Well, first thing first.
1. What does this warning mean?
This code is correct in terms of syntax and it will execute correctly returning the expected result: the ii
-th element of x
will contain the value foo( ii )
.
However, before this small piece of code runs, the variable x
is not defined. Now, when the loop starts, x(1)
is assigned the value foo( 1 )
, and so Matlab creates x
as a length-1 array. At the second iteration x(2)
is assigned the value foo( 2 )
and so Matlab needs to change x
to be of length 2, and so on: x
changes its length/size at each iteration.
2. Why is changing variable size every iteration is a bad thing?
Consider what happens in the background (in terms of memory allocation) when x
changes its size every iteration: At each iteration Matlab needs to find a free memory space to host the new size of x
. If you are lucky, there is enough free space right after x
so all that happens is a change to the amount of memory allocated to x
and writing the new value at the right spot.
However, if there is not enough free space just after x
, Matlab has to find a new spot for all the ii-1
elements already in x
, allocate this new space for x
, copy all ii-1
values already in x
to the new spot, and free the old spot x
used. This allocate-copy-free operations happening in the background can be extremely time consuming, especially when x
is large.
3. How can this problem be solved?
The simplest solution is to pre-allocate all the space x
needs before the loop:
x = zeros(1,n);
for ii=1:n
x( ii ) = foo( ii );
end
By pre-allocating we ascertain that x
is allocated all the memory it requires up-front, thus no costly memory allocation/copy is needed when the loop is executing.
An alternative cool solution to the problem
If you are too lazy (like me) and don’t want to pre-allocate you can simply:
for ii=n:-1:1
x( ii ) = foo( ii );
end
This way, the first time x
is assigned a value it is assigned to its n
-th element (the last one) and therefore Matlab immediately allocates room for all n
elements of x
.
Cool!
-
Direct link to this question
⋮
-
Direct link to this question
-
LPP.m
Hi to everybody i have a problem running these codes .its a linear segment price in power generation economic dispatch matlab code
in line 17 AND SOME OTHER LINES THIS ERROR COMES: Variable F appears to change size on every loop iteration I DONT KNOW HOW TO SOLVE THAT!
1 Comment
Direct link to this comment
⋮
-
Link
Direct link to this comment
Note that this is a warning, not an error. It hints that your code is less efficient than it could be. For large arrays the cost of growing the array can end up being a major slowdown.
Sign in to comment.
Answers (1)
-
Direct link to this answer
⋮
-
Direct link to this answer
The solution is to pre-allocate:
Do
results = ones(1,10)
for ii = 1:10
results(ii) = ii;
end
instead of:
for ii = 1:10
begin_for_error(ii) = ii;
end
I didn’t go through your code so you’ll have to modify to fit your needs.
5 Comments
Direct link to this comment
⋮
-
Link
Direct link to this comment
And replace
q=1;
for i=1:N % converts the column matrix «x» to a rectangular shape
% for easy row addition on next loop
for j=1:N
X(i,j)=x(q);
q=q+1;
end
end
by:
Learn to use vectorized expressions. The fact that they are faster might not matter in your case, but they are cleaner and leaner: Less chances for typos and easier to write and maintain.
Direct link to this comment
⋮
-
Link
Direct link to this comment
What if you don’t want to allocate specific values to the enteries of the array? I got the same one while I was trying to plot an animation. Is there a way of optimising it as the exceution of my code is noticably slow.
Direct link to this comment
⋮
-
Link
Direct link to this comment
Since the goal is to overwrite them anyway, does it matter what value it starts out with?
You could consider pre-allocating with NaN, as many functions will ignore those.
If you want more specific advice, you should post this as a separate question.
Direct link to this comment
⋮
-
Link
Direct link to this comment
Direct link to this comment
⋮
-
Link
Direct link to this comment
Sign in to comment.
See Also
Categories
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
An Error Occurred
Unable to complete the action because of changes made to the page. Reload the page to see its updated state.
-
juho
- Пользователь
- Сообщения: 39
- Зарегистрирован: Пт мар 28, 2014 8:53 am
Запись данных в столбец
Здравствуйте уважаемые форумчане. Немного стыдно задавать такой простой вопрос. Поэтому прошу не сердиться на меня. Этот вопрос уже второй день не могу решить. Спасибо. В моей задаче данные записываются в заранее созданный массив в столбец v, при этом возникает Warning — The variable v appears to change size on every loop iteration consider preallocating for speed. Так и не пойму, что же я делаю не так. Спасибо.
v = [];
for a = 0:1:5
b = a + 1;
v=[v; b];
end
-
Таусинов
- Пользователь
- Сообщения: 302
- Зарегистрирован: Пт фев 25, 2011 9:29 pm
Re: Запись данных в столбец
Сообщение Таусинов » Пт июл 15, 2016 5:33 pm
juho писал(а):
Код: Выделить всё
v = [];
for a = 0:1:5
b = a + 1;
v=[v; b];
end
Код: Выделить всё
v = zeros(1, 6);
for a = 0:1:5
b = a + 1;
v(a + 1) = b;
end
Матлаб просит указать итоговый размер вашего вектора, под который, по идее, заранее выделит память. В вашем же случае память будет выделяться динамически, это более сложный процесс, который будет занимать большее количество времени. Циклов так же следует стараться избегать.
-
juho
- Пользователь
- Сообщения: 39
- Зарегистрирован: Пт мар 28, 2014 8:53 am
Re: Запись данных в столбец
Сообщение juho » Пт июл 15, 2016 6:28 pm
Спасибо Вам большое. Теперь стало все понятно. Буду стараться избегать таких ошибок. Еще раз благодарю.
machet1k 2 / 2 / 1 Регистрация: 20.12.2011 Сообщений: 40 |
||||
1 |
||||
Моя матрица не хочет заполняться27.07.2013, 14:15. Показов 4402. Ответов 2 Метки нет (Все метки)
Друзья, подскажите, пожалуйста, почему на такой простецкий код
матлаб ругается вот такой ошибкой «the variable ‘Energy’ appears to change size on every loop iteration. Consider preallocating for speed»
0 |
Programming Эксперт 94731 / 64177 / 26122 Регистрация: 12.04.2006 Сообщений: 116,782 |
27.07.2013, 14:15 |
2 |
5224 / 3554 / 374 Регистрация: 02.04.2012 Сообщений: 6,460 Записей в блоге: 17 |
|
27.07.2013, 16:19 |
2 |
Это он еще не ругается, а предупреждает, мол размер переменной на каждом шагу цикла будет меняться (что замедляет вычисления) PS: создать такую нулевую матрицу можно так:
2 |
Sabbat 135 / 22 / 1 Регистрация: 19.10.2012 Сообщений: 42 |
||||||||
27.07.2013, 16:21 |
3 |
|||||||
Ругается т. к. меняется размер переменной Energy при каждой итерации
и в Матлабе нумерация с 1 а не с 0
2 |
iterationmatlabmemory-management
When writing the following Matlab code:
for ii=1:n
x(ii) = foo( ii ); % foo is some function of ii that cannot be vectorized.
end
I get the following m-lint warning:
The variable
x
appears to change size on every loop iteration
My question:
- What does that warning mean?
- Why is changing variable size every iteration is a bad thing?
- How can this problem be solved?
This question is not duplicate of this one, since it deals with more general aspects of preallocation, rather a specific instance of it.
Related Question