Changing data diagonally in matrix based on formula and table

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A Aln
A Aln 2019 年 3 月 23 日
コメント済み: A Aln 2019 年 3 月 23 日
Hello,
Let say I have this matrix:
1 1 1 0 1 0 0 0
1 1 0 1 0 1 0 0
1 0 1 1 0 0 1 0
0 1 1 1 0 0 0 1
1 0 0 0 1 1 1 0
0 1 0 0 1 1 0 1
0 0 1 0 1 0 1 1
0 0 0 1 0 1 1 1
and this table:
kxx kyy kzz poroh
______ ______ ______ _______
1 5 6 80 0.12
2 5 6 80 0.21
3 13 62 23 0.20
4 14 81 14 0.22
5 10 16 63 0.23
6 12 62 36 0.26
7 11 45 41 0.21
8 5 5 5 0.19
and this equation:
(1/visc)*((2*kzz2*kzz1*az2*az1)/(kzz2*az2*delz+kz1*az1*delz)
%% All the variables are constant in this equation except (kzz) < which I have in the table.
this equation represent the diagonal started from column 5, I want to change the ones to the result of the equation based on the table
For example (1) in first row and column 5;
the equation will be
(1/visc)*((2*63*80*az2*az1)/(63*az2*delz+80*az1*delz)
kzz5 = 63 and kzz1 = 80
so based on the location on matrix, the code should take the value from the table.
this is the code that I used to change the data diagonally but it changed it to a constant number
k = -(x*y);
d = diag(XX,k);
n = d;
n(n==1) = 2;
XX = XX - diag(d,k) + diag(n,k);
Sorry if the code is not consistent but this is part of big project

採用された回答

Sajeer Modavan
Sajeer Modavan 2019 年 3 月 23 日
unable to understand what you exactly looking. But this may help you
clc
clear
matrix = [ 1 1 1 0 1 0 0 0
1 1 0 1 0 1 0 0
1 0 1 1 0 0 1 0
0 1 1 1 0 0 0 1
1 0 0 0 1 1 1 0
0 1 0 0 1 1 0 1
0 0 1 0 1 0 1 1
0 0 0 1 0 1 1 1];
Table = [ 1 5 6 80 0.12
2 5 6 80 0.21
3 13 62 23 0.20
4 14 81 14 0.22
5 10 16 63 0.23
6 12 62 36 0.26
7 11 45 41 0.21
8 5 5 5 0.19];
kzz = Table(:,4);
visc = 1;
az1 = 1;
az2 = 1;
delz = 1;
for jj = 1:length(matrix(:,1))
for ii = 1:length(matrix(1,:))
Matrix(jj,ii) = (1/visc)*((2*kzz(ii)*kzz(jj)*az2*az1)/(kzz(ii)*az2*delz+kzz(jj)*az1*delz));
end
end
  5 件のコメント
Sajeer Modavan
Sajeer Modavan 2019 年 3 月 23 日
kzz = Table(:,4);
visc = 1; %Replace with original values
az1 = 1; %Replace with original values
az2 = 1; %Replace with original values
delz = 1; %Replace with original values
for jj = 1:4
ii = jj+4;
matrix(jj,ii) = (1/visc)*((2*kzz(ii)*kzz(jj)*az2*az1)/(kzz(ii)*az2*delz+kzz(jj)*az1*delz));
end
A Aln
A Aln 2019 年 3 月 23 日
thank you

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