Difference lower then 2 in one vector

So i have a sample data from simulink stored in a vector(yvars40), i have to find out if there are any elements next to each other with their difference lower than 2.If there are, value fvars40 types 1, else types 0, I came up with this but it doesn't work. Matlab says: 'Index exceeds the number of array elements.'

for i = length(yvars40)
   if yvars40(i) - yvars40(i+1) < 2
      fvars40 = 1;
   else
      fvars40 = 0;
   end
end

Answer 1

[the cyclist]

When you get to the last iteration of your loop

i == length(yvars40)

and therefore in your if statement, you try to compare the last element with the "next" element, which does not exist because you are at the end of the vector. Instead, maybe you need

for i = 1:(length(yvars40)-1)

[Frantisek Stemberk]

oh that makes sense, i’ll try that thank you!

[Frantisek Stemberk]

hello, so this script worked, thank you again

for i = 1:(length(yvars40)-1)
   if yvars40(i) - yvars40(i+1) < 2
      fvars40 = 1;
      disp('fvars40=1')
   else
      fvars40 = 0;
      disp('fvars40=0')
   end
end

Now it types 1 if true or zero is false into command window, how would you store these values into one vector ? The vector for storing being the value fvars40.

[Walter Roberson]

for i = 1:(length(yvars40)-1)
   if yvars40(i) - yvars40(i+1) < 2
      fvars40(i) = 1;
   else
      fvars40(i) = 0;
   end
end

... Or you could

fvars40 = yvars40(1:end-1) - yvars40(2:end) < 2;

with no loop.

Question: you talked about a "difference" of 2. But consider that for [-5 1] you would be testing -5 - 1 < 2 which would be -6 < 2 and you would say that is true. Is that really what you want?

[Frantisek Stemberk]

yeah right, that is not really what i want, i want to find if there’s difference of 2 between 2 adjacent vector values and if there isn’t, the fvars vector will store 1 in itself, if yes then 0

[Frantisek Stemberk]

and it has to work for negative values too, haven’t considered that, can i solve it by typing the if condition like <+-2 ?

[Walter Roberson]

fvars40 = abs(yvars40(1:end-1) - yvars40(2:end)) < 2;

To confirm, a difference of exactly 2 is not to be included, right?

[Frantisek Stemberk]

yes, it's only for < 2