How to remove the for loop?

2016 2 1
2 个回答
回答已采纳
更新时间：2016 2 8
6 次查看（30 天）

0 个投票

w=[1 -0.75 -0.09375 -0.0390625 -0.021972656 -0.014282227 -0.010116577 -0.007587433 -0.005927682 -0.004775077 -0.003939439 -0.00331271 -0.002829606 -0.002448698 -0.00214261 -0.001892639 -0.001685632 -0.001512111 -0.0013651 -0.001239367 -0.001130923 -0.001036679 -0.000954216 -0.000881613 -0.000817328 -0.000760115 -0.000708954 -0.000663003 -0.000621565 -0.000584057 -0.000549987 -0.000518939];
y=[0.841470985 0.963558185 0.999573603 0.946300088 0.808496404 0.598472144 0.33498815 0.041580662 -0.255541102 -0.529836141 -0.756802495 -0.916165937 -0.993691004 -0.982452613 -0.883454656 -0.705540326 -0.464602179 -0.182162504 0.116549205 0.404849921 0.656986599 0.850436621 0.967919672 0.998941342 0.940730557 0.798487113 0.584917193 0.319098362 0.024775425 -0.271760626 -0.544021111 -0.76768581]';
t=0:0.1:pi;
dy=zeros(32,1); %Initialization of dy
for i=2:length(t)
dy(i)= mtimes(w(1:i),y(i:-1:1))
end
%Expected value of dy is as follows
% dy=[0 0.332454946250000 0.198017059406250 0.0734163455546875 -0.0510670313701985 -0.168851648273860 -0.270865192726796 -0.348550483481562 -0.395200810620278 -0.406748398590689 -0.382201097182780 -0.323762604399792 -0.236650286304258 -0.128636329190141 -0.00935712148064122 0.110545636146184 0.220374543577469 0.310331564135441 0.372393027597958 0.401026261573602 0.393683796647517 0.351030956373549 0.276886467490983 0.177881407093338 0.0628669228432652 -0.0578763540568076 -0.173556593560093 -0.273834716498144 -0.349747909930254 -0.394510207344599 -0.404118623202944 -0.377710685470080]

6 个评论
显示 4更早的评论 隐藏 4更早的评论

Stephen23
Stephen23 2016-2-1
Why do you want to remove the loop? Just include preallocation of the output array and it will be neat and readable code. Trying to remove this loop will not make the code clearer!
Parag Patil
Parag Patil 2016-2-1
Thanks. But I want to vectorize the code for running it on GPU.
Stephen23
Stephen23 2016-2-1
编辑：Stephen23 2016-2-1
Sure, but you did not answer my question: why? Is the code slow? Are you preallocating the array before the loop? Not preallocating is the main cause of beginners complaining about slow code, just like in your example.
How big are the data arrays?
Syed Ajimal khan
Syed Ajimal khan 2016-2-1
What would be the value for dy(1).. you are alloting diractly as dy(2). so may i send in the same way?
Parag Patil
Parag Patil 2016-2-3
Stephen, the code is part of a GL-Derivative. After profiling the entire program, this part is taking the max time. So in order to reduce the execution time i am trying to implement it on GPU using gpuArray(). The data arrays depend on the time period t, which can be pi,2pi,3pi,..,npi. I have rephrased the question with the values of w & y for t=0:0.1:pi.
Syed, i am directly alloting dy(2) because its required that initially dy(1)=0.
Walter Roberson
Walter Roberson 2016-2-3
Note that
dy(i)= mtimes(w(1:i),y(i:-1:1))
is the same as
dy(i) = dot(w(1:i), y(i:-1:1))
for which it is not necessary that y be a column vector

请先登录，再进行评论。

请先登录，再回答此问题。

 采纳的回答

Andrei Bobrov
Andrei Bobrov 2016-2-3
编辑：Andrei Bobrov 2016-2-3

1 个投票

dy = sum(triu(w(:)*ones(1,numel(w))).*toeplitz([y(1);zeros(numel(y)-1,1)],y));
dy(1) = 0;

4 个评论
显示 2更早的评论 隐藏 2更早的评论

Walter Roberson
Walter Roberson 2016-2-3
Cute. But I suspect that might not be efficient ;-)
Joss Knight
Joss Knight 2016-2-3
Good answer, but seems more complicated than it needs to be. All you need is
dy = [w * triu(toeplitz(y))]';
dy(1) = 0;
Andrei Bobrov
Andrei Bobrov 2016-2-6
Hi Joss! Your comment has a better answer. Thank you!

请先登录，再进行评论。

更多回答（1 个）

Walter Roberson
Walter Roberson 2016-2-3

0 个投票

gpu_w = gpuArray(w(:));
gpu_yR = gpuArray( flipud(y(:)));
idx = gpuArray(2:length(t)).';
dy_2_onwards = arrayfun(@(K) dot(gpu_w(1:K), gpu_yR(end-K+1:end)), idx);
dy = [0; gather(dy_2_onwards)];
I also suggest comparing the timing of
gpu_w = gpuarray(w(:));
gpu_y = gpuarray(y(:));
idx = gpuarray(2:length(t));
dy_2_onwards = arrayfun(@(K) dot(gpu_w(1:K), gpu_y(K:-1:1)), idx);
dy = [0; gather(dy_2_onwards)];
In both of these, if you already know for sure that w and y are the same orientation then you could skip the (:) . dot on gpuArray might also be okay with vectors of different orientation; I do not have access to the documentation for it.
I would further compare to not using the GPU and instead using
dy = [0; arrayfun(@(K) dot(w(1:K), y(K:-1:1)), (2:length(t)).')]
as I suspect the overhead of using the GPU might not be worth the effort.
It also would not surprise me if a plain loop were faster than the arrayfun.

3 个评论
显示 1更早的评论 隐藏 1更早的评论

Joss Knight
Joss Knight 2016-2-3
This won't work on the GPU because gpuArray/arrayfun only supports scalar operations. You can't index any more than one element, you can't call transpose, and you can't call dot.
Walter Roberson
Walter Roberson 2016-2-3
编辑：Walter Roberson 2016-2-3
Ah. But dot is listed as allowed in http://www.mathworks.com/help/distcomp/run-built-in-functions-on-a-gpu.html so it is not obvious why you would not be able to call dot from gpuArray/arrayfun ?

请先登录，再进行评论。

类别

在 帮助中心 和 File Exchange 中查找有关 Loops and Conditional Statements 的更多信息

产品

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by