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How to use if-then with matrix indices

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Shan  Chu
Shan Chu am 3 Mai 2016
Beantwortet: Guillaume am 3 Mai 2016
Hi, I got a matrix n by n of a symbolic value a. And I want to change the value of a based on the matrix indices.
syms a
n=50;
z=ones(n)*a;
for i=1:1:n
for j=1:1:n
if j=1 or j=i then a=4
else a=8
Could you please help? thanks

Akzeptierte Antwort

Guillaume
Guillaume am 3 Mai 2016
You certainly don't need a loop to generate your a.
a = ones(n) * 8; %default value everywhere
[rows, cols] = ndgrid(1:n); %matrices of row and column indices
a(rows == 1 | rows == cols) = 4; %substitute when condition is met
Note that because your i and j variables are meaningless, it's not clear whether you meant it to be column 1 or row 1 that get sets to 4. I assume it was rows but it's obvious how to change it to columns. Using variable names with meaning makes the code much clearer. Moreover i and j are actually functions in matlab, so are not recommended as variable names anyway.

Weitere Antworten (2)

Andrei Bobrov
Andrei Bobrov am 3 Mai 2016
Bearbeitet: Andrei Bobrov am 3 Mai 2016
n = 50;
z = ones(n)*8;
z(:,1) = 4;
z(eye(n)>0) = 4;
  2 Kommentare
Shan  Chu
Shan Chu am 3 Mai 2016
sorry, actually a=4 when j=1 and j=i
Andrei Bobrov
Andrei Bobrov am 3 Mai 2016
corrected

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Torsten
Torsten am 3 Mai 2016
n = 50;
z = ones(n)*8;
z(:,1)=4;
z(eye(n)==1)=4;
Best wishes
Torsten.

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