How to use datenum with a date represented as a scaler?

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I have data I need to ingest that is date specific however, the data is represented as a scaler e.g. 20180924 with no spaces or hyphens. How do I get datenum to take this date in and separate it out into a vector so I can use it?
  3 Comments
Stephen23
Stephen23 on 24 Sep 2018
@Eric Metzger: do you need to convert one such integer, or multiple integers?

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Accepted Answer

Kelly Kearney
Kelly Kearney on 24 Sep 2018
Is your final goal to convert to a datenumber or an year-month-day array? As mentioned in the other answers, if you're using a recent version of Matlab, datetimes are more flexible than datenumbers. Once you've converted the value to a datetime, you can move between datenumbers and datevectors pretty easily:
x = [20180924];
>> t = datetime(num2str(x, '%08d'), 'inputFormat', 'yyyyMMdd')
t =
datetime
24-Sep-2018
>> datenum(t)
ans =
737327
>> datevec(t)
ans =
2018 9 24 0 0 0
  4 Comments
Kelly Kearney
Kelly Kearney on 24 Sep 2018
Specifically in the this case, it makes sure the answer is robust to dates earlier than year 1000 by padding any less-than-4-digit years with zeros. (Datetime is actually smart enough to figure things out even without the padding... but I like to be explicit just in case.) Note that this will fail if you're dealing with any BC dates, but I'm assuming you would have specified if that was the case.

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More Answers (3)

ANKUR KUMAR
ANKUR KUMAR on 24 Sep 2018
If A is numeric, then
A=20180924
a1=num2str(A)
datenum(str2double({a1(1:4),a1(5:6),a1(7:8)}));
If you wish to store date in vector, then
A1=[str2double({a1(1:4),a1(5:6),a1(7:8)})]
  6 Comments
Eric Metzger
Eric Metzger on 24 Sep 2018
Okay, this will break it up into three separate cell, but I get this error when I try to use datenum on it:
A1 =
1×3 cell array
{'2018'} {'09'} {'24'}
>> datenum(A1) Error using datenum (line 189) DATENUM failed.
Caused by: Error using datevec (line 281) Cannot parse date 2018.
I am not use to working with the {} brackets for arrays/matrices but the [] e.g [2018 9 24 18 15 56] where it is the yyyy mm dd hh mm ss. The "three cell array" is not working the way I normally use it. This is close to not quite the cigar yet.

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Walter Roberson
Walter Roberson on 24 Sep 2018
datetime(TheScalar, 'ConvertFrom', 'yyyymmdd')
You can convert the result to datenum if you insist: just double() the datetime
  3 Comments
Walter Roberson
Walter Roberson on 24 Sep 2018
It isn't clear what result is being looked for. If it is separated parts, with the implementation of sprintf -> str2num -> datenum -> datevec being considered, then my numeric code computes the pieces directly without needing any calls more expensive then mod.

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Peter Perkins
Peter Perkins on 1 Oct 2018
If possible, use datetime rather than datenum/datestr/datevec. As Walter showed, you can convert directly from numeric to datetime. It's also possible that you do not need to store the separate pieces, since datetime lets you get at them any time you want, as a property.

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