Reading large files

textscan

Assume you have formatted data in a large text file or string, e.g.

Data,2015-09-16,15:41:52;781,780.000000,0.0034,2.2345
Data,2015-09-16,15:41:52;791,790.000000,0.1255,96.5948
Data,2015-09-16,15:41:52;801,800.000000,1.5123,0.0043

one may use textscan to read this quite fast. To do so, get a file identifier of the text file with fopen:

fid = fopen('path/to/myfile');

Assume for the data in this example, we want to ignore the first column "Data", read the date and time as strings, and read the rest of the columns as doubles, i.e.

Data  ,  2015-09-16  , 15:41:52;801  , 800.000000  , 1.5123  ,  0.0043
ignore      string         string         double      double     double

To do this, call:

data = textscan(fid,'%*s %s %s %f %f %f','Delimiter',',');

The asterisk in %*s means "ignore this column". %s means "interpret as a string". %f means "interpret as doubles (floats)". Finally, 'Delimiter',',' states that all commas should be interpreted as the delimiter between each column.

To sum up:

fid = fopen('path/to/myfile');
data = textscan(fid,'%*s %s %s %f %f %f','Delimiter',',');

data now contains a cell array with each column in a cell.

Date and time strings to numeric array fast

Converting date and time strings to numeric arrays can be done with datenum, though it may take as much as half the time of reading a large data file.

Consider the data in example Textscan. By, again, using textscan and interpret date and time as integers, they can rapidly be converted into a numeric array.

I.e. a line in the example data would be interpreted as:

Data , 2015  - 09  -  16  ,  15  :  41  :  52  ;  801 , 800.000000 , 1.5123 , 0.0043
ignore double double double double double double double    double     double   double

which will be read as:

fid = fopen('path/to/myfile');
data = textscan(fid,'%*s %f %f %f %f %f %f %f %f %f %f','Delimiter',',-:;');
fclose(fid);

Now:

y = data{1};          % year
m = data{2};          % month
d = data{3};          % day
H = data{4};          % hours
M = data{5};          % minutes
S = data{6};          % seconds
F = data{7};          % milliseconds

% Translation from month to days
ms = [0,31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334]; 

n = length(y);        % Number of elements
Time = zeros(n,1);    % Declare numeric time array

% Algorithm for calculating numeric time array
for k = 1:n
    Time(k) = y(k)*365 + ms(m(k)) + d(k) + floor(y(k)/4)...
              - floor(y(k)/100) + floor(y(k)/400) + (mod(y(k),4)~=0)...
              - (mod(y(k),100)~=0) + (mod(y(k),400)~=0)...
              + (H(k)*3600 + M(k)*60 + S(k) + F(k)/1000)/86400 + 1;
end

Using datenum on 566,678 elements required 6.626570 seconds, whilst the method above required 0.048334 seconds, i.e. 0.73% of the time for datenum or ~137 times faster.