Operation with logical data. Which is better?

n=5;
a=[1 9 0 2 3];
a.*(a<n)

ans = 1×5

1 0 0 2 3

(a.*(a<5))>0  %A

ans = 1×5 logical array
   1   0   0   1   1

logical(a.*(a<5)) %B

ans = 1×5 logical array
   1   0   0   1   1

Is a better solution A or B? Or does another solution exist?

Answer 1

[Stephen23]

n = 5;
a = [1,9,0,2,3];
b = a<n & a

b = 1×5 logical array
   1   0   0   1   1

Answer 2

[Image Analyst]

Those are different things you're showing. "a.*(a<5)" is a double vector. Your A only shows positive values, while B shows non-zero elements, both positive and negative value locations. Note the difference in my modified code

n = 5;
a = [1, 9, 0, -2, 3];
theProduct = a.*(a<n)

theProduct = 1×5

1 0 0 -2 3

theProduct > 0  %A

ans = 1×5 logical array
   1   0   0   0   1

logical(theProduct) %B

ans = 1×5 logical array
   1   0   0   1   1

So it depends on what you want to do.

Most of the time, I just use the logical operation (like < or > etc.) to product a logical variable. I rarely call the logical function explicitly.

[pipor]

ok thank

Answer 3

[Riya]

Hello pipor,

As per my understanding, you want to know which function is better to use.

Please note that, in terms of space and time complexity, both Solution A and Solution B have the same complexity.

Space Complexity:

Both solutions require additional space to store the intermediate array `a.*(a<5)`. The space complexity for both solutions is O(n), where n is the length of array `a`. This is because the intermediate array will have the same size as the input array `a`.

Time Complexity:

Both solutions involve element-wise multiplication and comparison operations. The time complexity for both solutions is also O(n), where n is the length of array `a`. This is because both solutions require iterating over each element of the array once.

Therefore, in terms of space and time complexity, both Solution A and Solution B are equivalent. You can choose either solution based on your preference for code readability and simplicity.

To analyse the time taken by both solutions, you can use the `tic` and `toc` functions in MATLAB to measure the execution time. Here's an example of how you can use `tic` and `toc` to compare the time taken by Solution A and Solution B:

n = 5; 
a = [1 9 0 2 3]; 

% Solution A 
tic; 
resultA = (a.*(a<5)) > 0; 
timeA = toc; 

% Solution B 
tic; 
resultB = logical(a.*(a<5)); 
timeB = toc; 

disp("Solution A:"); 

Solution A:

disp(resultA); 

   1   0   0   1   1

fprintf("Time taken by Solution A: %.6f seconds\n", timeA); 

Time taken by Solution A: 0.007393 seconds

disp("Solution B:"); 

Solution B:

disp(resultB); 

   1   0   0   1   1

fprintf("Time taken by Solution B: %.6f seconds\n", timeB); 

Time taken by Solution B: 0.001997 seconds

By running this code, you will get the output showing the results of both solutions and the time taken by each solution. The time taken may vary depending on your hardware and other factors, but this will give you an idea of the relative time difference between the two solutions.

For more information regarding tic and toc function you can refer the following links:

• https://www.mathworks.com/help/matlab/ref/tic.html?searchHighlight=tic%20toc&s_tid=srchtitle_support_results_1_tic%20toc
• https://www.mathworks.com/help/matlab/ref/toc.html?searchHighlight=tic%20toc&s_tid=srchtitle_support_results_2_tic%20toc

I hope it helps!