Get back variables out of function inside a for loop

Hello everyone,

I have extracted the values stored in 2 different matrices and used them in order to create a line (plot). One point is always the same (I1,M1) where I1 is the x coordinate and M1 is the Y.

inside J and WZ matrices, I have, let's say, 3 values each (but they could be more or less) and I have used them in order to create 3 different lines with one point constant (I1, M1) while the others change. PLease note that, I have added 2500 to all the values stored inside WZ.

Running the for loop, ends up in having created a slope matrix, with 3 values inside, which reprenset 3 different slopes. I have found the minimun value among them (-0.0168), and now, I would like to getting back from that value (the minimum), and to call back the corresponding WZ and J values associated

J = [2.5 27 56];
WZ = [12.2 23.2 33];
M1 = 22

M1 = 22

I1 = 2600

I1 = 2600

for ii = 1:numel(WZ)
    slope(ii) = (J(ii)-M1)/(WZ(ii)+3750-I1);
    slope_1 = min(slope);    
end
disp(slope_1)

   -0.0168

   -0.0168    0.0043    0.0287

[Stephen23]

The simpler MATLAB approach is to write vectorized code and use the second output from MIN():

J  = [2.5,27,56];
WZ = [12.2,23.2,33];
M1 = 22;
I1 = 2600;
slope = (J-M1)./(WZ+3750-I1);
[slope_1,idx] = min(slope); 
WZ_1 = WZ(idx)

WZ_1 = 12.2000

J_1 = J(idx)

J_1 = 2.5000

[Enzo]

@Stephen23 yes, it works even better. Thanks a lot

Answer 1

[Alan Stevens]

Here's one way

J = [2.5 27 56];
WZ = [12.2 23.2 33];
M1 = 22;
I1 = 2600;

for ii = 1:numel(WZ)
    slope(ii) = (J(ii)-M1)/(WZ(ii)+3750-I1);
    slope_1 = min(slope);    
end
disp(slope_1)

   -0.0168

indx = find(slope==slope_1)

indx = 1

WZ_1 = WZ(indx)

WZ_1 = 12.2000

J_1 = J(indx)

J_1 = 2.5000

[Enzo]

@Alan Stevens you are the man! thanks so much

[Stephen23]

Using the second output from MIN() is simpler than calling superfluous FIND().

Answer 2

[Dyuman Joshi]

J = [2.5 27 56];
WZ = [12.2 23.2 33];
M1 = 22;
I1 = 2600;

slope=(J-M1)./(WZ+3750-I1)

slope = 1×3

-0.0168 0.0043 0.0287

[minval,minid]=min(slope)

minval = -0.0168

minid = 1

Jval=J(minid)

Jval = 2.5000

WZval=WZ(minid)

WZval = 12.2000