mar*_*tin 8 arrays string double matlab cell
map1 = containers.Map({'212','2','12','44'},[4,5,6,7]);
keyset = str2double(keys(map1));
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现在我对keyset进行一组操作,这些操作将返回
Keyset= [203,2,12,39];
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我厌倦了以下几点:
num2cell(num2str(keyset));
num2cell(num2str(keyset,1));
num2cell(num2str(keyset,'%11.0g'));
num2cell(num2str(keyset,3));
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以上所有结果都给出了最终细胞阵列的奇怪结果.我只需要将整数用作另一个容器映射的键.
Jon*_*nas 13
我提出了5个额外的解决方案,其中三个解决方案比目前提出的解决方案快4-5倍.从中吸取的教训是:
num2str 是慢的cellfun并且arrayfun可以增加显着的开销三种性能最高的解决方案在性能方面非常相似:
循环分配单元格元素
n4 = length(Keyset);
tmp4 = cell(n4,1);
for i4 = 1:n4
tmp4{i4} = sprintf('%i',Keyset(i4));
end
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将all转换为字符串并调用 textscan
tmp6 = textscan(sprintf('%i\n',Keyset'),'%s');
tmp6 = tmp6{1};
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将all转换为字符串并调用regexp.
tmp3 = regexp(sprintf('%i ',Keyset),'(\d+)','match');
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这是带有时序的完整测试代码:
function t = speedTest
t=zeros(7,1);
for ii=1:100,
Keyset=randi(1,10,100); % random keys
tic;
eval( [ 'tmp1 = { ', sprintf(' %d ', Keyset), ' }; '] );
t(1)=t(1)+toc;
tic;
tmp2=arrayfun(@num2str, Keyset, 'Uniform', false);
t(2)=t(2)+toc;
tic;
tmp3 = regexp(sprintf('%i ',Keyset),'(\d+)','match');
t(3) = t(3)+toc;
tic;
n4 = length(Keyset);
tmp4 = cell(n4,1);
for i4 = 1:n4
tmp4{i4} = sprintf('%i',Keyset(i4));
end
t(4) = t(4)+toc;
tic;
n5 = length(Keyset);
tmp5 = cell(n5,1);
for i5 = 1:n5
tmp4{i5} = num2str(Keyset(i5));
end
t(5) = t(5)+toc;
tic;
tmp6 = textscan(sprintf('%i\n',Keyset'),'%s');
tmp6 = tmp6{1};
t(6) = t(6)+toc;
tic;
tmp7 = num2cell(Keyset);
tmp7 = cellfun(@(x)sprintf('%i',x),tmp7,'uni',false);
t(7) = t(7)+toc;
end;
t
t =
1.7820
21.7201
0.4068
0.3188
2.2695
0.3488
5.9186
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怎么样:
arrayfun(@num2str, Keyset, 'Uniform', false)'
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这应该为您的示例产生一个4乘1的单元格数组:
ans =
'203'
'2'
'12'
'39'
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