Please your help is needed so that i can fprint results of a matrix(say matrix M) at each iteration of a for loop.

Please your help is needed so that i can fprint results of a matrix(say matrix M) at each iteration of a for loop. Because i want to work with a particular matrix of all the iterations and i want to know at what iteration that such matrix occurred. Many thanks in advance.

[Stephen23]

@Peju Oyewole: what is your question? What have you tried so far?

[Peju Oyewole]

I want to have full representation of each matrix at each iteration. I have used 'disp' which is not give me each matrix at each iteration Thanks.

[Peju Oyewole]

I am doing an operation to get a matrix M_C2 from If…elseif condition within a two for loop in parallel. The for loop will be running for 8760(for i=1:8760) for a specific value at each iteration. At some iterations based on the if condition, matrix M_C2 will be generated. I want any matrix generated at a particular iteration to be saved into a text file and at what iteration that matrix was generated, and I need to recall the text file to be again utilized as the primary data file for further operations within the loop. Also my two parallel for loop will be repeated for 1000 times. Also I want to know numbers of matrix generated at the end of the 8760times and save. Below is some of my code Tvals = 1:8760;%hourly iteration a4vals = [0.63 0.62 0.6 0.58 0.59 0.65 0.74 0.86.......] %hourly load demand for K = 1 : length(Tvals) T = Tvals(K); a4= a4vals(K); .........my working ........ U=rand % if conditions if U>=PRO1(J,2)+PRO1(J,3)+PRO1(J,4) counter = counter + 1 disp ('up state') elseif (PRO1(J,2)+PRO1(J,3)<=U) && (U<PRO1(J,2)+PRO1(J,3)+PRO1(J,4)) counter_1 = counter_1 + 1 M_C(J,:)=zeros(1,5)% To identify the failed component M_C2=M_C Many thanks.

Answer 1

[Image Analyst]

Just put the name of the matrix on its own line without a semicolon in your loop

for k = 1 : whatever
    matrixName
end

It will print to the command window. Or you can use fprintf():

for row = 1 : size(m, 1)
    fprintf('%f ', m(row, :));
    fprintf('\n');
end

[Peju Oyewole]

Please Above code is not working for me.

I am doing an operation to get a matrix M_C2 from If…elseif condition within a two for loop in parallel. The for loop will be running for 8760(for i=1:8760) for a specific value at each iteration. At some iterations based on the if condition, matrix M_C2 will be generated. I want any matrix generated at a particular iteration to be saved into a text file and at what iteration that matrix was generated, and I need to recall the text file to be again utilized as the primary data file for further operations within the loop. Also my two parallel for loop will be repeated for 1000 times. Also I want to know numbers of matrix generated at the end of the 8760times and save. Below is some of my code

Tvals = 1:8760;%hourly iteration
a4vals = [0.63 0.62 0.6	0.58 0.59 0.65 0.74 0.86.......] %hourly load demand      
for K = 1 : length(Tvals)
    T = Tvals(K);
   a4= a4vals(K);
.........my working
........
U=rand    % if conditions
        if U>=PRO1(J,2)+PRO1(J,3)+PRO1(J,4)
            counter = counter + 1
            disp ('up state')
        elseif (PRO1(J,2)+PRO1(J,3)<=U) && (U<PRO1(J,2)+PRO1(J,3)+PRO1(J,4))
            counter_1 = counter_1 + 1
            M_C(J,:)=zeros(1,5)% To identify the failed component
            M_C2=M_C

Many thanks.

[Image Analyst]

I don't see my code in there. No fprintf(). Either attach whichever matrixes you want in a .mat file

save('peju.mat', 'M_C2', 'M_C', 'PRO1');

and attach peju.mat, or give me code that actually runs. My code should work. You don't need a double for loop like mini suggested. I'll check later today (I'm off soon for a day of sightseeing in Beijing).

[Peju Oyewole]

The result on the command window is overwrite, please I need the matrix M_C2(70,5) generated based on the if condition to be saved on a text file(or better option) at the hour(iteration) the matrix is generated. I don't want any of the generated matrix to be overwrite. that is, if at 3rd iteration(hour) a matrix(70,5) is generated, I want it save for 3rd iteration(hour),if at 10th iteration(hour) a matrix(70,5) is generated, I want it save for 10th iteration(hour), because any matrix generated at each iteration is used for further operation within the for loop. Thanks.

[Image Analyst]

Then you need to index it. I don't know why you chose J as the index when it's K that is your loop index. You need to use the proper index.

[Peju Oyewole]

J is another for loop within the loop. The loop I'm working with is Tvals = 1:8760;%hourly iteration a4vals = [0.63 0.62 0.6 0.58 0.59 0.65 ......] %hourly load demand for K = 1 : length(Tvals) T = Tvals(K); a4= a4vals(K); ......some of working Bus_data=Bus_data1.*a4vals(K); M_C2=M_C; M_C2 Note; M_C is generated based on if conditional statement. Whenever M_C(70,5) is generated I want it written and saved in a text file for further operation within the for loop for that hour(iteration) and the result saved for that hour(iteration). Thanks.

[Image Analyst]

It might be easier if you just provided us with the full code rather than partial snippets. Attach any data that needs to be read in.

To format your code read this link

Before you paste in here, type control-a, control-i, in the MATLAB editor, to fix up your indenting.

[Peju Oyewole]

M_C=[1.0000 10.0000 0 10.0000 0; 1.0000 10.0000 0 10.0000 0; 1.0000 76.0000 14.1000 30.0000 -25.0000; 1.0000 76.0000 14.1000 30.0000 -25.0000; 2.0000 10.0000 0 10.0000 0; 2.0000 10.0000 0 10.0000 0; 2.0000 76.0000 7.0000 30.0000 -25.0000; 2.0000 76.0000 7.0000 30.0000 -25.0000; 7.0000 80.0000 17.2000 60.0000 0; 7.0000 80.0000 17.2000 60.0000 0; 7.0000 80.0000 17.2000 60.0000 0; 13.0000 95.1000 40.7000 80.0000 0; 13.0000 95.1000 40.7000 80.0000 0; 13.0000 95.1000 40.7000 80.0000 0; 15.0000 12.0000 0 6.0000 0; 15.0000 12.0000 0 6.0000 0; 15.0000 12.0000 0 6.0000 0; 15.0000 12.0000 0 6.0000 0; 15.0000 12.0000 0 6.0000 0; 15.0000 155.0000 0.0500 80.0000 -50.0000; 16.0000 155.0000 25.2200 80.0000 -50.0000; 18.0000 400.0000 137.4000 200.0000 -50.0000; 21.0000 400.0000 108.2000 200.0000 -50.0000; 22.0000 50.0000 -4.9600 16.0000 -10.0000; 22.0000 50.0000 -4.9600 16.0000 -10.0000; 22.0000 50.0000 -4.9600 16.0000 -10.0000; 22.0000 50.0000 -4.9600 16.0000 -10.0000; 22.0000 50.0000 -4.9600 16.0000 -10.0000; 22.0000 50.0000 -4.9600 16.0000 -10.0000; 23.0000 155.0000 31.7900 80.0000 -50.0000; 23.0000 155.0000 31.7900 80.0000 -50.0000; 23.0000 350.0000 71.7800 150.0000 -25.0000; 33.0000 1.0000 2.0000 0.0030 0.0140; 34.0000 1.0000 3.0000 0.0550 0.2110; 35.0000 1.0000 5.0000 0.0220 0.0850; 36.0000 2.0000 4.0000 0.0330 0.1270; 37.0000 2.0000 6.0000 0.0500 0.1920; 38.0000 3.0000 9.0000 0.0310 0.1190; 39.0000 3.0000 24.0000 0.0020 0.0840; 40.0000 4.0000 9.0000 0.0270 0.1040; 41.0000 5.0000 10.0000 0.0230 0.0880; 42.0000 6.0000 10.0000 0.0140 0.0610; 43.0000 7.0000 8.0000 0.0160 0.0610; 45.0000 8.0000 9.0000 0.0430 0.1650; 46.0000 8.0000 10.0000 0.0430 0.1650; 47.0000 9.0000 11.0000 0.0020 0.0840; 48.0000 9.0000 12.0000 0.0020 0.0840; 49.0000 10.0000 11.0000 0.0020 0.0840; 50.0000 10.0000 12.0000 0.0020 0.0840; 51.0000 11.0000 13.0000 0.0060 0.0480; 52.0000 11.0000 14.0000 0.0050 0.0420; 53.0000 12.0000 13.0000 0.0060 0.0480; 54.0000 12.0000 23.0000 0.0120 0.0970; 55.0000 13.0000 23.0000 0.0110 0.0870; 57.0000 14.0000 16.0000 0.0050 0.0590; 58.0000 15.0000 16.0000 0.0020 0.0170; 59.0000 15.0000 21.0000 0.0060 0.0490; 60.0000 15.0000 21.0000 0.0060 0.0490; 61.0000 15.0000 24.0000 0.0070 0.0520; 62.0000 16.0000 17.0000 0.0030 0.0260; 63.0000 16.0000 19.0000 0.0030 0.0230; 64.0000 17.0000 18.0000 0.0020 0.0140; 65.0000 17.0000 22.0000 0.0140 0.1050; 66.0000 18.0000 21.0000 0.0030 0.0260; 67.0000 18.0000 21.0000 0.0030 0.0260; 68.0000 19.0000 20.0000 0.0050 0.0400; 69.0000 19.0000 20.0000 0.0050 0.0400; 70.0000 20.0000 23.0000 0.0030 0.0220; 71.0000 20.0000 23.0000 0.0030 0.0220; 72.0000 21.0000 22.0000 0.0090 0.0680]; Tvals = 1:8760;%hourly iteration a4vals = [0.63 0.62 0.6 0.58 0.59 0.65 .......] %hourly load demand for K = 1 : length(Tvals) T = Tvals(K); a4= a4vals(K); for J=1:70 J L=Component_data(:,1); U=Component_data(:,2);

        syms 's' 'l' 'u' 't'
        CompMat=[7*l -4*u -2*u -1*u;-4*l 1*l+6*u -2*l -1*u;-2*l -2*u 3*l+2*u -1*u;-1*l -1*l -1*l 3*u];
        diag_s=[s 0 0 0; 0 s 0 0; 0 0 s 0; 0 0 0 s];
        CompMat_2=diag_s+CompMat;
        D_CompMat_2=det(CompMat_2);
        F_D_CompMat_2 = factor(D_CompMat_2);
        P=inv(CompMat_2)*[1 0 0 0]';

        for i=1:3
            i
            [n, d] = numden(P(i));
            F=simplifyFraction(P(i));
            E_n = expand(n);
            F_n = factor(n);
            F_d=factor(d);
            syms s l u A B C D real
            T_deno=F_d';
            K_1= expand(A*T_deno(2)*T_deno(3));
            K_2=expand((B)*(T_deno(1))*(T_deno(3)));
            K_3=expand(((C*s)+D)*T_deno(1)*T_deno(2));
            K_1_S=subs(K_1,s,0);%When s=0 to solve for A
            K_2_S=subs(K_2,s,0);
            K_3_S=subs(K_3,s,0);
            E_n_S=subs(E_n,s,0);
            F_E_n_S=factor(E_n_S);
            K_T_S=K_1_S+K_2_S+K_3_S;
            F_K_T_S=factor(K_T_S,A);
            Mat_F_K_T_S=F_K_T_S';
            P_F_A=simplify(E_n_S/Mat_F_K_T_S(1));%Value of A
            pretty(P_F_A);
            P_F_A_D=double(subs(P_F_A,[l,u],[L(J),U(J)]));
            K_1_S_B=subs(A*T_deno(2)*T_deno(3),s,(-l-u));%When s=-l-u to solve for B
            K_2_S_B=subs(B*T_deno(1)*T_deno(3),s,(-l-u));
            K_2_S_BB=simplify(K_2_S_B);
            K_3_S_B=subs(((C*s)+D)*T_deno(1)*T_deno(2),s,(-l-u));
            E_n_SB=simplify(subs(E_n,s,(-l-u)));
            P_F_B=simplify((E_n_SB/K_2_S_BB)*B);%Value of B
            %pretty(P_F_B);
            P_F_B_D=double(subs(P_F_B,[l,u],[L(J),U(J)]));
            J_1=expand((P_F_A)*(T_deno(2)*T_deno(3)));
            J_2=expand((P_F_B)*(T_deno(1)*T_deno(3)));
            J_3=expand(((C*s)+D)*(T_deno(1)*T_deno(2)));
            J_1_C = collect(J_1,s);
            J_1_C_coeff=coeffs(J_1_C,s);
            J_1_C_coeff_Flip=fliplr(J_1_C_coeff);
            J_TT=(J_1)+(J_2)+(J_3);
            J_TT_Coll = collect(J_TT,s);
            J_TT_coeff=coeffs(J_TT_Coll,s);
            J_TT_coeff_Flip=fliplr(J_TT_coeff);
            J_TT_coeff_Flip_Transpose=J_TT_coeff_Flip';
            E_n_Coll = collect(E_n,s);
            E_n_coeff=coeffs(E_n_Coll,s);
            E_n_coeff_Flip=fliplr(E_n_coeff);
            E_n_coeff_Flip_Transpose=E_n_coeff_Flip';
            Value_C=(-1*((J_TT_coeff_Flip_Transpose(1))-C));%Value of C
            Value_C_F=simplify(Value_C);%Value of C
            Value_C_F_D=double(subs(Value_C_F,[l,u],[L(J),U(J)]));
            pretty(Value_C_F);
            Q_D=subs(J_TT_coeff_Flip_Transpose(2),C,Value_C);
            Q_D_1=(-1*((Q_D)-D))+E_n_coeff_Flip_Transpose(2);%Value of D_1
            Q_D_1_F=simplify(Q_D_1);%Value of D_1
            pretty(Q_D_1_F);
            Q_D_1_F_D=double(subs(Q_D_1_F,[l,u],[L(J),U(J)]));
            Q_D_2=(J_TT_coeff_Flip_Transpose(3))-((D*l) + (D*u));
            Q_DD_2=(-1*(Q_D_2))+E_n_coeff_Flip_Transpose(3);%Value of D_2
            Q_DD_2_F=simplify(Q_DD_2);%Value of D_2
            %pretty(Q_DD_2_F);
            Q_DD_2_F_D=subs(Q_DD_2_F,[l,u],[L(J),U(J)]);
            Final_P_L_1=ilaplace(A*(1/s));
            Final_P_L_1_D=double(subs(Final_P_L_1,A,P_F_A_D));
            Final_P_L_2=ilaplace(B*1/T_deno(2));
            Final_P_L_2_S=subs(Final_P_L_2,B,P_F_B_D);
            Final_P_L_2_S_D=double(subs(Final_P_L_2_S,[l,u,t],[L(J),U(J),T]));
            %Final_P_L_2_S_D=double(subs(Final_P_L_2,[l,u,t],[L(J),U(J),T]))
            Final_P_L_33_S=subs((C*s+D)/T_deno(3),[l,u,C,D],[L(J),U(J),Value_C_F,Q_D_1_F]);
            Final_P_L_33_S_L=ilaplace(Final_P_L_33_S);
            Final_P_L_33_S_L_S=subs(Final_P_L_33_S_L,[l,u,],[L(J),U(J)]);
            Final_P_L_33_S_L_S_D=double(subs(Final_P_L_33_S_L_S,t,T));
            Final_P_L_3=ilaplace((C*s+D)/T_deno(3));

            Final_P_L_3_D=subs(Final_P_L_3,[l,u,C,D],[L(J),U(J),Value_C_F_D,Q_D_1_F_D]);
            Final_P_L_3_W=rewrite(Final_P_L_3,'exp');
            Final_P_L_3_W_D=subs(Final_P_L_3_W,[l,u,C,D],[L(J),U(J),Value_C_F_D,Q_D_1_F_D]);
            Final_P_L_3_W_D_D=double(subs(Final_P_L_3_W_D,t,T));
            %Final_P_L_321=(Final_P_L_1)+(Final_P_L_2)+(Final_P_L_3);
            Final_P_3_S=subs(((C*s)+D)/T_deno(3),[C,D],[Value_C_F,Q_D_1_F]);
            Final_P_3_S_L=ilaplace(Final_P_3_S);
            Final_P_3_S_L_W=rewrite(Final_P_3_S_L,'exp');
            %pretty(Final_P_3_S_L_W)
            Final_P_3_S_L_W_S=subs(Final_P_3_S_L_W,[l,u,t],[L(J),U(J),T]);
            %=simplify(Final_P_3_S_L_W_S)
            Final_P_3_S_L_W_S_D=double(subs(Final_P_3_S_L_W,[l,u,t],[L(J),U(J),T]));
            For_Disp=Final_P_L_1+Final_P_L_2+Final_P_L_3;
            PRO(J,i)=(Final_P_L_1_D)+(Final_P_L_2_S_D)+(Final_P_L_3_W_D_D);

        end
    Final_P_1_L=ilaplace(P(4));
    ST_PR_4(J)=double(subs(Final_P_1_L,[l,u,t],[L(J),U(J),T]));
    ST_PR_4';
    PRO;
    PRO1=[PRO,ST_PR_4'];
    PROB_UP=PRO1(J,1);
    PROB_SMART=PRO1(J,2);
    PROB_DERATED=PRO1(J,3);
    PROB_FaILED=PRO1(J,4);
    FOR_S_UP=PRO1(J,1)+PRO1(J,2)+PRO1(J,3)+PRO1(J,4);
    FOR_UP=PRO1(J,2)+PRO1(J,3)+PRO1(J,4);
    FOR_DETD=PRO1(J,2)+ PRO1(J,3);
    FOR_SMT=PRO1(J,2);

    U=rand
        if U>=PRO1(J,2)+PRO1(J,3)+PRO1(J,4)
            counter = counter + 1
            disp ('up state')
        elseif (PRO1(J,2)+PRO1(J,3)<=U) && (U<PRO1(J,2)+PRO1(J,3)+PRO1(J,4))
            counter_1 = counter_1 + 1
            M_C(J,:)=zeros(1,5)% To identify the failed component

        elseif (PRO1(J,2)<=U)&&(U<PRO1(J,2)+ PRO1(J,3))
            counter_2 = counter_2 + 1
            disp ('smart')

        elseif (0<=U)&&(U<PRO1(J,2))
            counter_3 = counter_3 + 1 
            disp ('derated')

        end

     end
   M_C2=M_C;
   M_C2(K)
 indices1 = find(M_C2(:,2)==0)
   Line_INF=M_C2(33:end,:)
    % To get the Line data 
    Line_data=xlsread('myfileD.xlsx','sheet20','A2:B39')%Calling other Line data
    Flinedata=[Line_INF,Line_data]   %Horizontally Concatenating the matrix
    Flinedata(:,1)=1:38 %Replacing the column 1 with 1-38
    %Deleting any line that is not available
    indices = Flinedata(:,2)==0 %To check for any column with zeros
    Flinedata(indices,:) = []          %To remove any row(s)with zeros
    T_Line = array2table(Flinedata, 'VariableNames', {'Line_No', 'From_Bus', 'To_Bus','R_pu', 'X_pu','B_pu','Xmer_Tap'})
    Line_No = Flinedata(:,1);
    From_Bus = Flinedata(:,2);
    To_Bus = Flinedata(:,3);
    R = Flinedata(:,4); %Resistance between buses (transmission lines or power transformers)
    X = Flinedata(:,5); %Reactance between buses (transmission lines or power transformers)
    B_TL = Flinedata(:,6)/2;     %Susceptance of transmission lines
    a = Flinedata(:,7);          %Xmer_Tap
    Z = R + 1i*X; 
    y = 1./Z; %%To calculate inverse of each element
    size(y);
    y_2 = 1i*B_TL;
    Bus_number = max(max(From_Bus),max(To_Bus)); %Maximum number of buses
    Branch_num = length(From_Bus); %Number of Branches
    Y = zeros(Bus_number, Bus_number); %Creation of an empty admittance matrix

            %Off-diagonal elements of the admittance matrix formation

                for i=1:Branch_num

                    Y(From_Bus(i),To_Bus(i)) = Y(From_Bus(i),To_Bus(i)) -y(i)/a(i);
                    Y(To_Bus(i),From_Bus(i)) = Y(From_Bus(i),To_Bus(i));
                end    

            %Diagonal elements of the admittance matrix formation

      for m=1:Bus_number
          for n=1:Branch_num
              if From_Bus(n)==m
                 Y(m,m) = Y(m,m) + y(n)/(a(n)^2) + y_2(n);
              elseif To_Bus(n)==m
                     Y(m,m) = Y(m,m) + y(n) + y_2(n);
             end
          end
      end
 Bus_data=Bus_data1.*a4vals(K);

end Apologies for any inconveniences.

Answer 2

[Mini]

Hope Peju is aware of file operation commands. Pls find the attached sample file. Check the ab.op file for the results.

[Peju Oyewole]

Please Above code is not working for me. I am doing an operation to get a matrix M_C2 from If…elseif condition within a two for loop in parallel. The for loop will be running for 8760(for i=1:8760) for a specific value at each iteration. At some iterations based on the if condition, matrix M_C2 will be generated. I want any matrix generated at a particular iteration to be saved into a text file and at what iteration that matrix was generated, and I need to recall the text file to be again utilized as the primary data file for further operations within the loop. Also my two parallel for loop will be repeated for 1000 times. Also I want to know numbers of matrix generated at the end of the 8760times and save. Below is some of my code Tvals = 1:8760;%hourly iteration a4vals = [0.63 0.62 0.6 0.58 0.59 0.65 0.74 0.86.......] %hourly load demand for K = 1 : length(Tvals) T = Tvals(K); a4= a4vals(K); .........my working ........ U=rand % if conditions if U>=PRO1(J,2)+PRO1(J,3)+PRO1(J,4) counter = counter + 1 disp ('up state') elseif (PRO1(J,2)+PRO1(J,3)<=U) && (U<PRO1(J,2)+PRO1(J,3)+PRO1(J,4)) counter_1 = counter_1 + 1 M_C(J,:)=zeros(1,5)% To identify the failed component M_C2=M_C Many thanks.

Answer 3

[Edik Manoukian]

Problem 1 Consider the feeder in the following figure. The substation transformer is connected to an infinite bus. The infinite bus voltages (i.e. the substation primary voltages) are balanced and being held at 69 kV for all power-flow problems. The substation transformer ratings are: 5000 kVA, 69 kV delta – 13.8 kV grounded Y, Z = 1.7 + j8.5 % The phase impedance matrix for a four-wire wye line between Node 2-3 is The four-wire wye feeder between Node 2-3 is 0.75 miles long. An unbalanced wye-connected load is located at node 3 and has the following values: Phase a: 650 kVA at 0.85 lagging power factor Phase b: 500 kVA at 0.90 lagging power factor Phase c: 950 kVA at 0.95 lagging power factor Assuming that the regulators are in the neutral position: 1) Determine the forward and backward sweep matrices for the substation transformer and the line segment. 2) Use the modified ladder technique to determine the line-to-ground voltages at all nodes. Use a tolerance of 0.0001 per unit. 3) Calculate all node voltages in actual values in volts and on a 120-V base.

Answer 4

[Edik Manoukian]

Three type B step-voltage regulators are installed in a wye connection at the substation in order to hold the load voltages (node 3) at a voltage level of 121 V and a bandwidth of 2 V. 1) Compute the actual equivalent line impedance between nodes 2 and 3. 2) Determine a potential transformer ratio and current transformer ratio given that the compensator circuit ratings are 120V and 5A. Determine the R and X compensator settings calibrated in volts and Ohms. The settings must be the same for all three regulators. 3) For the load conditions of Problem 1 with the regulators in the neutral position, compute the voltages across the voltage relays in the compensator circuits. 4) Determine the appropriate tap settings for the three regulators to hold the node 3 voltages at 121 V in a bandwidth of 2 V. 5) With the regulators taps set, compute the actual load voltages in volts and in per unit.