Time dependent variable in plot.

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Brian
Brian 2025-2-5
评论： Brian 2025-2-6
Hello,
Fairly new to Matlab, so probably a simple question/answer.
I want to make a plot where my y-axis value depends on the timezone (x-axis). For example:
When x >= 0 & x < 5
y = 10;
When x >= 5 & x < 10
y = 10 + x*0.5;
when x >= 10 & < 15
y = -5;
The values above are just for the example.
What is the best (most efficient) way to make a plot that shows this?
Thanks in advance.

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采纳的回答

Torsten
Torsten 2025-2-5
Ran in:
The simplest method is to define x, compute y in a loop and plot:
x = 0:0.1:15;
for i = 1:numel(x)
if x(i) >= 0 & x(i) < 5
y(i) = 10;
elseif x(i) >= 5 & x(i) < 10
y(i) = 10 + x(i)*0.5;
elseif x(i) >= 10 & x(i)<= 15
y(i) = -5;
end
end
plot(x,y)
●
  2 个评论
Steven Lord
Steven Lord 2025-2-5
Ran in:
Another approach is to use logical indexing to create "masks" for each section of the x and y vectors that fall into each of your regions and use the appropriate section of the x vector to populate the corresponding section of the y vector. We'll start off with your x data.
x = 0:0.1:15;
We can create a y vector the same size as x with some "default" data. I'll use the zeros function but you could use the ones, inf, nan, or any of a number of functions.
y = zeros(size(x));
Now let's select the appropriate elements of x for your first region.
% When x >= 0 & x < 5
case1 = (x >= 0) & (x < 5);
Use case1 to select which elements in y are to be set to 10.
% y = 10;
y(case1) = 10;
We can do the same thing for case 2:
% When x >= 5 & x < 10
case2 = (x >= 5) & (x < 10);
But this time we're not setting elements of y to a constant, we're setting them to a function of the corresponding elements in x. So in this case we need to index using case2 both in the assignment to y and in the referencing from x. [People often miss the indexing into x on the right side of the equals sign, in which case they'd get an error that the things on the two sides of the equals sign are not the same size. (*)]
% y = 10 + x*0.5;
y(case2) = 10 + x(case2)*0.5;
And the same for the third case. But I'll let you finish that.
% when x >= 10 & < 15
% y = -5;
Now if we call plot with x and y, the first two parts ought to match Torsten's plot. The third part won't (note the limits on the Y axis) since I left that as an exercise to the reader :)
plot(x, y)
●
(*) See?
fprintf("The expression y(case2) refers to %d elements in y.\n", nnz(case2))
The expression y(case2) refers to 50 elements in y.
fprintf("But x has %d elements.\n", numel(x))
But x has 151 elements.
y(case2) = 10 + x*0.5; % no x(case2) gives an error
Unable to perform assignment because the left and right sides have a different number of elements.
Brian
Brian 2025-2-6
Both approaches work! Thank you both for answering so quick and clear.

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