obtaining an array containing arrays
1 view (last 30 days)
Show older comments
hello, suppose i am generating an array with every loop. I want a new array to store all the values of the array together but seperately as different elements. If I have A = [1 2 3 4] in the 1st loop and A = [2 4 6 8] when it completes the second loop i want an array B that will have B= [1 2 3 4 ; 2 4 6 8]. If anyone can help me with this, I will be grateful. Thank you.
0 Comments
Accepted Answer
James Tursa
on 20 Jun 2017
Edited: James Tursa
on 20 Jun 2017
E.g.,
m = number of loop iterations
B = zeros(m,4);
for k=1:m
A = results of current iteration
B(k,:) = A;
end
4 Comments
James Tursa
on 20 Jun 2017
Edited: James Tursa
on 20 Jun 2017
You could set things up so the first iteration expands the matrix appropriately. E.g.,
m = number of loop iterations
B = zeros(m,0); % <-- Unknown column size, so start with 0
for k=1:m
A = results of current iteration
B(k,1:size(A,2)) = A; % <-- 1st iteration will set the column size
end
More Answers (1)
Jan
on 20 Jun 2017
Edited: Jan
on 20 Jun 2017
nLoop = 5;
nValue = 4;
Result = zeros(nLoop, nValue); % Pre-allocate!
for k = 1:nLoop
A = randi(10, 1, nValue) % Example data
Result(k, :) = A;
end
disp(Result)
3 Comments
Jan
on 20 Jun 2017
Edited: Jan
on 20 Jun 2017
See James' answer. If nValue i changing between the elements, use a cell array:
nLoop = 5;
Result = cell(nLoop, 1); % Pre-allocate!
for k = 1:nLoop
A = randi(10, 1, randi(5)) % Example data
Result{k} = A;
end
An "implicite pre-allocation" works also, if you create the last element at first:
nLoop = 5;
for k = nLoop:-1:1 % Backwards for implicite pre-allocation
A = randi(10, 1, 4) % Example data
Result(k, :) = A;
end
It is essential, that the output does not grow iteratively, because this requires an exponentially growing number of resources. With 5 iterations this cannot be measured, but the rule is to care about a pre-allocation in general as a good programming practice.
See Also
Categories
Find more on Loops and Conditional Statements in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!