Use Textscan or sscanf?

My data :)

1      764.2981300   3     0  25   103
1      764.2981300   3     0  25   103
1      764.2981300   3     0  25   103
* Gp# 5[2,1,3,4,6] x =    51.26, y =    149.1, z =     43.3, q = 0.885 dT[     1,       18,    24,    59] Src Amplitude = 38.3
1      764.2981300   3     0  25   103
1      764.2981300   3     0  25   103
* Gp# 5[2,1,3,4,6] x =    51.26, y =    149.1, z =     43.3, q = 0.885 dT[     1,       18,    24,    59] Src Amplitude = 38.3

Hi all,

Thanks in advance for your help.

I am loading text files with lines similar to these. The first 3 lines (Set 1) will always have this structure (different numbers). Another set (* Gp#) will be similar to the ones shown. I used fgetl to grab each line and place it in a cell{k,1} because I do not know how many times Set 1 will appear between (* Gp#) lines.

To read the first line I have tried

jj = sscanf('%d%c%d%c%d%c%d%c%d%c%d',string), resulting in jj = ' '

Is this sscanf-able?

I thought about textscan(fid), but since the lines change randomly between one format and another, I am not sure how to do so.

Any general advice?

Regards,

John

Answer 1

[Cedric]

Assuming that lines starting with a '*' are some sort of block footers, here is one way to extract data blocks:

 fid  = fopen('myData.txt', 'r') ;
 data = {[]} ;                               % Init cell array and 1st block.
 dId  = 1 ;                                  % Data block ID/counter.
 while ~feof(fid)
    line = fgetl(fid) ;
    if line(1) == '*'                        % Block end => new block+init.
        dId = dId+1 ;
        data{dId} = [] ;
        continue ;
    end
    data{dId} = [data{dId}; sscanf(line, '%f %f %f %f %f %f').'] ;
 end
 data(end) = [] ;                            % Remove last, empty block.
 fclose(fid) ;

Here is another solution based on a regexp split:

 data = regexp(fileread('myData.txt'), '\*.+?e = [\-\d\.]+', 'split') ;
 data(end) = [] ;
 for dId = 1 : numel(data)
    data{dId} = reshape(sscanf(data{dId}, '%f'), 6, []).' ;
 end

Running either on your example outputs:

 >> data
 data = 
    [3x6 double]    [2x6 double]

 >> data{1}
 ans =
    1.0000  764.2981    3.0000         0   25.0000  103.0000
    1.0000  764.2981    3.0000         0   25.0000  103.0000
    1.0000  764.2981    3.0000         0   25.0000  103.0000

 >> data{2}
 ans =
    1.0000  764.2981    3.0000         0   25.0000  103.0000
    1.0000  764.2981    3.0000         0   25.0000  103.0000

[John]

Thank you for your response. You have helped me out tremendously.

John