I have a 6×37 matrix of values ranging from 1-6 which are classes. stability_class is attached.
I want to assign values to four matrices a,b,d,e to be the same dimension with stability_class and have different entries when stability_class changes from 1-6 across rows and columns.
a=zeros(size(X));
b=zeros(size(X));
d=zeros(size(X));
e=zeros(size(X));
[c,f]=find(stability==1);
ind=[c,f];
i=ind(1:end,:);
j=ind(:,1:end);
for k=1:size(c,1)
a(i(k,1),j(k),2)=0.32;
b(i(k,1),j(k,2))=0.0004;
d(i(k,1),j(k,2))=0.24;
e(i(k,1),j(k,2))=0.0001;
end
[c, f]=find( stability==2);
ind=[c,f];
i=ind(1:end,:);
j=ind(:,1:end);
for k=1:size(c,1)
a(i(k,1),j(k,2))=0.32;
b(i(k,1),j(k,2))=0.0004;
d(i(k,1),j(k,2))=0.24;
e(i(k,1),j(k,2))=0.0001;
end
[c,f]=find(stability==3) ;
ind=[c,f];
i=ind(1:end,:);
j=ind(:,1:end);
for k=1:size(c,1)
a(i(k,1),j(k,2))=0.16;
b(i(k,1),j(k,2))=0.0004;
d(i(k,1),j(k,2))=0.12;
e(i(k,1),j(k,2))=0.00;
end
[c,f]= find(stability==4);
ind=[c,f];
i=ind(1:end,:);
j=ind(:,1:end);
for k=1:size(c,1)
a(i(k,1),j(k,2))=0.16;
b(i(k,1),j(k,2))=0.0004;
d(i(k,1),j(k,2))=0.12;
e(i(k,1),j(k,2))=0.00;
end
[c,f]=find(stability==5);
ind=[c,f];
i=ind(1:end,:);
j=ind(:,1:end);
for k=1:size(c,1)
a(i(k,1),j(k,2))=0.16;
b(i(k,1),j(k,2))=0.0004;
d(i(k,1),j(k,2))=0.14;
e(i(k,1),j(k,2))=0.0003;
end
[c,f]=find(stability==6);
ind=[c,f];
i=ind(1:end,:);
j=ind(:,1:end);
for k=1:size(c,1)
a(i(k,1),j(k,2))=0.11;
b(i(k,1),j(k,2))=0.0004;
d(i(k,1),j(k,2))=0.08;
e(i(k,1),j(k,2))=0.0015;
end
But I get a 3D array(x:y:z) for a with uncompleted matrices for b,d,e.

2 Kommentare

KSSV
KSSV am 28 Aug. 2020
Question is not clear. What output you expect?
Aduloju Oluwatobi
Aduloju Oluwatobi am 28 Aug. 2020
I want q 6×37 matrix for a,b,d,e

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Bruno Luong
Bruno Luong am 28 Aug. 2020

0 Stimmen

a=zeros(size(stability));
b=zeros(size(stability));
d=zeros(size(stability));
e=zeros(size(stability));
ind=find(stability==1);
a(ind)=0.32;
b(ind)=0.0004;
d(ind)=0.24;
e(ind)=0.0001;
ind=find( stability==2);
a(ind)=0.32;
b(ind)=0.0004;
d(ind)=0.24;
e(ind)=0.0001;
ind=find(stability==3) ;
a(ind)=0.16;
b(ind)=0.0004;
d(ind)=0.12;
e(ind)=0.00;
ind= find(stability==4);
a(ind)=0.16;
b(ind)=0.0004;
d(ind)=0.12;
e(ind)=0.00;
ind=find(stability==5);
a(ind)=0.16;
b(ind)=0.0004;
d(ind)=0.14;
e(ind)=0.0003;
ind=find(stability==6);
a(ind)=0.11;
b(ind)=0.0004;
d(ind)=0.08;
e(ind)=0.0015;

3 Kommentare

If you want more compact code put values in the Lookup Table rather than hard code it.
LUT=[...
0.32, 0.0004, 0.24, 0.0001;
0.32,0.0004,0.24,0.0001;
0.16,0.0004,0.12,0.00;
0.16,0.0004,0.12,0.00;
0.16,0.0004,0.14,0.0003;
0.11,0.0004,0.08,0.0015 ...
];
LUTfun = @(col) reshape(LUT(stability,col),size(stability));
a=LUTfun(1);
b=LUTfun(2);
c=LUTfun(3);
d=LUTfun(4);
Stephen23
Stephen23 am 28 Aug. 2020
find is not required, would be simpler and more efficient without.
Bruno Luong
Bruno Luong am 28 Aug. 2020
I know I just want to change little to OP's code so he can still follow.

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