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我有一个文本文件，其中包含许多时间步长的结果。对于每个时间步，第一行中保存了一些基本信息，然后是一个包含此步骤的其他数据的矩阵。矩阵大小在每个时间步都可能不同，并且不是预定义的。

如何基于这样的文本文件创建结构？

提前致谢！

PS 文本文件中的结果如下所示：

 time=   4.3750000000000001E-004           3           7           4           1   4.9999989999999998E-004
  0.0000E+00  0.0000E+00  0.0000E+00  0.0000E+00  0.0000E+00  0.0000E+00  0.0000E+00
  0.0000E+00  0.0000E+00  5.6569E+08  7.5717E+08  5.6569E+08  0.0000E+00  0.0000E+00
  0.0000E+00  0.0000E+00  0.0000E+00  0.0000E+00  0.0000E+00  0.0000E+00  0.0000E+00
  0.0000E+00  0.0000E+00  0.0000E+00  0.0000E+00  0.0000E+00  0.0000E+00  0.0000E+00

 time=   5.0000000000000001E-004           3           5           3           0   4.9999989999999998E-004
  0.0000E+00  0.0000E+00  0.0000E+00  0.0000E+00  0.0000E+00
  0.0000E+00  0.0000E+00  2.3593E+08  0.0000E+00  0.0000E+00
  0.0000E+00  0.0000E+00  0.0000E+00  0.0000E+00  0.0000E+00

 time=   1.8125000000000001E-003           3           3           3           1   1.8749999000000001E-003
  0.0000E+00  0.0000E+00  0.0000E+00
  0.0000E+00  3.9138E+07  0.0000E+00
  0.0000E+00  0.0000E+00  0.0000E+00
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1 回答 1

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假设每个新的数据集都time=以间隔行开头和结尾，您可以使用以下代码示例： 

%Open text file for reading (assume file name is 'Data.txt').
f = fopen('Data.txt', 'r');

%Initialize main (store data) to empty matrix.
main = [];

%Initialize index to 1
i = 1;

while (~feof(f))
    %Read single line from text file (as long string).
    S = fgets(f);

    if (strfind(S, 'time') > 0)
        %Remove 'time= ' from the beginning of S.
        S = strrep(S, 'time=', '');

        %Convert string to array of numbers.
        T = sscanf(S, '%f ');

        %Store vector T to main(i).sub1
        main(i).sub1 = T';

        %Set A to empty matrix - prepare for filling with new data.
        A = [];

        %Read next line from text file (as long string).
        S = fgets(f);        
    end    

    %Convert string to array of numbers.
    L = sscanf(S, '%f ');

    if (isempty(L) || feof(f))
        %Store matrix A to main(i).sub2
        main(i).sub2 = A;

        %Advance i (data index) by 1.
        i = i + 1;
    else
        %In case A is not empty, concatenate T to bottom of A.
        A = [A; L'];
    end
end

%Close file.
fclose(f);

结果： 

>>main(1)

sub1 =

   4.3750e-04   3.0000e+00   7.0000e+00   4.0000e+00   1.0000e+00   5.0000e-04

sub2 =

           0           0           0           0           0           0           0
           0           0   565690000   757170000   565690000           0           0
           0           0           0           0           0           0           0
           0           0           0           0           0           0           0


>>main(2)

sub1 =

   0.00050   3.00000   5.00000   3.00000   0.00000   0.00050

sub2 =

           0           0           0           0           0
           0           0   235930000           0           0
           0           0           0           0           0


>>main(3)

sub1 =

   0.0018125   3.0000000   3.0000000   3.0000000   1.0000000   0.0018750

sub2 =

          0          0          0
          0   39138000          0
          0          0          0
于 2017-03-22T18:01:49.477 回答