Convert from Microsoft's DateTime Structure to readable date

I have a dataset with collection intervals marked with numeric identifiers such as the following:

- 635364711700000000 - 635364711800000000 -635364711900000000

These IDs are from Microsoft's DateTime Structure (see link below). Does anybody know if Matlab has functionality to convert from the DateTime Structure to a timestamp in human readable format?

http://msdn.microsoft.com/en-us/library/system.datetime.aspx

Answer 1

[José-Luis]

No, I don't think there is. But if you don't care about leap seconds it is not too complicated:

epoch = datenum(1,1,0);
matlab_time = epoch + MicrosoftTime / (86000 * 10^7); %1 unit Microsoft time = 100 nanoseconds

e.g., taking care of the fact that MicrosftTime should not be stored as double:

datestr(double(datenum(1,1,0) + int64(635364711800000000) / (86400 * 10^7)))

[Tim]

Thanks, just a quick followup before I accept this. I'm interested in the timestamp down to the "seconds" level of detail, how would you modify this command to accomplish that?

datestr(double(datenum(1,1,0) + int64(635364711800000000) / (86400 * 10^7)))

[José-Luis]

What do you mean?

 datevec(double(datenum(1,1,0) + int64(635364711800000000) / (86400 * 10^7)))

[Tim]

For example, the below command using "635364711800000000" as the input returns May 23rd, 2014. However, there's no finer time information such as hours, minutes, seconds, which all show 0.

>> datevec(double(datenum(1,1,0) + int64(635364711800000000) / (86400 * 10^7)))

ans =

        2014           5          23           0           0           0

[José-Luis]

My bad, didn't remember that you get rid of the fractional part when you divide an integer. Please try:

 numDays = double(int64(635364711800000000) / int64(86400 * 10^7))
 remainder = double(mod(int64(635364711800000000) , int64(86400*10^7)))
 datevec(datenum(1,1,0) +  numDays + remainder / (10^7*86400))

[Tim]

That looks great, thanks!

[José-Luis]

My pleasure.