How to test multiple values in my for and if statement.

9 Ansichten (letzte 30 Tage)
Nathan Downes
Nathan Downes am 1 Mär. 2023
Bearbeitet: Jan am 2 Mär. 2023
I am trying to run this for discontinutiy functions and I want to know how to test all values for delfection of a beam from 1-35 instead of having to individually type in each number.
for x = 5
if x < 7
x = (.349*x^3 - 237.71*x + 237.36)/(10400000*.0013021);
elseif x >= 7 && x < 23
x = (.349*x^3 -.2833*(x-7)^3 - 237.71*x + 237.36)/((10400000*.0013021));
elseif x >= 23 && x < 34
x = (.349*x^3 -.2833*(x-7)^3 - .3833*(x-23)^3 - 237.71*x + 237.36)/((10400000*.0013021));
else x = 0;
end
end

Antworten (2)

Cameron
Cameron am 1 Mär. 2023
Bearbeitet: Cameron am 1 Mär. 2023
I'm not sure why you were using x to loop through your variable, but this is how I'd do it.
x = (1:35)';
coeff = zeros(length(x),2);
coeff(x >= 7) = -0.2833;
coeff(x >= 23 & x < 34,2) = -0.3833;
y = (.349*x.^3 + coeff(:,1).*(x-7).^3 + coeff(:,2).*(x-23).^3 - 237.71*x + 237.36)./(10400000*.0013021);
y(x>=34) = 0;
plot(x,y,'-ok')

Jan
Jan am 1 Mär. 2023
Bearbeitet: Jan am 2 Mär. 2023
x = 1:35;
y = zeros(size(x));
c = 10400000 * 0.0013021; % Leaner code
for k = 1:numel(x)
if x(k) < 7
y(k) = (0.349*x(k)^3 - 237.71*x(k) + 237.36) / c;
elseif x(k) < 23 % x(k) >= 7 was tested before already
y(k) = (0.349*x(k)^3 - 237.71*x(k) + 237.36 - 0.2833*(x(k)-7)^3) / c;
elseif x(k) < 34 % x(k) >= 23 was tested before already
y(k) = (0.349*x(k)^3 - 237.71*x(k) + 237.36 - 0.2833*(x(k)-7)^3 - ...
0.3833*(x(k)-23)^3) / c;
% else, y(k) = 0 is the default value already
end
end
[EDITED] Alternatively:
x = 1:35;
y = zeros(size(x));
q = x < 34;
y(q) = 0.349 * x(q).^3 - 237.71 * x(q) + 237.36;
p = q & x >= 7;
y(p) = y(p) - 0.2833 * (x(p) - 7).^3;
p = q & x >= 23;
y(p) = y(p) - 0.3833 * (x(p) - 23).^3;
y = y / (10400000 * 0.0013021);
Or shorter:
x = 1:35;
y = (0.349 * x.^3 - 237.71 * x + 237.36 ...
- 0.2833 * (max(0, x - 7)).^3 ...
- 0.3833 * (max(0, x - 23)).^3) .* (x < 34) ...
/ (10400000 * 0.0013021);

Kategorien

Mehr zu Testing Frameworks finden Sie in Help Center und File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by