merging column elements into one

I have a matrix of 4 columns containing year in 1st column, month in 2nd, day in 3rd, and time in 4th. Is there a way to combine them into 1 column to get 'dd-mmm-yyyy HH:MM:SS' ?

Thanks

[Walter Roberson]

What is the datatype of your matrix? If it is a cell, then what is the format of each of the columns at present?

[MF]

double. For example:

year     month   day   hour

 2015     2      1      1

Answer 1

[Azzi Abdelmalek]

Look at this example

v={'2016' '03' '02' '12:10:00';'2016' '03' '02' '13:20:00'}
out=arrayfun(@(x) [v{x,1} '-' v{x,2} '-'  v{x,3}  ' ' v{x,4}] , (1:size(v,1))','un',0)

[Azzi Abdelmalek]

v=[2015 2 1 1;2016 2 10 2]
w=[v zeros(size(v,1),2)]
out=datestr(w,'yyyy-mm-dd HH:MM:SS')

Answer 2

[Walter Roberson]

v={'2016' '03' '02' '12:10:00';'2016' '03' '02' '13:20:00'}
strcat(v(:,1),{'-'},v(:,2),{'-'},v(:,3), {' '}, v(:,4))

but only if v happens to be in the correct format already, which is something we do not know yet as you have not replied about what the format of your array is.

[MF]

my v is a 9834x5 double matrix

Answer 3

[Stephen23]

Use datstr and cellstr:

>> M = [2016,03,28,12;2015,2,1,1]
M =
   2016      3     28     12
   2015      2      1      1
>> M(:,6) = 0;
>> C = cellstr(datestr(M,'dd-mmm-yyyy HH:MM:SS'));
>> C{:}
ans = 28-Mar-2016 12:00:00
ans = 01-Feb-2015 01:00:00

You can read the documentation and pick the date format that suits your needs best.

Answer 4

[Roger Stafford]

You can easily compress seconds, minutes, hours, days of the month, months, and years into one double precision floating point number. It has more than enough capacity. Let T be a vector with six elements: [year number, month number, day of month number, hour, minutes, seconds]. To convert to a single 'double':

d = T(6)+64*(T(5)+64*(T(4)+64*(T(3)+64*(T(2)+64*T(1)))));

(There is room here to go up past the year 8000000.) To convert back to T vector:

function T = dec2base64(d)
 T = zeros(1,6);
 for k = 6:-1:2
  q = floor(d/64);
  T(k) = d - 64*q;
  d = q;
 end
 T(1) = d;
return