可以下面的循环以量化?

Question

我有一个用于环它执行的下列功能：

采取M8矩阵，并且：

1. 它分割成块的大小512元(含义X8矩阵==512，而元件的数目可128,256,512,1024,2048)
2. 改造的框入1 512(数量元素)的矩阵。
3. 采取的最后1/4的矩阵，把它放在前面，
   例如 Data = [Data(1,385:512),Data(1,1:384)];

以下是我的代码:

for i = 1 : NumOfBlock  
    if i == 1  
        Header = tempHeader(1:RowNeeded,:);  
        Header = reshape(Header,1,BlockSize); %BS  
        Header = [Header(1,385:512),Header(1,1:384)]; %CP  
        Data = tempData(1:RowNeeded,:);  
        Data = reshape(Data,1,BlockSize); %BS  
        Data = [Data(1,385:512),Data(1,1:384)]; %CP  
        start = RowNeeded + 1;  
        end1 = RowNeeded * 2;  
    else  
        temp = tempData(start:end1,:);  
        temp = reshape(temp,1,BlockSize); %BS  
        temp = [temp(1,385:512),temp(1,1:384)]; %CP  
        Data = [Data, temp];  
    end

    if i <= 127 & i > 1
        temp = tempHeader(start:end1,:);
        temp = reshape(temp,1,BlockSize); %BS
        temp = [temp(1,385:512),temp(1,1:384)]; %CP
        Header = [Header, temp];
    end

    start = end1 + 1;
    end1=end1 + RowNeeded;  
end

运行这种循环有5万元将超过1小时。我需要尽可能快(在sec)。这是循环能够量化?

Answer 1

再一次我要谢谢Amro给我一个想法就如何解决我的问题。对不起对不会让自己清楚在这个问题。

这是我解决我的问题：

%#BS CDMA, Block size 128,512,1024,2048  
  BlockSize = 512;  
  RowNeeded = BlockSize / 8;  
  TotalRows = size(tempData);  
  TotalRows = TotalRows(1,1);  
  NumOfBlock = TotalRows / RowNeeded;  
  CPSize = BlockSize / 4;  

%#spilt into blocks  
  Header = reshape(tempHeader',[RowNeeded,8, 128]);  
  Data = reshape(tempData',[RowNeeded,8, NumOfBlock]);  
  clear tempData tempHeader;  

%#block spread & cyclic prefix  
    K = zeros([1,BlockSize,128],'single');  
    L = zeros([1,BlockSize,NumOfBlock],'single');  
       for i = 1:NumOfBlock  
           if i <= 128  
              K(:,:,i) = reshape(Header(:,:,i),[1,BlockSize]);  
              K(:,:,i) = [K((CPSize*3)+1:BlockSize),K(1:CPSize*3)];
           end  
           L(:,:,i) = reshape(Data(:,:,i),[1,BlockSize]);  
           L(:,:,i) = [L((CPSize*3)+1:BlockSize),L(1:CPSize*3)];
        end

Answer 2

根据你的描述，这就是我想出了：

M = 320;           %# M must be divisble by (numberOfElements/8)
A = rand(M,8);     %# input matrix

num = 512;         %# numberOfElements
rows = num/8;      %# rows needed

%# equivalent to taking the last 1/4 and putting it in front
A = [A(:,7:8) A(:,1:6)];

%# break the matrix in blocks of size (x-by-8==512) into the third dimension
B = permute(reshape(A',[8 rows M/rows]),[2 1 3]);

%'# linearize everything
B = B(:);

该图可能有助于了解上述：

Answer 3

矢量可能或不可能的帮助。什么会有帮助的是知道 哪里 瓶颈。使用探查所概述的在这里：

http://blogs.mathworks.com/videos/2006/10/19/profiler-to-find-code-bottlenecks/

Answer 4

这将是好的，如果你会告诉你们试图要做到(我的猜测是，一些模拟在动态系统，但是很难告诉).

是的，当然它可以量化:你的每一个街区实际上是四个子块；用你的(非常非标准)指数：

1...128, 129...256, 257...384, 385...512

每一个内线是什么-没有-你叫它矢量化应该做到以下几点：

i=threadIdx之间0和127 temp=data[1+i] 数据[1+i]=data[385+i] 数据[385+i]=data[257+i] 数据[257+i]=data[129+i] 数据[129+i]=温度

你应该当然还并行块，不仅矢量.