MATLAB Answers


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

Asked by Eric Metzger on 24 Sep 2018
Latest activity Answered by Peter Perkins
on 1 Oct 2018
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?


Is the date numeric or character vector?
Are you using R2013a or earlier, or do you have an imposed requirement to use datenum? This is easier with datetime
Neither. It is a numeric scaler. I am using R2018a.
@Eric Metzger: do you need to convert one such integer, or multiple integers?

Sign in to comment.

4 Answers

Answer by Kelly Kearney
on 24 Sep 2018
 Accepted Answer

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 =
>> datenum(t)
ans =
>> datevec(t)
ans =
2018 9 24 0 0 0


Show 1 older comment
Do have one question though, what does the "%08d" do and "inputFormat"?
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.

Sign in to comment.

Answer by ANKUR KUMAR on 24 Sep 2018

If A is numeric, then
If you wish to store date in vector, then


"I want to store it as a vector [2018 09 24]"
You can never store 09 as a matrix element because it reads as 09 as 9. But if you want to store as 09, then you must to move to class char.
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.

Sign in to comment.

Answer by Walter Roberson
on 24 Sep 2018

datetime(TheScalar, 'ConvertFrom', 'yyyymmdd')
You can convert the result to datenum if you insist: just double() the datetime


Year=floor(TheScalar /10000);
Day=mod(TheScalar, 100);
Month=mod((TheScalar - Day) / 100, 100);
I would use the ymd function or the datevec function on the datetime object.
theScalar = 20180924
dt = datetime(theScalar, 'ConvertFrom', 'yyyymmdd')
[y, m, d] = ymd(dt)
DV = datevec(dt)
You can even go from the date vector back to a datetime.
dt2 = datetime(DV)
isequal(dt, dt2)
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.

Sign in to comment.

Answer by 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.


Sign in to comment.