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?
最佳答案
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.
关于string - Matlab整数字符串解码...速度优化,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/47263760/