How can I get a numeric vector of time from two string vectors containing date and time?

I have data from Keysight benchvue and I tried first to export it to Matlab format but then the timestamp vector has values like

>> format long
>> Time_s(1)

ans =

       1.312396739146183e+17

which is far from what I would expect. This data was taken at 2016-11-18 19:29:51:461 and converting that to a numeric gives

>> datenum(2016,11,18,19,29,51.461)

ans =

       7.366528124011690e+05

I then tried to export it to csv-format instead and reading it with textscan. In that case I get two cell arrays, one containing the date and one containing the time. I don't understand how to use datevec to get the time as a numeric vector. I have tried

>> datevec(theContent{1} theContent{2})

   datevec(theContent{1} theContent{2})
                         ↑
  Error: Unexpected MATLAB expression.

where

>> theContent{1}(1)

ans = 

      '2016-11-18'

and

>> theContent{2}(1)

ans = 

      '19:29:51.461'

I also tried to concatenate the two arrays but then I get no blank space between them

>> strcat(theContent{1}(1),theContent{2}(1))

ans = 

      '2016-11-1819:29:51.461'

I guess (and hope) that this has a simple solution that some of You knows about.

Any suggestions are highly appreciated!

Thanks in advance!

Emanuel

[Walter Roberson]

https://community.keysight.com/message/73027

"Our hdf5 files use IEEE-1588 timestamps. IEEE-1588 is the “Precision Time Protocol”, which is referenced to midnight, Jan. 1, 1970."

[Emanuel Gunnarsson]

Hi Walter! I'm sorry, I don't understand how this helps me.

[Walter Roberson]

This shows that the timestamps you are seeing are potentially in IEEE-1588 format. Unfortunately in the time I spent investigating, I did not figure out what the format is, exactly. No obvious conversion suggested itself.

[Emanuel Gunnarsson]

Hi again Walter! Thanks for spending time on it! I accually saw the post You linked to yesterday, but as You say: one does not get any wiser from that.

Answer 1

[Preethi]

hi,

use datenum and datevec

>> datevec(7.366528124011690e+05)
ans =
       2016           11           18           19           29       51.461
>> datetime(ans)
ans = 

   18-Nov-2016 19:29:51

hope this is useful.

[Emanuel Gunnarsson]

Hi Preethi! Thanks, but the problem is that I do not have the kind of serial date numbers I would expect. As I wrote in the beginning my serial date numbers are in the range

1.312396739146183e+17

which is far from

7.366528124011690e+05

[Preethi]

try

x = sprintf([theContent{1}(1),' ' ,theContent{2}(1)])

[Emanuel Gunnarsson]

Hi again Preethi!

Yeah!! Thank You so much!

Apart from that the paranthesis should be changed to curly brakcets, this really solved it for me! Now I can do what You suggested:

>> x = sprintf([theContent{1}{1},' ' ,theContent{2}{1}])

x =

2016-11-18 19:29:51.461

>> datevec(x)

ans =

                      2016                        11                        18                        19                        29          51.4610000000102

Great!