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

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Nathan Downes
Nathan Downes 2023-3-1
编辑： Jan 2023-3-2
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

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回答（2 个）

Cameron
Cameron 2023-3-1
编辑：Cameron 2023-3-1
Ran in:
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 2023-3-1
编辑：Jan 2023-3-2
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);

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