How to make this simple script faster?

Each of the Sta{i}.test is a one-column data with unknown number of rows. As you know, to keep expanding Variable X is not the most efficient way of doing this.

Is there a way I can get my X values faster without writing a loop?

X = [];
for i = a:b
    if isfield(Sta{i}, 'test') 
         X = [X;  Sta{i}.test];
    end
end

Many thanks!

[Image Analyst]

Are you saying that sometimes the structure inside the cell might not have a field called "test"? What if you just allocated X as, say, 10 million elements or way more than you ever expect to have, then crop to the proper length after the loop. See my Answer below (scroll down).

Answer 1

[Image Analyst]

Does this work:

X = zeros(10000000, 1);
a = 1;
b = length(Sta);
lastIndex = 0
for i = a:b
	if isfield(Sta{i}, 'test')
		t = Sta{i}.test;
		lt = length(t);
		X(lastIndex+1:lastIndex+lt) = t;
		lastIndex = lastIndex + lt;
	end
end
X = X(1:lastIndex);

What is the size of Sta? How many elements does it have?

[Leon]

There are dozens of elements for each Sta. The size is on the thousands.

One question, if test is always available. Is it possible to do things like this without writing a loop? For example, is there something like this within Matlab?

Sta([a:b]).test

Similar to

Data([a:b],5);

[Image Analyst]

It might be. I experimented with doing things like (:) and {:} and putting stuff in brackets and using vertcat() but I just couldn't find the exact syntax that would do it. You'd have to experiment around some.

Can you store them in a more convenient way in the first place?

[Leon]

Any recommendation on what's the best way to store them in order to access them in the most convenient way?

In a matrix named A like this:

Sta#, Var1, Var2
1     35    127
1     21    256
1     11    222
2     15    188
2     12    236
3     27    333
4     35    467
5     11    222
5     23    376

Then how do I quickly get all the rows with a Sta# between 2 and 4, for example?

Ind = A(:,1)>=2 & A(:,1)<=4;

B = A(Ind, :);

Is there a better setup? Many thanks!

Answer 2

[David Hill]

You could preallocate X with nan to the largest expected and then delete the excess nan's at the end.

X = nan(10000000,1);
count=1;
for i = a:b
    if isfield(Sta{i}, 'test') 
        temp=count+length(Sta{i}.test); 
        X(count:temp-1) = Sta{i}.test;
        count=temp;
    end
end
X=X(~isnan(X));

[Leon]

Many thanks for the solution, David! So bad I can only accept one answer.