read data from csv file & fast processing

1 visualizzazione (ultimi 30 giorni)
shaz
shaz il 7 Nov 2012
i have a data like which is read from csv file like
10;20;2;3;45;56;87;56;988;434;10;20;2;3;45;56;87;56;988;434
i have 9lakhs of data like this what is best way to separate by ; & form a matrix
because now it is consuming to much of time
Thanks in advance
  12 Commenti
shaz
shaz il 8 Nov 2012
Modificato: Walter Roberson il 8 Nov 2012
a;b;c;d;f;g;i;p;t;r;l;e;r;y;u % Headers
10;20;2;3;4.5;56;87;56;988;434;10;20;2;3;45;56;87;56;988;434
10;20;2;3;4.5;56;87;56;988;434;10;20;2;3;45;56;87;56;988;434
10;20;2;3;45;56;87;56;988;434;10;20;2;3;45;56;87;56;988;434
....................% data
.....................% data
like wise we have 9 lakh rows(900000 rows) of data & one row of header
now how to get the headers in cell format(1st row) & rest data in matrix format
Jan
Jan il 27 Nov 2012
@shaz: Although it looks obvious, that you neither read the comments nor care about them: Please stop using the term "lakh" without any further explanations, because it is not part of the English language such that it might confuse the readers. It is very friendly that Per explained this, but this has been your your turn actually.
Can you imagine the effects, when you do not care about questions for clarifications?

Accedi per commentare.

Risposta accettata

Jan
Jan il 7 Nov 2012
Modificato: Jan il 8 Nov 2012
fid = fopen(FileName, 'r');
if fid == -1, error('Cannot open file!'); end
header = fgetl(fid); % Read header line
pos = ftell(fid); % Store position [EDITED: fseek -> ftell]
line1 = fgetl(fid); % first data line
data = sscanf(line1, '%g;', Inf);
ncol = length(data);
fmt = repmat('%g; ', 1, ncol);
fseek(fid, pos, 'bof'); % Restore file position
data = fscanf(fid, fmt, Inf);
data = transpose(reshape(data, ncol, []));
fclose(fid);
Please check the reshape and transpose, if it satisfies your needs.
  3 Commenti
Jan
Jan il 27 Nov 2012
Does this comment contain a question?
shaz
shaz il 10 Dic 2012
@simon: thanks for the support...great

Accedi per commentare.

Più risposte (0)

Tag

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by