string - Matlab整数字符串解码...速度优化

Question

我有以下问题:

我需要通过以下关联将整数序列“c”解码为字符字符串消息“m”:

  numpos = 10 % ( = size(c,2)/2)
  c = [3 4 1 1 4 2 5 2 3 3,1 1 1 1 2 2 2 3 3 3]

每一行“c”代表2*numpos个整数，其中第一个numpos参数编码位置

types = {'a' 'b@2' 'c@6' 'd@10' 'e@11'} 

和第二个 numpos 参数仅在类型包含字符“@”时应用，如下所示:

  m = ' c:1@6 d:1@10 a a d:2@10 b:2@2 e:2@11 b:3@2 c:3@6 c:3@6' 

我目前的解决方案如下:

  function m = c2m(c,types)

  numpos = size(c,2)/2;

  F = cellfun(@(f) [' ' f], strrep(types,'@',':%d@'),'unif',0);
  m = arrayfun(@(f,k) sprintf(f{1},k),F(c(:,1:numpos)),c(:,numpos+(1:numpos)),'unif', 0);
  m = arrayfun(@(i) horzcat(m{i,:}), (1:numlines)', 'unif', 0)

  end

测试代码如下:

  numlines = 10;
  c = repmat([3 4 1 1 4 2 5 2 3 3,1 1 1 1 2 2 2 3 3 3],numlines,1);
  types = {'a' 'b@2' 'c@6' 'd@10' 'e@11'};

  m = c2m(c,types);

  m =

    10×1 cell array

      {' c:1@6 d:1@10 a a d:2@10 b:2@2 e:2@11 b:3@2 c:3@6 c:3@6'}
      {' c:1@6 d:1@10 a a d:2@10 b:2@2 e:2@11 b:3@2 c:3@6 c:3@6'}
      {' c:1@6 d:1@10 a a d:2@10 b:2@2 e:2@11 b:3@2 c:3@6 c:3@6'}
      {' c:1@6 d:1@10 a a d:2@10 b:2@2 e:2@11 b:3@2 c:3@6 c:3@6'}
      {' c:1@6 d:1@10 a a d:2@10 b:2@2 e:2@11 b:3@2 c:3@6 c:3@6'}
      {' c:1@6 d:1@10 a a d:2@10 b:2@2 e:2@11 b:3@2 c:3@6 c:3@6'}
      {' c:1@6 d:1@10 a a d:2@10 b:2@2 e:2@11 b:3@2 c:3@6 c:3@6'}
      {' c:1@6 d:1@10 a a d:2@10 b:2@2 e:2@11 b:3@2 c:3@6 c:3@6'}
      {' c:1@6 d:1@10 a a d:2@10 b:2@2 e:2@11 b:3@2 c:3@6 c:3@6'}
      {' c:1@6 d:1@10 a a d:2@10 b:2@2 e:2@11 b:3@2 c:3@6 c:3@6'}

代码对我来说仍然太慢，我正在寻找任何加速。在这种情况下，大部分 CPU 时间花在了内置函数“sprintf”上。

问题的典型实际规模是:

   numpos ~ 30 ... 60
   numlines ~ 1e4 ... 1e5

有什么想法吗？

Answer 1

在 16b 中，MATLAB 提供了一些新的文本函数，让这一切变得简单。同样在 16b 中，MATLAB 提供了新的字符串数据类型，使它变得更快。

 function m = c2m_new(c,types, numlines)

     types = string(types);

     num_values = size(c,2)/2;

     a = c(:,1:num_values);
     b = c(:,(num_values+1):end);

     m = types(a);
     m = insertBefore(m,"@", ":" + b);
     m = join(m,2);
 end

>> numlines = 10;
>> c = repmat([3 4 1 1 4 2 5 2 3 3,1 1 1 1 2 2 2 3 3 3],numlines,1);
>> types = {'a' 'b@2' 'c@6' 'd@10' 'e@11'};
>> c2m_new(c,types,numlines)

ans = 

  10×1 string array

    "c:1@6 d:1@10 a a d:2@10 b:2@2 e:2@11 b:3@2 c:3@6 c:3@6"
    "c:1@6 d:1@10 a a d:2@10 b:2@2 e:2@11 b:3@2 c:3@6 c:3@6"
    "c:1@6 d:1@10 a a d:2@10 b:2@2 e:2@11 b:3@2 c:3@6 c:3@6"
    "c:1@6 d:1@10 a a d:2@10 b:2@2 e:2@11 b:3@2 c:3@6 c:3@6"
    "c:1@6 d:1@10 a a d:2@10 b:2@2 e:2@11 b:3@2 c:3@6 c:3@6"
    "c:1@6 d:1@10 a a d:2@10 b:2@2 e:2@11 b:3@2 c:3@6 c:3@6"
    "c:1@6 d:1@10 a a d:2@10 b:2@2 e:2@11 b:3@2 c:3@6 c:3@6"
    "c:1@6 d:1@10 a a d:2@10 b:2@2 e:2@11 b:3@2 c:3@6 c:3@6"
    "c:1@6 d:1@10 a a d:2@10 b:2@2 e:2@11 b:3@2 c:3@6 c:3@6"
    "c:1@6 d:1@10 a a d:2@10 b:2@2 e:2@11 b:3@2 c:3@6 c:3@6"

查看性能:

>> numlines = 1E4;
>> c = repmat([3 4 1 1 4 2 5 2 3 3,1 1 1 1 2 2 2 3 3 3],numlines,1);
>> types = {'a' 'b@2' 'c@6' 'd@10' 'e@11'};

% My solution
>> tic; for i = 1:10; c2m_new(c,types, numlines); end; toc
Elapsed time is 0.669311 seconds.

% michalkvasnicka's solution
>> tic; for i = 1:10; c2m(c,types, numlines); end; toc
Elapsed time is 23.643991 seconds.

% gnovice's solution
>> tic; for i = 1:10; c2m_gnovice(c,types, numlines); end; toc
Elapsed time is 8.960392 seconds.