Changing data diagonally in matrix based on formula and table

Question

A Aln 2019-3-23

1
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/451848-changing-data-diagonally-in-matrix-based-on-formula-and-table

评论： A Aln 2019-3-23

采纳的回答： Sajeer Modavan

在 MATLAB Online 中打开

Hello,

Let say I have this matrix:

   1     1     0     1     0     0     0
   1     0     1     0     1     0     0
   0     1     1     0     0     1     0
   1     1     1     0     0     0     1
   0     0     0     1     1     1     0
   1     0     0     1     1     0     1
   0     1     0     1     0     1     1
   0     0     1     0     1     1     1

and this table:

            kxx       kyy       kzz       poroh 
          ______    ______    ______    _______
       5        6        80       0.12
       5        6        80       0.21
      13       62        23       0.20
      14       81        14       0.22
      10       16        63       0.23
      12       62        36       0.26
      11       45        41       0.21
       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

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Answer 1

Sajeer Modavan 2019-3-23

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/451848-changing-data-diagonally-in-matrix-based-on-formula-and-table#answer_366963

在 MATLAB Online 中打开

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
    

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A Aln 2019-3-23

编辑：A Aln 2019-3-23

在 MATLAB Online 中打开

Thank you for your time, but this is not what I want, maybe I explained it in a wrong way.

X

   1     1     0     1     0     0     0
   1     0     1     0     1     0     0
   0     1     1     0     0     1     0
   1     1     1     0     0     0     1
   0     0     0     1     1     1     0
   1     0     0     1     1     0     1
   0     1     0     1     0     1     1
   0     0     1     0     1     1     1
     

there are 7 diagonals in this matrix, each diagonal represented by equation, For the diagonal just below the 'X', the equation as follow:

(1/visc)*((2*kzz2*kzz1*az2*az1)/(kzz2*az2*delz+kz1*az1*delz)

the variables are kzz2 and kzz1, the values of those variables in the table,

for the first (1) in the diagonal, in row number 1 and column 5 (just below the 'X'), the code should refer to the table:

in the table, row 1 and column kzz = 80 = kzz1 in the eqn

row 5 and column kzz = 63 = kzz2 in the eqn

for the next (1) << at row 2 and column 6 in the matrix

in the table, row 2 and column kzz = 80 = kzz1 in the eqn

row 6 and column kzz = 36 = kzz2 in the eqn

So each (1) in the diagonal, must be replaced by this equation

hope it's clear, thank you again !

Sajeer Modavan 2019-3-23

在 MATLAB Online 中打开

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