how to use "randi" in specific case?

Hi!

i want to ask you if it is possible to use "randi" in this case:

i have a binary matrix A (m*n), each row has just "1" and other elements are set to "0", i want to put randomly in the "1" position a value from this range for example [10,15,20,25].

i'm asking how can i remove the case of seeing the same value in the same column or two same values in two successive columns.

for example:

0	0	10	0	0	0	0	0	0	0
0	0	10	0	0	0	0	0	0	0
0	0	0	15	0	0	0	0	0	0
0	0	0	20	0	0	0	0	0	0
0	0	0	10	0	0	0	0	0	0
0	0	0	0	25	0	0	0	0	0
0	0	0	0	0	0	20	0	0	0
0	0	0	0	0	0	0	15	0	0
0	0	0	0	0	0	0	0	10	0
0	0	0	0	0	0	0	0	10	0
0	0	0	0	0	0	0	0	0	15
0	0	0	0	0	0	0	0	0	25

as you see the third column i have the same value "10" ,also in the fourth i have the same value of previous column, i want to remove this case.

i'll appreciate any help, thank you very much.

[Torsten]

What means "random" in this case if you impose the above conditions on the "randomness" ?

[Majid]

Hi !@Torsten

could you please see the modification done above.

[Torsten]

If your binary matrix had, e.g. (five 1's in some column) or (four 1's in some column and at least one 1 in the preceeding or next column), you couldn't do what you try to do. And then ?

[Majid]

@Torsten i want just a loop while, if the value exist in the column for more than one time or in just the previous or next column i will do another "randi"

Answer 1

[Voss]

Maybe something like this (the example A requires at least 5 values, e.g., [10 15 20 25 30]):

A = [ ...
    0	0	1	0	0	0	0	0	0	0; ...
    0	0	1	0	0	0	0	0	0	0; ...
    0	0	0	1	0	0	0	0	0	0; ...
    0	0	0	1	0	0	0	0	0	0; ...
    0	0	0	1	0	0	0	0	0	0; ...
    0	0	0	0	1	0	0	0	0	0; ...
    0	0	0	0	0	0	1	0	0	0; ...
    0	0	0	0	0	0	0	1	0	0; ...
    0	0	0	0	0	0	0	0	1	0; ...
    0	0	0	0	0	0	0	0	1	0; ...
    0	0	0	0	0	0	0	0	0	1; ...
    0	0	0	0	0	0	0	0	0	1];

vals = [10 15 20 25 30];

[m,n] = size(A);
n_vals = numel(vals);

is_used = false(1,n_vals);
is_used_previous = false(1,n_vals);

for jj = 1:n
    idx = find(A(:,jj));
    for ii = 1:numel(idx)
        if all(is_used | is_used_previous)
            continue
        end
        unused_val_idx = find(~is_used & ~is_used_previous);
        new_val_idx = randi(numel(unused_val_idx));
        A(idx(ii),jj) = vals(unused_val_idx(new_val_idx));
        is_used(unused_val_idx(new_val_idx)) = true;
    end
    is_used_previous = is_used;
    is_used(:) = false;
end

disp(A);

     0     0    15     0     0     0     0     0     0     0
     0     0    10     0     0     0     0     0     0     0
     0     0     0    25     0     0     0     0     0     0
     0     0     0    20     0     0     0     0     0     0
     0     0     0    30     0     0     0     0     0     0
     0     0     0     0    10     0     0     0     0     0
     0     0     0     0     0     0    10     0     0     0
     0     0     0     0     0     0     0    15     0     0
     0     0     0     0     0     0     0     0    10     0
     0     0     0     0     0     0     0     0    20     0
     0     0     0     0     0     0     0     0     0    25
     0     0     0     0     0     0     0     0     0    30

[Majid]

@Voss it works, thank you very much!

[Voss]

You're welcome!

Answer 2

[David Hill]

n=10;%length of matrix
l=[10 15 20 25];
m=zeros(length(l),n);
r1=randperm(length(l));r2=randperm(n,length(l));r3=randperm(length(l));
for k=1:length(l)
  m(r1(k),r2(k))=l(r3(k));
end
m

m = 4×10

0 0 0 0 10 0 0 0 0 0 0 0 0 0 0 0 0 0 0 25 0 0 0 0 0 0 0 0 20 0 0 0 0 0 0 0 0 15 0 0

[Majid]

@David Hill thank you for your quick reply! but i forgot to put that i already have a binary matrix which has just "1" per row, so the values i want to generate should take the position of each "1" , i did some modification above.

[David Hill]

m=20;n=10;
A=zeros(m,n);
for k=1:m
  A(k,randi(n))=1;
end
l=[10 15 20 25];
A(A==1)=l(randi(length(l),1,nnz(A)))

A = 20×10

0 0 0 0 0 0 10 0 0 0 0 15 0 0 0 0 0 0 0 0 0 15 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 25 0 0 0 0 0 20 0 0 0 0 0 0 0 0 25 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 15 0 0 0 0 0 0 0 10 0 0 0 20 0 0 0 0 0 0 0 0 0 15 0 0 0 0 0 0 0 0 0

h=histc(A,l);
[r,c]=find(h>1);
[R,C]=find(movsum(h>=1,2,2,'endpoints','discard')>1);
C=C+1;
idx=[];
for k=1:length(r)
    f=find(A(:,c(k))==l(r(k)));
    idx=[idx;f(2:end)];
end
for k=1:length(R)
    idx=[idx;find(A(:,C(k))==l(R(k)))];
end
idx=unique(idx);
A(idx,:)=[]

A = 9×10

0 0 0 0 0 0 10 0 0 0 0 0 0 0 0 0 0 25 0 0 0 0 0 20 0 0 0 0 0 0 0 0 0 0 0 0 0 0 15 0 20 0 0 0 0 0 0 0 0 0 15 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 10 0 0 25 0 0 0 0 0 0 0 0 0 0 0 0 0 20 0 0 0 0

[Majid]

@David Hill the code you provided eliminate definitely the same value but i need to change it with other value.

[David Hill]

You keep changing the problem. What are you trying to do? What is the big picture? I think you can figure it out from here.

[Majid]

@David Hill this is an eg when i did a sample "randi" exactly as you mentioned, but as you see the value "10" exists two times in column 3 also it exists in the next column so i want to change its value with another , its like another randi until having different values .

0	0	10	0	0	0		
0	0	10	0	0	0	
0	0	0	15	0	0		
0	0	0	20	0	0	
0	0	0	10	0	0	
0	0	0	0	25	0	
0	0	0	0	0	0	
0	0	0	0	0	0	
0	0	0	0	0	0	

this is an e.g of required result:

0	0	10	0	0	0		
0	0	25	0	0	0	
0	0	0	15	0	0		
0	0	0	20	0	0	
0	0	0	25	0	0	
0	0	0	0	10	0	
0	0	0	0	0	0	
0	0	0	0	0	0	
0	0	0	0	0	0	

[Voss]

@Majid: 25 exists in columns 3 and 4. Is that allowed?

[Majid]

@Voss no, because it is a e.g i just focus in one value"10", here just i take l=[10 15 20 25]; for my real case the range of values is large.