How can I create vectors using the IF function?

I have five, 8760x1 vectors and I need to create a new vector for each of the five vectors that gives the number 0 for each row if it is not the minimum number of the 5 vectors or if it is the minimum number of the five vectors puts in a number from a separate 8760x1 vector. Thanks for your help.

[James Tursa]

Are your five vectors separate variables? Or contained within a cell array? Or columns of a 2D matrix?

[James Tursa]

I am not concerned what the data represents. I simply want to know if the data is in separately named variables, or cells in a cell array, or columns of a 2D matrix, or something else.

[Erik Hausen]

separately named variables.

Answer 1

[Walter Roberson]

t = horzcat(Vec1, Vec2, Vec3, Vec4, Vec5);
mt = min(t,2);
eq1 = Vec1 == mt;
eq2 = Vec2 == mt;
eq3 = Vec3 == mt;
eq4 = Vec4 == mt;
eq5 = Vec5 == mt;
Vec1out = 0 * Vec1; %same size and class
Vec2out = 0 * Vec2;
Vec3out = 0 * Vec3;
Vec4out = 0 * Vec4;
Vec5out = 0 * Vec5;
Vec1out(eq1) = Vec6(eq1);
Vec2out(eq2) = Vec6(eq2);
Vec3out(eq3) = Vec6(eq3);
Vec4out(eq4) = Vec6(eq4);
Vec5out(eq5) = Vec6(eq5);

[Stephen23]

@Erik Hausen: this answer, while being a good and correct solution, illustrates why putting data into lots of separate variables is a bad idea. If all of this data was placed into one matrix, then this whole task would boil down to about three lines of code (using the same concept!).

Careful planning of how data is structured and arranged makes your programming and work much easier.

[Guillaume]

"Careful planning of how data is structured and arranged makes your programming and work much easier."

I could not agree more with Stephen. @Erik, follow this advice (and avoid overuse of for and if) and you'll find that your code is much simpler. In this particular, if instead of five 8760 x 1 vectors you had one 8760 x 5 matrix, the code would simply be:

m = randi([1 5], 8760, 5); %for demo
othervector = randi([10 20], 8760, 1); replacement vector for demo

out = zero(size(m));
isrowmin = bsxfun(@eq, m, min(m, [], 2))
replacement = repmat(othervector, 1, 5);
out(isrowmin) = replacement(isrowmin);

The code is only four lines. One line to prepare the return variable, one line to find which elements of each row is the minimum, and two lines to replace the minimums by your replacement vector.

[Erik Hausen]

Greatly appreciated! I will make it into 1 matrix.