obtaining an array containing arrays
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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.
采纳的回答
James Tursa
2017-6-20
编辑：James Tursa
2017-6-20
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 个评论
Jan
2017-6-20
:-) Obviously the solution is obvious. +1
shru s
2017-6-20
James Tursa
2017-6-20
编辑：James Tursa
2017-6-20
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
shru s
2017-6-22
更多回答（1 个）
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 个评论
shru s
2017-6-20
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.
shru s
2017-6-22
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