# Get back variables out of function inside a for loop

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Enzo on 9 Dec 2022
Answered: Dyuman Joshi on 9 Dec 2022
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
##### 2 CommentsShowHide 1 older comment
Enzo on 9 Dec 2022
@Stephen23 yes, it works even better. Thanks a lot

Alan Stevens on 9 Dec 2022
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
##### 2 CommentsShowHide 1 older comment
Stephen23 on 9 Dec 2022
Using the second output from MIN() is simpler than calling superfluous FIND().

Dyuman Joshi on 9 Dec 2022
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

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