Asked by Jo Betteley
on 7 Feb 2018

I have a large matrix which has RPM in the first column and Torques populating the rest of the matrix. I need help finding a way to search the first column for a set rpm and then search that row for value of torque, from that i need the position of that value in the matrix to then find an efficiency value in an identical sized matrix.

I have so far:

RPM_val =1000; [row] = find(T_T==RPM_val)

This gives me the row where the 1000rpm is, how do i then search just that row for the value of torque i am after,or the nearest value to it.

Thank you

Answer by John D'Errico
on 7 Feb 2018

Edited by John D'Errico
on 7 Feb 2018

Accepted Answer

Time to learn about matrix indexing?

[row] = find(T_T==RPM_val)

First of all, this searches the ENTIRE matrix for a value equal to RPM_val. If it turns out that another column happens to hit that value, you will get the wrong answer. While that might be a spectacularly unlikely event, or so you think, something as silly as this is the source of many a bug. Good code avoids obvious bugs by careful planning.

So instead, do this:

[row] = find(T_T(:,1)==RPM_val);

That searches only the first column. Be careful though. What is there is no exact match? find will return empty then. Do you want to find the closest?

Once you do have the row, then consider what this does:

T_T(row,2:end)

That extracts a vector of elements only in the indicated row, excluding the first column. How might you use that in a secondary search using find?

John D'Errico
on 7 Feb 2018

If all you have are a list of values, and you want to find the nearest, then that seems quite reasonable.

You can always go into more depth. But more complexity often brings problems that you will need to solve. If they are a well behaved monotonic sequence, and you wanted to use interpolation to find an intermediate value, then interp1 could help you. Even more complexity (and SOMETIMES more accuracy) would come from a spline interpolant of some sort. But you need to decide how much you want to invest into a problem.

Jo Betteley
on 8 Feb 2018

It will work to the closest and in this instance i think it should be fine, the gaps efficiency it finally relates to are very small so for now i think this will work.

As for the interpolation, if i interp the rpm values i think it will quickly start going above my head, for now i will keep pushing forward with what you have shown me and see how far i get. Thank you again for the help!, it is greatly appreciated

John D'Errico
on 8 Feb 2018

Sign in to comment.

Opportunities for recent engineering grads.

Apply Today
## 0 Comments

Sign in to comment.