Not expected results using datevec funcion

Dear colleagues,

I am writing from Brazil in order to get some help about datavec funciton.

I am using this fucntion to convert a date string data to vector format ([yy mm dd HH MM SS])

In Brazil date format is defined as follows: dd/mm/yy. I have a set of date data in two main formats. one without timestamp and another with timestamp, for instance: '13/12/2013' , '13/12/2013 00:15:00', respectively.

I get an error when I call the function to get the conversion:

   Situation 1)   result = datevec('13/12/2013', 'dd/mm/yyyy HH:MM:SS');

When I use the function with data with timestamp, as follows

   Situation 2) result = datevec('13/12/2013 00:00:00', 'dd/mm/yyyy HH:MM:SS');

or

   Situation 3)   result = datevec('13/12/2013 00:15:00', 'dd/mm/yyyy HH:MM:SS');

I get correct answer:

Anybody has some hint to handle situation 1?

I appreciate any help. Thanks all in advance.

Best regards, Luis.

Answer 1

[Azzi Abdelmalek]

Use this

   result = datevec(datestr('13/12/2013', 'dd/mm/yyyy HH:MM:SS'))

datestr will set the missed time to 00:00:00

[Luis Fernando]

Azzi, thank you for your rapid reply.

However, the command you have recommended results " >> result = datevec(datestr('13/12/2013', 'dd/mm/yyyy HH:MM:SS'))

result =

    19     5     6     0     0     0

>> "

an answer that not match with the expected result: [2013 12 13 0 0 0] which is the desired answer.

Anyway, I thank you again for your atention. See ya, mate.

[Azzi Abdelmalek]

Try this

d={'13/12/2013','26/12/2013 19:20:05'}
dd=cellfun(@(x) [x  ' 00:00:00'],d,'un',0)
out=datevec(dd,'dd/mm/yyyy HH:MM:SS')

[Luis Fernando]

Azzi, it works as a charm! Thank you for your help, my friend. It is gonna help a lot! Have a good night or day! :-)

[Jan]

Instead of the slow and complicated cellfun method to join strings, strcat is niocer and faster:

dd = strcat(d, ' 00:00:00')

[Luis Fernando]

Jan,

   Good hint as well! Thank you!