Basic question about initializing variables
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What's the difference between
a = [ 1 2 3 ];
and
a = zeros(1,3);
a = [ 1 2 3 ];
Performance? Logic? I'm just interested to know. Thanks for your help!
采纳的回答
Jan
2017-4-20
In
a = zeros(1,3);
a = [ 1 2 3 ];
Matlab reserves memory for 3 doubles and fills then with zeros in the first line. In the second line this memory is freed and a new vector is created. This needs about the double work compared to creating [1, 2, 3] directly and in consequence is a waste of time only.
This is not a pre-allocation. A pre-allocation would mean, that the reserved memory is re-used later:
a = zeros(1, 3);
a(1) = 1;
a(2) = 2;
a(3) = 3;
or
a = zeros(1, 3);
a(:) = 1:3;
The "(:)" is essential here, because it tells Matlab to overwrite the contents of the formerly existing memory.
6 个评论
Ahmed Hossam
2017-4-20
编辑：Ahmed Hossam
2017-4-20
Ahmed Hossam
2017-4-20
编辑：Ahmed Hossam
2017-4-20
@Ahmed Hossam: You are confusing different topics together.
You should preallocate arrays whenever the array would otherwise be repeatedly expanded (for example before any loop): this significantly improves performance (but it is not a fatal error without it). This is what that help page is explaining, which its very first line clearly states: "for and while loops that incrementally increase the size of a data structure each time through the loop can adversely affect performance and memory use"
However this has nothing to do with your question (which does not mention loops, and does not incrementally expand any array). And as such Jan Simon correctly answered your question (which is not about array preallocation required before loops).
Jan
2017-4-20
a = [];
for k = 1:1e6
a(k) = k;
end
This is allowed in Matlab. Even if a is removed by clear a before the loop, Matlab runs fine. But the iterative growing of the arrays is very expensive: In the first iteration, Matlab reserves memory for 1 double (8 bytes) and assigns the value to the first element. In the next iteration, memory for 1 doubles is reserved, the first element is copied, the former array is released and the new value is copied. Finally, Matlab has reverved and copies not 1e6*8 bytes (8 MB), but sum(1:1e6)*8 bytes: 4 TB! This takes time. With a proper pre-allocation only 8MB are used:
a = zeros(1, 1e6);
for k = 1:1e6
a(k) = k;
end
Ahmed Hossam
2017-4-20
Ahmed Hossam
2017-4-20
编辑：Ahmed Hossam
2017-4-20
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