/Rt_denum_p; % la misura di R0 negativo 1x1

Rp(n)=norm(R0_p(n));

R_P{i,j}=Rp;

KCX_n=K+C*(R0_n); % il calcolo di (K+C||x||) per il valore di R0 positivo che è una matrice 2x1

KCX_p(:,n)=K+C*(R0_p(:,n)); % il calcolo di (K+C||x||) per il valore di R0 negativo che è una matrice 2x1

x_n(:,n)= invATA_AT*KCX_n(:,n); %è una matrice 3x1 che è il calcole della localizzazione del besaglio per il valore di R0 positivo

X_n{i,j}=x_n;

R0n_st(n)=sqrt(x_n(1,n)^2+x_n(2,n)^2);

x_p(:,n)= invATA_AT*KCX_p(:,n);% è una matrice 3x1 che è il calcole della localizzazione del besaglio per il valore di R0 negativo

X_p{i,j}=x_p;

R0p_st(n)=sqrt(x_p(1,n)^2+x_p(2,n)^2);

R_s{i,j}=[Rn ; Rp];

R_S{i,j}=min(Rn,Rp);

Rm{i,j}=zeros(2,8);

[R_S{i,j}, idx{i,j}] = min([R_N{i,j};R_P{i,j}]);

Rm{i,j} = nan(2, length(R_S ));

Rm(:,idx==1) = x_n(:,idx==1 )

Rm(:,idx==2) = x_p(:,idx==2)

end

end

end

[mina massoud]

i tried also in this way but still not working

Rm{i,j}(:,idx==1) = x_n{i,j}(:,idx==1 )

Rm{i,j}(:,idx==2) = x_p{i,j}(:,idx==2)

[mina massoud]

sorry i have to write it in this way

Rm{i,j}(:,idx==1) = X_n{i,j}(:,idx==1 )

Rm{i,j}(:,idx==2) = X_p{i,j}(:,idx==2)

but still not working

[Adam Danz]

This works

A=10*randn(2,8);
B=10*randn(2,8);
MA=mean(A,1);
MB=mean(B,1);

[E, idx] = min([MA;MB]); 
AB = nan(1, length(E)); 
AB(idx==1) = A(idx==1 ); 
AB(idx==2) = B(idx==2);

I haven't read through your code but if there's a specific part that you're having trouble implementing, share that part of the code and format it the same way I format the code so it's readable.

[mina massoud]

%this is the part of the code that doesn't work

for i=1:300

for j=1:100

A{i,j}=10*randn(2,8);

B{i,j}=10*randn(2,8);

MA{i,j}=mean(A{i,j},1);

MB{i,j}=mean(B{i,j},1);

[E{i,j}, idx{i,j}] = min([MA{i,j};MB{i,j}]);

AB{i,j} = nan(2, length(E{i,j}));

AB{i,j}(idx==1) = A{i,j}(idx==1 );

AB{i,j}(idx==2) = B{i,j}(idx==2);

end

end

% thank you adam

[mina massoud]

% i tried also this , but still not working

for i=1:300

for j=1:100

A{i,j}=10*randn(2,8);

B{i,j}=10*randn(2,8);

MA=mean(A{i,j},1);

MB=mean(B{i,j},1);

[E, idx] = min([MA;MB]);

AB = nan(2, length(E));

AB(idx==1) = A(:,idx==1);

AA{i,j}=AB

AB(idx==2) = B(:,idx==2);

AA{i,j}=AB

end

end