Help with this tiple nested for loop

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Hi there, I have the following five matrices:
A1 = ones(5,5)*1;
A2 = ones(5,5)*2;
A3 = ones(5,5)*3;
A4 = ones(5,5)*4;
A5 = ones(5,5)*5;
I want a total of 96 seperate matrices and want to repeat A1,A2,A3,A4 and A5 after every 32 loops, Hence, I will have at the end 96 of 5 x 5 matrices consiting of A1, A2, A3, A4 and A5. So, the first 6 matrices (1-6) are of A1, the second lot of (7-12) matrices are A2, the third lot (13-18) are A3, the fouth (19 - 24) are A4 and A5 (25 - 32). After the 32nd matrix I want to go back to A1 and repeat as before till we get 96 matrices.
So, I have attempted the code below:
for i = 1:96
for j = 1:3
for k = 1:32
U(:,:,k,j) = A1;
Ux(:,:,i) = U(:,:,k,j);
if k > 6
U(:,:,k,j) = A2;
Ux(:,:,i) = U(:,:,k,j);
end
if k > 12
U(:,:,k,j) = A3;
Ux(:,:,i) = U(:,:,k,j);
end
if k > 18
U(:,:,k,j) = A4;
Ux(:,:,i) = U(:,:,k,j);
end
if k > 24
U(:,:,k,j) = A5;
Ux(:,:,i) = U(:,:,k,j);
end
end
end
end
Could someone help me acheive this please?
Many thanks in advance, guys!
,,

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 采纳的回答

Matt J
Matt J about 22 hours 前

0 个投票

I don't know what you're trying to do with Ux, but U is just,
U = cat(3, ...
repelem(cat(3,A1,A2,A3,A4),1,1,6), ...
repelem(A5,1,1,8));
U = repmat(U,1,1,1,3);
The reason the intent of Ux is not clear in the current code is because Ux(:,:,i) is overwritten for every j and k. Its final value is always the value at k = 32, namely A5.

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Scott Banks
Scott Banks about 1 hour 前
Hi Matt, the Ux is "supposed" to be where I store each of the 96 matrices that consist of the A1,A2,A3,A4 and A5.
I copy and pasted your code and its only giving me 32 matrices. I need 96, but that repeat in the pattern and order of the 32. So, like 3 lots of 32.
Scott Banks
Scott Banks about 1 hour 前
I'm getting my desired out come doing this:
A1 = ones(5,5)*1;
A2 = ones(5,5)*2;
A3 = ones(5,5)*3;
A4 = ones(5,5)*4;
A5 = ones(5,5)*5;
for j = 1:3
for k = 1:32
U(:,:,k,j) = A1;
if k > 6
U(:,:,k,j) = A2;
end
if k > 12
U(:,:,k,j) = A3;
end
if k > 18
U(:,:,k,j) = A4;
end
if k > 24
U(:,:,k,j) = A5;
end
end
end
Torsten
Torsten 40 minutes 前
A1 = ones(5)*1;
A2 = ones(5)*2;
A3 = ones(5)*3;
A4 = ones(5)*4;
A5 = ones(5)*5;
U = zeros(5,5,32,3);
for j = 1:3
for k = 1:6
U(:,:,k,j) = A1;
end
for k = 7:12
U(:,:,k,j) = A2;
end
for k = 13:18
U(:,:,k,j) = A3;
end
for k = 19:24
U(:,:,k,j) = A4;
end
for k = 25:32
U(:,:,k,j) = A4;
end
end
Matt J
Matt J 33 minutes 前
编辑：Matt J 31 minutes 前
Ran in:
Ran in:
I'm getting my desired out come doing this:
That gives identical results to what I posted:
A1 = ones(5,5)*1;
A2 = ones(5,5)*2;
A3 = ones(5,5)*3;
A4 = ones(5,5)*4;
A5 = ones(5,5)*5;
isequal(version1(A1,A2,A3,A4,A5), ...
version2(A1,A2,A3,A4,A5))
ans = logical
1
function U=version1(A1,A2,A3,A4,A5)
for j = 1:3
for k = 1:32
U(:,:,k,j) = A1;
if k > 6
U(:,:,k,j) = A2;
end
if k > 12
U(:,:,k,j) = A3;
end
if k > 18
U(:,:,k,j) = A4;
end
if k > 24
U(:,:,k,j) = A5;
end
end
end
end
function U=version2(A1,A2,A3,A4,A5)
U = cat(3, ...
repelem(cat(3,A1,A2,A3,A4),1,1,6), ...
repelem(A5,1,1,8));
U = repmat(U,1,1,1,3);
end
Scott Banks
Scott Banks 17 minutes 前
移动：Torsten 15 minutes 前
Hi guys,
this was part of a much bigger task, and I am hoping if it's okay to ask a follow up?
So I have this section of code from my script:
%% Transformation from local to global matrix
T = cat(3, Tcx, Tbx, Tdx1, Tdx2);
for i = 1:352
for j = 5
for k = 32
Kmc(:,:,k,j) = T(:,:,i)'*Kc(:,:,1)*T(:,:,i);
if k > 6
Kmc(:,:,k,j) = T(:,:,i)'*Kc(:,:,2)*T(:,:,i);
end
if k > 12
Kmc(:,:,k,j) = T(:,:,i)'*Kc(:,:,3)*T(:,:,i);
end
if k > 18
Kmc(:,:,k,j) = T(:,:,i)'*Kc(:,:,4)*T(:,:,i);
end
if k > 24
Kmc(:,:,k,j) = T(:,:,i)'*Kc(:,:,5)*T(:,:,i);
end
end
end
if i > 160
Km(:,:,i) = T(:,:,i)'*Kb*T(:,:,i);
end
if i > 288
Km(:,:,i) = T(:,:,i)'*Kd*T(:,:,i);
end
if i > 320
Km(:,:,i) = T(:,:,i)'*Kd*T(:,:,i);
end
end
I don't get any errors, and I end up with a 495 x 495 x 32 x 5 for Kmc which is what I want. However, the Kmc matrices are just filled with zeros. There are no values in them at all.
Hence, I am wondering why this is?
Many thanks, guys!
Torsten
Torsten 10 minutes 前
编辑：Torsten 7 minutes 前
Kmc(:,:,32,5) is changed here:
if k > 24
Kmc(:,:,k,j) = T(:,:,i)'*Kc(:,:,5)*T(:,:,i);
end
but we don't know the right-hand side
T(:,:,i)'*Kc(:,:,5)*T(:,:,i);
Matt J
Matt J about 4 hours 前
this was part of a much bigger task, and I am hoping if it's okay to ask a follow up?
Yes, but does that mean the original question has been answered. If so, please Accept-click the answer.
Scott Banks
Scott Banks about 7 hours 前
移动：Matt J about 3 hours 前
@Torsten
Hi Torsten, it quite a large amount of code, but all this should give you what the right-hand side should equal.
E = 199947961.502;
Ic = [29997.921/100^4; 22528.963/100^4; 17510.628/100^4; 14268.001/100^4; 11407.468/100^4];
Ib = 4.82E-04;
Ac = [0.0212858; 0.0168; 0.0136; 0.0113; 0.0093];
Ab = 0.0115528;
Ad = 0.0254469;
L = 6;
h = 3.5;
alpha = atand(3.5/6);
beta = atand(6/3.5);
d = sqrt(h^2 + L^2);
H = 112;
ns = 112/h;
n = (ns+1)*5*3
%% Local stiffness matrices
for i = 1:5
Kc(:,:,i) = [E*Ac(i)/h 0 0 -E*Ac(i)/h 0 0;
0 12*E*Ic(i)/h^3 6*E*Ic(i)/h.^2 0 -12*E*Ic(i)/h^3 6*E*Ic(i)/h^2;
0 6*E*Ic(i)/h^2 4*E*Ic(i)/h 0 -6*E*Ic(i)/h^2 2*E*Ic(i)/h;
-E*Ac(i)/h 0 0 E*Ac(i)/h 0 0;
0 -12*E*Ic(i)/h^3 -6*E*Ic(i)/h^2 0 12*E*Ic(i)/h^3 -6*E*Ic(i)/h^2;
0 6*E*Ic(i)/h^2 2*E*Ic(i)/h 0 -6*E*Ic(i)/h^2 4*E*Ic(i)/h];
end
Kb = [E*Ab/L 0 0 -E*Ab/L 0 0;
0 12*E*Ib/L^3 6*E*Ib/L^2 0 -12*E*Ib/L^3 6*E*Ib/L^2;
0 6*E*Ib/L^2 4*E*Ib/L 0 -6*E*Ib/L^2 2*E*Ib/L;
-E*Ab/L 0 0 E*Ab/L 0 0;
0 -12*E*Ib/L^3 -6*E*Ib/L^2 0 12*E*Ib/L^3 -6*E*Ib/L^2;
0 6*E*Ib/L^2 2*E*Ib/L 0 -6*E*Ib/L^2 4*E*Ib/L]
Kd = zeros(6,6);
Kd(1,1) = E*Ad/d;
Kd(1,4) = -E.*Ad/d;
Kd(4,1) = -E.*Ad/d;
Kd(4,4) = E*Ad/d
%% Transformation matrices
Tc1 = zeros(6,18);
Tc1(2,1) = -1;
Tc1(1,2) = 1;
Tc1(3,3) = 1;
Tc1(5,16) = -1;
Tc1(4,17) = 1;
Tc1(6,18) = 1;
Tb1 = zeros(6,6);
Tb1(1,1) = 1;
Tb1(2,2) = 1;
Tb1(3,3) = 1;
Tb1(4,4) = 1;
Tb1(5,5) = 1;
Tb1(6,6) = 1;
Td1 = zeros(6,21);
Td1(1,1) = cosd(alpha);
Td1(1,2) = sind(alpha);
Td1(4,19) = cosd(alpha);
Td1(4,20) = sind(alpha);
Td2 = zeros(6,15);
Td2(1,1) = cosd(beta + 90);
Td2(1,2) = sind(beta + 90);
Td2(4,13) = cosd(beta + 90);
Td2(4,14) = sind(beta + 90);
%% Counters for member nodes for columns
vecC = 1:5;
countC = 0:ns;
vecN1 = vecC + 5.*countC'
nM = size(vecN1,2) * (size(vecN1,1)-1);
nC = 3 * max(vecN1(:));
Tcx = zeros(size(Tc1,1),nC,nM);
j = 0
% For loop to complete the transformation matrices for columns
for i = 1:5
R = vecN1(:,i);
for k = 1:ns
j = j + 1;
C = (3*R(k) - 2):(3*R(k+1));
Tcx(:,C,j) = Tc1;
end
end
%% Counters for member nodes for beams
vecB = 6:10;
countB = 0:ns-1;
vecN2 = [vecB + 5.*countB']';
nM = 128;
nC = 3 * max(vecN2(:));
Tbx = zeros(size(Tb1,1),nC,nM);
j = 0
% For loop to complete the transformation matrices for beams
for i = 1:size(vecN2,2)
R = vecN2(:,i);
for k = 1:4
j = j + 1;
C = (3*R(k) - 2):(3*R(k+1));
Tbx(:,C,j) = Tb1;
end
end
%% counters for member nodes for diagonal braces
countD = 0:ns;
vecD1 = [2 + 5.*countD]'
vecD2 = [3 + 5.*countD]'
vecD3 = [4 + 5.*countD]'
for i = 1:length(vecD1)-1
vecN3(:,i) = [vecD1(i);vecD2(i+1)];
vecN4(:,i) = [vecD3(i);vecD2(i+1)];
end
% For loop to complete the transformation matrices for diagonal braces
nM = 32;
nC = n;
Tdx1 = zeros(size(Td1,1),nC,nM);
Tdx2 = zeros(size(Td2,1),nC,nM);
j = 0
for i = 1:size(vecN3,2)
R = vecN3(:,i);
D = vecN4(:,i);
for k = 1
j = j + 1;
C = (3*R(k) - 2):(3*R(k+1));
B = (3*D(k) - 2):(3*D(k+1));
Tdx1(:,C,j) = Td1;
Tdx2(:,B,j) = Td2;
end
end
%% Transformation from local to global matrix
T = cat(3, Tcx, Tbx, Tdx1, Tdx2);
for i = 1:352
for j = 5
for k = 32
Kmc(:,:,k,j) = T(:,:,i)'*Kc(:,:,1)*T(:,:,i);
if k > 6
Kmc(:,:,k,j) = T(:,:,i)'*Kc(:,:,2)*T(:,:,i);
end
if k > 12
Kmc(:,:,k,j) = T(:,:,i)'*Kc(:,:,3)*T(:,:,i);
end
if k > 18
Kmc(:,:,k,j) = T(:,:,i)'*Kc(:,:,4)*T(:,:,i);
end
if k > 24
Kmc(:,:,k,j) = T(:,:,i)'*Kc(:,:,5)*T(:,:,i);
end
end
end
if i > 160
Km(:,:,i) = T(:,:,i)'*Kb*T(:,:,i);
end
if i > 288
Km(:,:,i) = T(:,:,i)'*Kd*T(:,:,i);
end
if i > 320
Km(:,:,i) = T(:,:,i)'*Kd*T(:,:,i);
end
end
Many thanks
Scott Banks
Scott Banks about 1 hour 前
All done, Matt.
Matt J
Matt J about 3 hours 前
编辑：Matt J about 3 hours 前
However, the Kmc matrices are just filled with zeros. There are no values in them at all. Hence, I am wondering why this is?
Because k is not changing throughout the 'loop'. It is set to a fixed value of 32, which means that the conditions k>6,..., k>24 are never met.
In any case, you should not be using loop structures like this. Everything you are currently doing can be done looplessly with pagemtimes, cat(), and repmat().
Torsten
Torsten about 1 hour 前
@Scott Banks
If you add the lines
M = Kmc(:,:,32,5)
M = M(:)
norm(M)
at the end of your code, you will see that Kmc is not all zero.
Scott Banks
Scott Banks about 1 hour 前
@Matt J
Ah, that's a typo. J should be 1:5 and k should be 1:32.
I am still getting all zeros, and @Torsten, that gives me a column vector of 245025 x 1.
Torsten
Torsten 40 minutes 前
编辑：Torsten 38 minutes 前
Ran in:
Ran in:
I am still getting all zeros, and @Torsten, that gives me a column vector of 245025 x 1.
Yes, and the fact that the norm of M is not zero shows that not all elements of Kmc(:,:,32,5) can be zero.
M is just Kmc(:,:,32,5), written as a column vector:
495*495
ans = 245025
So you don't get all zeros.

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Scott Banks
Scott Banks
about 23 hours 前

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Torsten
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