Which of these two assignments is more efficient

Please have a look at the function below and let me know which of the sections labelled (1) and (2) is more efficient.

 function pw =realtime_func(c)
 persistent c_mem;
 if isempty(c_mem)
     c_mem=zeros(30,1);
 end

 %%(1) Is this faster
 c_mem=[c_mem(2:end);c];

 %%(2) or is this faster
 c_mem(1)=[];
 c_mem(end+1)=c;

 % output
 pw=hasTrumpCrashedTheEconomyYet(c_mem);
 end

[KSSV]

You can run a profiler to check your self or you can use tic, toc to check the timings. By the way is your label (2) working? It will throw error.

Give details, what is c and what you want to do.

Answer 1

[Jan]

Most likely method 1 is faster, because 2 must allocate 2 arrays. But Matlab's JIT acceleration might recognize this and optimize the code. So the only reliable answer is to measure it. Use timeit or run a tic toc and a loop with perhaps 1 million iterations.

Note that asking in the forum needs some time also. So you will have to run the faster version trillions of times to get the invested time back. Is this piece of code really the bottleneck of the complete program? If not: Avoid "premature optimization" (search this term in the net in case of doubts).

[bethel o]

Thanks Jan, I have ran tic toc test and the average time after 1M iteration is:

 label 1: 7.1861s
 label 2: 7.3500s

So I will use the label 1. Regarding premature optimization, thanks for pointing that out; its just that I have a near-realtime code which require a lot of these assignments and the samplerate my be increased so I guess these will at some point have an impart when the samplerate is very high

Answer 2

[James Tursa]

Another method to try that might be faster for your application:

%%(3) or is this faster
 c_mem(1:end-1)=c_mem(2:end);
 c_mem(end)=c;

[bethel o]

Thanks James for this. Here is the result including label 3 from James using 1M iterations:

 label 1: 7.1384s
 label 2: 7.3883s
 label 3: 6.5292s

This very interesting for me because I was feeling that label 1 will be faster than label 3 because it comprises one line of code. I was thinking that interpreted language works by compiling the code line by line and therefore will be faster for lesser line of code; but I was obviously wrong. I am not sure why label 3 is the fastest?

[Jan]

It seems like [c_mem(2:end);c] creates the vector c_mem(2:end) explicitely and a new vector, when the last element is added. James solution moves the values inside the existing array whithout allocating a new one.