matlab mat文件 太大,MATLAB .mat文件中的开销过大

这对我来说似乎是个错误.解决方法是以块的形式写入预分配的数组.

从预先分配开始：

fid = fopen('01_hit12.par', 'r');

data = fread(fid, inf, 'uint8');

nlines = nnz(data == 10) + 1;

fclose(fid);

matObj.moleculeNumber = zeros(1,nlines,'uint8');

matObj.isotopeologueNumber = zeros(1,nlines,'uint8');

matObj.vacuumWavenumber = zeros(1,nlines,'double');

matObj.lineIntensity = zeros(1,nlines,'double');

matObj.airWidth = zeros(1,nlines,'single');

matObj.selfWidth = zeros(1,nlines,'single');

matObj.lowStateE = zeros(1,nlines,'single');

matObj.tempDependWidth = zeros(1,nlines,'single');

matObj.pressureShift = zeros(1,nlines,'single');

然后以10000块的形式写入,我修改了你的代码如下：

... % your code plus pre-alloc first

bs = 10000;

while ischar(hitranTemp)

if abs(hitranTemp(1)) == 32;

hitranTemp(1) = '0';

end

for ii = 1:bs,

hitran{ii} = textscan(hitranTemp,'%2u%1u%12f%10f%10f%5f%5f%10f%4f%8f%15c%15c%15c%15c%6u%2u%2u%2u%2u%2u%2 u%1c%7f%7f','delimiter','','whitespace','');

hitranTemp = fgetl(fidr);

if hitranTemp==-1, bs=ii; break; end

end

% this part really ugly, sorry! trying to keep it compact...

matObj.moleculeNumber(1,k:k+bs-1) = uint8(builtin('_paren',cellfun(@(c)c{1},hitran),1:bs));

matObj.isotopeologueNumber(1,k:k+bs-1) = uint8(builtin('_paren',cellfun(@(c)c{2},hitran),1:bs));

matObj.vacuumWavenumber(1,k:k+bs-1) = builtin('_paren',cellfun(@(c)c{3},hitran),1:bs);

matObj.lineIntensity(1,k:k+bs-1) = builtin('_paren',cellfun(@(c)c{4},hitran),1:bs);

matObj.airWidth(1,k:k+bs-1) = single(builtin('_paren',cellfun(@(c)c{5},hitran),1:bs));

matObj.selfWidth(1,k:k+bs-1) = single(builtin('_paren',cellfun(@(c)c{6},hitran),1:bs));

matObj.lowStateE(1,k:k+bs-1) = single(builtin('_paren',cellfun(@(c)c{7},hitran),1:bs));

matObj.tempDependWidth(1,k:k+bs-1) = single(builtin('_paren',cellfun(@(c)c{8},hitran),1:bs));

matObj.pressureShift(1,k:k+bs-1) = single(builtin('_paren',cellfun(@(c)c{9},hitran),1:bs));

k = k + bs;

fprintf('.');

end

fclose(fidr);

磁盘上的最终大小为21,393,408字节.用法分解为,

>> S = whos('-file','01_hit12.mat');

>> fileBytes = sum([S.bytes]);

>> T = dir(which('01_hit12.mat'));

>> diskBytes = T.bytes; ratio = diskBytes/fileBytes;

>> fprintf('%10d whos\n%10d disk\n%10.6f\n',fileBytes,diskBytes,ratio)

8531608 whos

21389582 disk

2.507099

效率仍然相当低,但并非失控.

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