Unable to perform assignment because the left and right sides have a different number of elements.

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Inmer Jimenez
Inmer Jimenez 2024-3-7
回答： Chunru 2024-3-7
Ran in:
I am getting that error when trying to run the code.
error: Unable to perform assignment because the left and right sides have a different number of elements.
Here is the code I am trying to run:
% Define the functions g1, g2, g3, g4
g1 = @(t) -5 * tri((2 - t) / 6);
g2 = @(t) 7 * tri(3 * t) - 4 * tri(t - 4);
g3 = @(t) tri(t + 2) - 4 * tri((t + 4) / 3);
g4 = @(t) -5 * tri(t) .* tri(t - 1/2);
% Define a range of time values
t_values = linspace(-10, 10, 1000);
% Initialize arrays to store results
nonzero_times = zeros(1, 4);
back_to_zero_times = zeros(1, 4);
max_values = zeros(1, 4);
min_values = zeros(1, 4);
% Calculate values for each function
for i = 1:4
% Find the first nonzero time
nonzero_indices = find(g1(t_values) ~= 0, 1);
nonzero_times(i) = t_values(nonzero_indices);
% Find the last time to go back to zero and stay there
nonzero_indices = find(g1(t_values) ~= 0);
back_to_zero_indices = find(g1(t_values(nonzero_indices)) == 0, 1, 'last');
back_to_zero_times(i) = t_values(nonzero_indices(back_to_zero_indices));
% Find the maximum value
max_values(i) = max(g1(t_values));
% Find the minimum value
min_values(i) = min(g1(t_values));
end
Unrecognized function or variable 'tri'.

Error in solution>@(t)-5*tri((2-t)/6) (line 2)
g1 = @(t) -5 * tri((2 - t) / 6);
% Display results
disp('Nonzero times:');
disp(nonzero_times);
disp('Back to zero and stay times:');
disp(back_to_zero_times);
disp('Maximum values:');
disp(max_values);
disp('Minimum values:');
disp(min_values);

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回答（1 个）

Chunru
Chunru 2024-3-7
The following line is problematic:
back_to_zero_indices = find(g1(t_values(nonzero_indices)) == 0, 1, 'last');
First, the function g1 return floating point value and the comparison with 0 is generally not working.
You may consider to use min (if you know there exist one point to be zero, or close to zero).
[~, back_to_zero_indices] = min(abs(g1(t_values(nonzero_indices))));
You can also consider something like:
back_to_zero_indices = find(abs(g1(t_values(nonzero_indices))) < 1e-5, 1, 'last');
% need to take care of empty indices case

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