4 Day temperature comparison, including current day
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How can I include the current day in the 3 day temperature comparison.
[datetime('today')-days(0),datetime('today')], gives an error
6 个评论
Omer Yasin Birey
2019-2-5
Hi Mark, in this array you only have current day twice. You should explain more for us to understand.
Guillaume
2019-2-5
[datetime('today')-days(0),datetime('today')] gives an error
It doesn't for me. If you get an error, always give us the full text of the error message so we now what the problem is (for all we know you're using an old version of matlab without datetime).
And yes, as Omer said, give more details of what you want.
Mark Clark
2019-2-5
Mark Clark
2019-2-5
Mark Clark
2019-2-5
Guillaume
2019-2-5
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.
采纳的回答
Omer Yasin Birey
2019-2-5
编辑：Omer Yasin Birey
2019-2-5
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]
4 个评论
Guillaume
2019-2-5
Subtracting 0 (days or anything) never does much, so I really don't see the point of datetime('today') - days(0). You may as well write datetime('today') and avoid puzzling the reader why you're subtracting nothing.
Assuming you're not running the code at exactly midnight
[datetime('today'), datetime]
will give you two different datetime on the same day. No idea if that's what desired.
Omer Yasin Birey
2019-2-6
编辑：Omer Yasin Birey
2019-2-6
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.
Guillaume
2019-2-6
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.
Yair Altman
2019-2-6
编辑：Yair Altman
2019-2-6
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.
更多回答（3 个）
KSSV
2019-2-5
thedates = (datetime('today')-days(3):datetime('today'))'
Mark Clark
2019-2-6
0 个投票
This is what Im trying to do. The today trace in the graph below.
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