Need to Produce a different matrix after each loop
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    DARLINGTON ETAJE
 il 25 Giu 2021
  
    
    
    
    
    Commentato: DARLINGTON ETAJE
 il 7 Lug 2021
            Hello Friends,
I need to produce a different set of deli after each loop but it remains the same. Help me out please...see code below.
deli is suppose to be different set of 12 by 1 after each loop
L1=[1 33 34 2 35 36	3 37 38	4 39 40];
L2=[3	37	38	4	39	40	5	41	42	6	43	44];
L3=[5	41	42	6	43	44	7	45	46	8	47	48];
L4=[7	45	46	8	47	48	9	49	50	10	51	52];
L5=[9	49	50	10	51	52	11	53	54	12	55	56];
L6=[11	53	54	12	55	56	13	57	58	14	59	60];
L7=[13	57	58	14	59	60	15	61	62	16	63	64];
L8=[15	61	62	16	63	64	17	65	66	18	67	68];
L9=[17	65	66	18	67	68	19	69	70	20	71	72];
L10=[19	69	70	20	71	72	21	22	23	24	25	26];
L11=[21	22	23	24	25	26	27	28	29	30	31	32];
L=[L1;L2;L3;L4;L5;L6;L7;L8;L9;L10;L11];
delu=[-107.9277784;-473.7489066;-456.4464814;-298.1506906;-38.38102414;169.6616599;218.8447121;105.0260144;-78.47051372;-205.707;-1.1961463;701.7499816;-2084.55571;-1229.851404;-0.390010302;39.67805573;-68.88297022;2231.216584;-5473.430599;-3189.887351;-0.390010302;41.85037402;-72.64553587];
delr=zeros(49,1);
del=[delu;delr];
deli=zeros(12,1);
delbar=([]);
mbar=repmat({zeros(12,1)},n,1);
for i=1:n
    for p=1:12
        deli(p,1)=del((L(i,p)),1);
    end
    mbar{i}=(kg{i}*deli)+fembar{i};
end
3 Commenti
  Walter Roberson
      
      
 il 26 Giu 2021
				It looks to me as if deli is indeed different after each for i loop. You do not ask to record the value for each i value though.
L1=[1 33 34 2 35 36	3 37 38	4 39 40];
L2=[3	37	38	4	39	40	5	41	42	6	43	44];
L3=[5	41	42	6	43	44	7	45	46	8	47	48];
L4=[7	45	46	8	47	48	9	49	50	10	51	52];
L5=[9	49	50	10	51	52	11	53	54	12	55	56];
L6=[11	53	54	12	55	56	13	57	58	14	59	60];
L7=[13	57	58	14	59	60	15	61	62	16	63	64];
L8=[15	61	62	16	63	64	17	65	66	18	67	68];
L9=[17	65	66	18	67	68	19	69	70	20	71	72];
L10=[19	69	70	20	71	72	21	22	23	24	25	26];
L11=[21	22	23	24	25	26	27	28	29	30	31	32];
L=[L1;L2;L3;L4;L5;L6;L7;L8;L9;L10;L11];
delu=[-107.9277784;-473.7489066;-456.4464814;-298.1506906;-38.38102414;169.6616599;218.8447121;105.0260144;-78.47051372;-205.707;-1.1961463;701.7499816;-2084.55571;-1229.851404;-0.390010302;39.67805573;-68.88297022;2231.216584;-5473.430599;-3189.887351;-0.390010302;41.85037402;-72.64553587];
delr=zeros(49,1);
del=[delu;delr];
deli=zeros(12,1);
delbar=([]);
n = 2; kg = {3;5}; fembar = {2; 103};
mbar=repmat({zeros(12,1)},n,1);
for i=1:n
    for p=1:12
        deli(p,1)=del((L(i,p)),1);
    end
    deli
    mbar{i}=(kg{i}*deli)+fembar{i};
end
Risposta accettata
  Jayant Gangwar
      
 il 7 Lug 2021
        It is my understanding that you want to produce and store the set of deli after each iteration. In your current code the deli produced after each set is correct you just need to store it as well after each iteration. In the code below each column of delstore represents the set of deli after the corresponding iteration.
L1=[1 33 34 2 35 36	3 37 38	4 39 40];
L2=[3	37	38	4	39	40	5	41	42	6	43	44];
L3=[5	41	42	6	43	44	7	45	46	8	47	48];
L4=[7	45	46	8	47	48	9	49	50	10	51	52];
L5=[9	49	50	10	51	52	11	53	54	12	55	56];
L6=[11	53	54	12	55	56	13	57	58	14	59	60];
L7=[13	57	58	14	59	60	15	61	62	16	63	64];
L8=[15	61	62	16	63	64	17	65	66	18	67	68];
L9=[17	65	66	18	67	68	19	69	70	20	71	72];
L10=[19	69	70	20	71	72	21	22	23	24	25	26];
L11=[21	22	23	24	25	26	27	28	29	30	31	32];
L=[L1;L2;L3;L4;L5;L6;L7;L8;L9;L10;L11];
delu=[-107.9277784;-473.7489066;-456.4464814;-298.1506906;-38.38102414;169.6616599;218.8447121;105.0260144;-78.47051372;-205.707;-1.1961463;701.7499816;-2084.55571;-1229.851404;-0.390010302;39.67805573;-68.88297022;2231.216584;-5473.430599;-3189.887351;-0.390010302;41.85037402;-72.64553587];
delr=zeros(49,1);
del=[delu;delr];
deli=zeros(12,1);
delbar=([]);
delstore=[];
n = 2; kg = {3;5}; fembar = {2; 103};
mbar=repmat({zeros(12,1)},n,1);
for i=1:n
    for p=1:12
        deli(p,1)=del((L(i,p)),1);
    end
    delstore=[delstore,deli];
    mbar{i}=(kg{i}*deli)+fembar{i};
end
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