Merge two columns of a cell array into one

Hi,

I have two columns of cell values:

I want to merge these two columns to produce one third column in the format yyyy-mm-dd hh-mm-ss.

Thanks for any help!

Answer 1

[KSSV]

c1 = {'2104-04-14'; '2014-04-14'} ;
c2 = {'13:17:08'; '13:16:45'} ;
datetime(strcat(c1,{' '},c2))

[BOB]

Thank you!

[KSSV]

Thanks is accepting the answer..:)

[Jan]

No, KSSV, only accepting is accepting the answer. ;-)

@BOB: If your problem is solved, please accept the answer. Then other readers know, that you do not need help anymore. In addition the author of the answer will take this as additional thank.

[John BG]

Hi Bob

this is John BG <mailto:[e-mail] [e-mail]>

why is that when reading the resulting merged columns, the format YYYY-MM-DD is reversed?

c1 = {'2104-04-14'; '2014-04-14'} ;
c2 = {'13:17:08'; '13:16:45'} ;
B=datetime(strcat(c1,{' '},c2))

   B(1)
  ans = 
    datetime
     14-Apr-2104 13:17:08
  >> B(2)
   = 
    datetime
     14-Apr-2014 13:16:45

Is this ok? or you really want the columns YYYY-MM-DD and HH:MM:SS merged as YYY-MM-DD HH:MM:SS?

wat if there's an empty HH:MM:SS?

[Jan]

@John BG: It is not useful to invent problems, that the OP does not have.

 why is that when reading the resulting merged columns, the format
 YYYY-MM-DD is reversed?

This is just the default format of the datetime object for the output, not for the internal storage, as explained in the documentation. If you want to keep the output in the original format:

c1 = {'2104-04-14'; '2014-04-14'} ;
c2 = {'13:17:08'; '13:16:45'} ;
B  = datetime(strcat(c1, {' '}, c2))
B.Format = 'yyyy-MM-dd hh:mm:ss'

 B = 
  2×1 datetime array
   2104-04-14 01:17:08
   2014-04-14 01:16:45

[John BG]

well, if you add the line

B.Format = 'yyyy-MM-dd hh:mm:ss'

then it complies with the question requirement:

YYYY-MM-DD

If you do not add that line, then you are supplying years for days and days for years, it would be kind of odd, wouldn't it?

the order matters, if one goes to cash machine and asks for

£301.- (YYYY-MM-DD)

but only gets

£103.- (DD-MM-YYYY)

wouldn't one complain?

Jan, no inventions, just reading the question.

[Jan]

@John BG: Bob said "Thank you!" already, which sounds like the problem was solved. Then datetime works already. Maybe you did not understand, how datetime stores its data internally, but it does not matter what you set as display format. A calculation like "£301.- (YYYY-MM-DD)" is nonsense. If you need the value of a datetime object, you would e.g. use:

301 - day(B(1))

and for the output of day(B) it does not matter, if the date is displayed as YYYY-MM-DD or DD-MM-YYYY. This is the idea of datetime compared to the old serial date numbers or date vectors, where the display and the meaning of the dates and times were related.

[Giuseppe Degan Di Dieco]

Dear KSSV, your solution it really helped me in concatenating two cell arrays of character vectors into one cell array. Also, the cell arrays belonged to a data table. Thank you.