F = @(v)find(any(v==B(:,1),2)&~any(v==B(:,2),2));
C = cellfun(F,A,'uni',0)
gives C=[5;7;11;14] for A{1}.
I want to seperate [5;7;11;14], as such
mes_true=mes_in(1)-mes_between(5)=12-9=3
mes_true=mes_in(2)-mes_between(7)-mes_between(11)=18-4-6=8
mes_true=mes_in(4)-mes_between(14)=14-7.8=6.2