divide cell array into date and time columns

I have a cell array with 6 columns and 10000 rows. The first column contains date and time in the format eg. '31/12/2001 12:57:03'. I want to seperate the first column into two seperate columns, one for date and the other for time. Then I want to convert the date to dateserial number.

thanks

Answer 1

[Walter Roberson]

temp = regexp(YourCellArray(:,1), '\s+', 'split');
datecol = cellfun(@(C) C{1}, temp, 'uniform', 0);
timecol = cellfun(@(C) C{2}, temp, 'uniform', 0)
serialdates = datenum(datecol);

If you do not actually need the time column, then this can be abbreviated down to

serialdates = datenum( regexprep(YourCellArray(:,1), '\s.*', '') );

[AA]

I want to display the time column as well but it i get the error message: index exceeds matrix dimensions. I do not know what the problem is. My aim is the convert the time as minutes after midnight with the following formula:

>> [60,1]*sscanf( '12:48', '%2d:%2d') 
    ans =
       768

[Walter Roberson]

YourCellArray{1,1} = '31/12/2001 12:57:03';
YourCellArray{2,1} = '30/12/2001 13:18:03';

then

temp = regexp(YourCellArray(:,1), '[\s:]+', 'split');
datecol = cellfun(@(C) C{1}, temp, 'uniform', 0);
hourcol = cellfun(@(C) C{2}, temp, 'uniform', 0);
mincol = cellfun(@(C) C{2}, temp, 'uniform', 0);
since_midnight = str2double(hourcol) * 60 + str2double(mincol);

... or,

temp = datevec(YourCellArray(:,1));
since_midnight = temp(:,4) * 60 + temp(:,5);

Answer 2

[Peter Perkins]

Think about using datetime and duration instead of datenum:

>> dt = datetime({'30/12/2015 15:54:30';'30/12/2015 15:54:30';'30/12/2015 15:54:30'},'InputFormat','dd/MM/yyyy HH:mm:ss')
dt = 
  3×1 datetime array
   30-Dec-2015 15:54:30
   30-Dec-2015 15:54:30
   30-Dec-2015 15:54:30
>> d = dateshift(dt,'start','day')
d = 
  3×1 datetime array
   30-Dec-2015 00:00:00
   30-Dec-2015 00:00:00
   30-Dec-2015 00:00:00
>> t = timeofday(dt)
t = 
  3×1 duration array
   15:54:30
   15:54:30
   15:54:30