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Translating vector into locations in a matrix

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Mark
Mark 2013-3-9
I'm attempting to set up an n row by 4 column matrix using data from column vectors Z1, Z2, & Z3. The problem is I need to place the random values from these column vectors into random locations in the n row by 4 column matrix. How can I specifically put certain values from the column vectors into certain locations in the K matrix? I know the K matrix will be always 4 columns wide, but could be any length. The rows of the K matrix will repeat every 4 utilizing a different value from one of the Z1, Z2, & Z3 column vectors. (I'm trying to come up with a local stiffness matrix that I can later transform into a global stiffness matrix). In the following code I'm attempting to set up a structure where the every 4th row in column 1 of Matrix K is row 1 - Z2(1,1), row 5 - Z2(5,1), row 9 - Z2(9,1), and so on. Any help would be appreciated.
K = zeros(4*member_count,4);
for r = 1:4:4*member_count;
for i = 1:Z2_count;
s(i,r) = Z2(i,1);
for j = 1:r
K(j,1) = s(i,r)
end
end
end

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Cedric
Cedric 2013-3-9
编辑：Cedric 2013-3-9
If you need to generate random indices that can appear multiple times, you'll be interested in RANDI. Ex.:
>> randi(10, 1, 10)
ans =
1 9 10 7 8 8 4 7 2 8
you see here that 8 appears three times, and 5 doesn't appear.
If you need to shuffle indices randomly so you essentially have a random filling of K but indices must be unique, you'll be interested in RANDPERM. Ex.:
>> randperm(10)
ans =
2 7 3 10 4 1 9 5 6 8
you see here that all integers between 1 and 10 are present, but they were permuted randomly.
I don't fully understand what you want to do, but here one way to perform the last bit of your statement without a loop:
>> Z2 = (10:10:100).' % Simple example.
Z2 =
10
20
30
40
50
60
70
80
90
100
>> idx = 1:4:numel(Z2)
idx =
1 5 9
>> K = zeros(numel(Z2), 4)
K =
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
>> K(idx,1) = Z2(idx)
K =
10 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
50 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
90 0 0 0
0 0 0 0
  2 个评论
Mark
Mark 2013-3-10
Thank you Cedric. This absolutely answered my question. I ended up using it multiple times in my situation.
K = zeros(4*member_count,4);
idx1 = 1:4:4*member_count;
K(idx1,1) = Z2;
K(idx1,2) = Z3;
K(idx1,3) = -Z2;
K(idx1,4) = -Z3;

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