How to avoid "NaT" when converting a cell into datetime ? (Or: how to add missing parts of datetime ?)

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How to avoid "NaT" when converting a cell into datetime ?
(Or: how to add missing parts of datetime ?)
a = [
{'14-Jul-2021 17:00:00'}
{'15-Jul-2021 12:00:00'}
{'16-Jul-2021 21:00:00'}
{'17-Jul-2021' }
{'18-Jul-2021 06:00:00'}
{'19-Jul-2021 04:00:00'}
{'20-Jul-2021 17:00:00'}
{'21-Jul-2021 08:00:00'}]
a = 8×1 cell array
{'14-Jul-2021 17:00:00'} {'15-Jul-2021 12:00:00'} {'16-Jul-2021 21:00:00'} {'17-Jul-2021' } {'18-Jul-2021 06:00:00'} {'19-Jul-2021 04:00:00'} {'20-Jul-2021 17:00:00'} {'21-Jul-2021 08:00:00'}
datetime(a)
ans = 8×1 datetime array
14-Jul-2021 17:00:00 15-Jul-2021 12:00:00 16-Jul-2021 21:00:00 NaT 18-Jul-2021 06:00:00 19-Jul-2021 04:00:00 20-Jul-2021 17:00:00 21-Jul-2021 08:00:00
As workaround, how to add a representative hour (e.g. '12:00:00') where it is missing, in order to avoid a "NaT" ?

Accepted Answer

Stephen23
Stephen23 on 12 Aug 2022
Edited: Stephen23 on 12 Aug 2022
"How to avoid "NaT" when converting a cell into datetime ?"
Convert each cell individually. Certainly not as efficient as one DATETIME call, but it is fairly simple:
C = {...
'14-Jul-2021 17:00:00'
'15-Jul-2021 12:00:00'
'16-Jul-2021 21:00:00'
'17-Jul-2021'
'18-Jul-2021 06:00:00'
'19-Jul-2021 04:00:00'
'20-Jul-2021 17:00:00'
'21-Jul-2021 08:00:00'}; % one cell array :)
D = cellfun(@datetime,C)
D = 8×1 datetime array
14-Jul-2021 17:00:00 15-Jul-2021 12:00:00 16-Jul-2021 21:00:00 17-Jul-2021 00:00:00 18-Jul-2021 06:00:00 19-Jul-2021 04:00:00 20-Jul-2021 17:00:00 21-Jul-2021 08:00:00
  5 Comments
Stephen23
Stephen23 on 12 Aug 2022
"but the thing is that the position can change"
Of course, you can use any indexing you want, you are not restricted to just using index 4. For example, if you want to detect all character vectors shorter than 12 characters and add 12 hours to the corresponding datetime:
C = {...
'14-Jul-2021 17:00:00'
'15-Jul-2021 12:00:00'
'16-Jul-2021 21:00:00'
'17-Jul-2021'
'18-Jul-2021 06:00:00'
'19-Jul-2021 04:00:00'
'20-Jul-2021 17:00:00'
'21-Jul-2021 08:00:00'}; % one cell array :)
D = cellfun(@datetime,C);
X = cellfun(@numel,C)<12; % or whatever logic you want to use (which you did not explain)
D(X) = D(X) + hours(12)
D = 8×1 datetime array
14-Jul-2021 17:00:00 15-Jul-2021 12:00:00 16-Jul-2021 21:00:00 17-Jul-2021 12:00:00 18-Jul-2021 06:00:00 19-Jul-2021 04:00:00 20-Jul-2021 17:00:00 21-Jul-2021 08:00:00
Sim
Sim on 12 Aug 2022
Edited: Sim on 12 Aug 2022
Cool, thanks!! :-)
The logic of indexing the missing parts of datetime elements by using the "characters length of those datetime elements" (i.e. 11 characters in '17-Jul-2021'), works well in my case, many thanks !

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