How do split columns of dates in the format 'dd mm yyyy hh:mm:ss' to get just hh:mm?

I have a variable containing a column vector of dates in the format 'dd mm yyyy hh:mm:ss' and want to create another column vector next to it or variable in the format hh:mm

Answer 1

[David Young]

I assume from the formats that your variable is a cell vector of strings, and that you want the result to also be a cell vector of strings.

% test data
datestrings = {'20 07 1872 13:22:16'; '01 03 2020 05:42:06'; '28 02 1999 12:00:00'};

% column vector with times only in format HH:MM
timestrings = cellstr(datestr(datenum(datestrings, 'dd mm yyyy HH:MM:SS'), 'HH:MM'))

[Edit: added call to cellstr after seeing Guillaume's answer.]

Answer 2

[Guillaume]

Option 1: Convert to datenum (or datevec or the new datetime) and back to string:

myvar = {'31 01 2015 15:24:59'; '02 02 2013 01:02:03'};

dnmyvar = datenum(myvar, 'dd mm yyyy HH:MM:SS');
mynewvar = [myvar num2cell(datestr(dnmyvar, 'HH:MM'), 2)]

Option 2: use regexp:

myvar = {'31 01 2015 15:24:59'; '02 02 2013 01:02:03'};

mynewvar = [myvar regexp(myvar, '\d\d:\d\d', 'match', 'once')] %regexp could be more robust if wanted

Option 3: hardcode the range to extract (not as good):

myvar = {'31 01 2015 15:24:59'; '02 02 2013 01:02:03'};
mynewvar = [myvar cellfun(@(d) d(12:16), myvar, 'UniformOutput', false)]

Answer 3

[Peter Perkins]

If you have access to R2014b, try this:

>> datestrings = {'20 07 1872 13:22:16'; '01 03 2020 05:42:06'; '28 02 1999 12:00:00'}
datestrings = 
    '20 07 1872 13:22:16'
    '01 03 2020 05:42:06'
    '28 02 1999 12:00:00'
>> dt = datetime(datestrings,'Format','dd MM yyyy HH:mm:ss')
dt = 
   20 07 1872 13:22:16
   01 03 2020 05:42:06
   28 02 1999 12:00:00
>> t = timeofday(dt)
t = 
   13:22:16
   05:42:06
   12:00:00
>> tstr = cellstr(t,'hh:mm')
tstr = 
    '13:22'
    '05:42'
    '12:00'