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4 Day temperature comparison, including current day

Asked by Mark Clark on 5 Feb 2019
Latest activity Edited by Yair Altman on 6 Feb 2019
How can I include the current day in the 3 day temperature comparison.
[datetime('today')-days(0),datetime('today')], gives an error

  6 Comments

Day 0 (current day) does not start at 00:00
I can get it to work with the past 3 days, but I would also like to add the current day.
Thank You for the replies.
'DateRange' value must have an end date and time later than the start date and time.
Error message when I use [datetime('today')-days(0),datetime('today')]
Is that the full text of the error message. Usually it also states which line(s) is responsible for the error. This is essential information we need, particularly since I cannot see anywhere in your code where you use [datetime('today')-days(0),datetime('today')].
On the other hand the error message is clear. The two dates you pass must be different, with the second one later than the first. Of course, subtracting 0 days from a date doesn't change that data, so [datetime('today')-days(0),datetime('today')] is twice the same date.
It's unclear what you expected to do with that.

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4 Answers

Answer by Omer Yasin Birey on 5 Feb 2019
Edited by Omer Yasin Birey on 5 Feb 2019
 Accepted Answer

You can try to use the last minute of the current day, since it says starting date must be before than ending date
todayDate = datetime('today');
todayDate = todayDate + minutes(1439);
[datetime('today')-days(0),todayDate]

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Well actually, datetime('today') and datetime('today')-days(0) does differ. As far as I can understand OP wanted to display dates with the format of 'dd,MMMM, yyyy HH:mm:ss' and on this matter to show hours, we must write
datetime('today','Format','dd,MMMM, yyyy HH:mm:ss')
Firstly, it is a longer line and to be honest I wouldn't want that in my array it can be quite confusing. Secondly, it takes more time to process. While this line is taking 0.017166 seconds, days(0) version takes only 0.003238 seconds. It is over 5 times faster.
Well ok, it does create a datetime with a different display format. If that's the whole purpose of subtracting 0, then I would say it's a bad idea. First, you're relying on undocumented behaviour, so that may change in a future release without notice. Secondly, if you want a specific format you'd be better be explicit about the format you want. So even though, 'Format', 'dd, MMM, yyyy HH:mm:ss' might be longer, it's at least very clear what the intent of the code is. Subtracting 0 just to alter the format, with no comment explaining what the purpose is, is just asking for a bug to be introduced later on when a different maintainer notice that 0 subtraction and remove it.
In this particular case, where the array is just passed to another function, the format does not matter at all.
I echo Guillaume's comment.
However, just for the record, it is typically faster to subtract a pure number rather than days(number). For example,
datetime('today')-1
is faster than
datetime('today')-days(1)
and yet has the same meaning/result. Just note that days(1) might be more readable.

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Answer by KSSV
on 5 Feb 2019

thedates = (datetime('today')-days(3):datetime('today'))'

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Answer by Mark Clark on 6 Feb 2019

This is what Im trying to do. The today trace in the graph below.
CaptureAGT.PNG

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Answer by Mark Clark on 6 Feb 2019

See https://thingspeak.com/channels/569794

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