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

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Eric Metzger
Eric Metzger 2018-9-24
回答： Peter Perkins 2018-10-1
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?
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Stephen23
Stephen23 2018-9-24
@Eric Metzger: do you need to convert one such integer, or multiple integers?

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采纳的回答

Kelly Kearney
Kelly Kearney 2018-9-24
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
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Kelly Kearney
Kelly Kearney 2018-9-24
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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ANKUR KUMAR
ANKUR KUMAR 2018-9-24
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)})]
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Eric Metzger
Eric Metzger 2018-9-24
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 2018-9-24
datetime(TheScalar, 'ConvertFrom', 'yyyymmdd')
You can convert the result to datenum if you insist: just double() the datetime
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Steven Lord
Steven Lord 2018-9-24
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)
Walter Roberson
Walter Roberson 2018-9-24
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 2018-10-1
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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