How to make an efficient if-statement-expression that when identifying a 1x3 vector consisting only of either 6's,7's,8's or 9's gives the same statement?

I could you run following if statement

vec=[9 9 9]; if vec(1:3)==9 | vec(1:3)==8 | vec(1:3)==7 | vec(1:3)==6 disp('Identified') end Identified

But is there not a more efficient way to accomplish this in stead of writing all these 'or-signs' in the if statement? With vectorized code? I have tried with a for loops, but got problems when writing it under an if statement with more than one evaluation option like elseif

Answer 1

[Matt J]

if ismember(vec,6:9)

[John]

The identified 1x3 vector should include the same number in each element. That means it should only allow vectors like [9 9 9] or [8 8 8] and give the same statement. I hope my question is clear now.

[Matt J]

Then you should edit your question, since the code you posted there isn't equivalent to what you've now described.

The modification of my approach that you need is

if ismember(vec,(6:9).'*[1 1 1],'rows')

[John]

I have tried the same code on a case where the if-statement-expression should (again) identify a 1x3 vector but with two identical elements ranging from 1:10. I was forced to split the 1x3 vector into three 1x2 vectors as following

vec=[1 2 2];
if ismember(vec(1:2),(1:10).'*[1 1],'rows') || ismember(vec(2:3),(1:10).'*[1 1],'rows') || ismember(vec(1:2:3),(1:10).'*[1 1],'rows')
display('Identified')
end
Identified

Is there a method to accomplish this in less steps?

[Matt J]

 if isequal(  sort(nonzeros(histc(vec,1:10)))' , [1 2])

Answer 2

[Andrei Bobrov]

x = unique(vec);
out = numel(x) == 1 && ismember(x, 6:9);