Datestr problem with Hour

Question

Harold Kidder 2014-6-15

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/134542-datestr-problem-with-hour

评论： dpb 2014-6-16

采纳的回答： Star Strider

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I wrote a test routine to set a start time stamp and step through hours.

function HourlyDateTEST2()

   dvec(1)=2014;
   dvec(2)=06;
   dvec(3)=15;
   dvec(4)=21;
   dvec(5)=00;
   dvec(6)=00;
   dateVal=datenum(dvec);
   for ii = 1:7
       datestr(dateVal, 'mm/dd/yyyy HH:MM:SS')
       dateVal = dateVal + 1/24;
   end;
end

The output looks like the following:

 06/15/2014 21:00:00
 06/15/2014 22:00:00
 06/15/2014 23:00:00
 06/15/2014 00:00:00
 06/16/2014 01:00:00
 06/16/2014 02:00:00
 06/16/2014 03:00:00

Notice that the 4th line has the wrong date. Should be 06/16/2014 00:00:00.

What am I doing wrong?

I am using MATLAB R2014a.

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Answer 1

Star Strider 2014-6-15

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https://ww2.mathworks.cn/matlabcentral/answers/134542-datestr-problem-with-hour#answer_141026

That’s easiest with the addtodate function.

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Harold Kidder 2014-6-16

在 MATLAB Online 中打开

My application requires a loop. I tried two approaches:

Approach 1 -- Offset from start date.

 function HourlyDateTEST2()
 dvec(1) = 2014;
 dvec(2) = 06;
 dvec(3) = 15;
 dvec(4) = 21;
 dvec(5) = 00;
 dvec(6) = 00;
 dateStart = datenum(dvec);
 for ii = 0:6
     dateVal = addtodate(dateStart, ii, 'hour');
     datestr(dateVal, 'mm/dd/yyyy HH:MM:SS')
 end;
 end

Result:

 06/15/2014 21:00:00
 06/15/2014 22:00:00
 06/15/2014 23:00:00
 *06/16/2014 00:00:00*
 06/16/2014 01:00:00
 06/16/2014 02:00:00
 06/16/2014 03:00:00

Approach 2 – Update dateVal on each loop

 function HourlyDateTEST3()
 dvec(1) = 2014;
 dvec(2) = 06;
 dvec(3) = 15;
 dvec(4) = 21;
 dvec(5) = 00;
 dvec(6) = 00;
 dateVal = datenum(dvec);
 for ii = 0:6
     datestr(dateVal, 'mm/dd/yyyy HH:MM:SS')
     dateVal = addtodate(dateVal, 1, 'hour');    
 end;
 end

Result:

 06/15/2014 21:00:00
 06/15/2014 22:00:00
 06/15/2014 23:00:00
 *06/15/2014 00:00:00*
 06/16/2014 01:00:00
 06/16/2014 02:00:00
 06/16/2014 03:00:00

Approach 1 worked but Approach 2 gave the same result as using dateVal + 1/24. Must still be some round off error.

I'll use Approach 1.

dpb 2014-6-16

编辑：dpb 2014-6-16

在 MATLAB Online 中打开

The error will only be ~1E-15 or so on the datenum; since the day in the whole part, if truncate ( fix instead of round as example), the day would still be the preceding; only when the fractional part is 0+ instead of 9- will the whole portion have yet turned over. So, the datenum value isn't off by a whole day in absolute value, only that it hasn't yet turned over the new leaf.

I've mentioned earlier I've had cases where I could see the effects when using floating point outside the various functions and then utilizing those values inside. Your case here is the only time I've ever observed it within where up until now the rounding has never shown up in anything I've ever caught it at, anyway.

I was going to demonstrate the value difference but interestingly this morning I can't duplicate it...I just reran OPs test here and received--

>> for ii = 1:7
datestr(dateVal, 'mm/dd/yyyy HH:MM:SS')
dateVal = dateVal + 1/24;
end
ans =
06/15/2014 21:00:00
ans =
06/15/2014 22:00:00
ans =
06/15/2014 23:00:00
ans =
06/16/2014 00:00:00
...

Now that one I can't explain...that's Twilight Zone stuff, there.

ADDENDUM

Just for grins I closed and restarted Matlab and can't reproduce it there, either. Yet last night got the OPs results. Wishing now had save the intermediaries then but they were already in the bit bucket....

MOST peculiar...

Star Strider 2014-6-16

I reported this a a bug to TMW yesterday. I got an e-mail a few minutes ago that said that they were able to reproduce it in R2014a, and will fix it â€˜in a future releaseâ€™. Whether this means an update or wait until R2014b Iâ€™m not sure. I sent them the link to this thread, so they have all the information they need.

Congratulations to Harold for discovering it!

dpb 2014-6-16

I'd still be curious if you can generate the two cases to see the actual datenum values. You running 32- or 64-bit version?

I'm still greatly puzzled how can't now seem to reproduce it here when was certain saw it in the original loop version. I suppose I could have mistakenly thought I saw what I didn't.

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Answer 2

dpb 2014-6-15

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/134542-datestr-problem-with-hour#answer_141013

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What am I doing wrong?

for ii = 1:7
  datestr(dateVal, 'mm/dd/yyyy HH:MM:SS')
  dateVal = dateVal + 1/24;
  ...

Using floating point arithmetic; the rounding in the accumulation of the 1/24 factor caused it.

Use

>> datestr(datenum(dvec(1),dvec(2),dvec(3),dvec(4)+[0:7].',dvec(5),dvec(6)))
ans =
15-Jun-2014 21:00:00
15-Jun-2014 22:00:00
15-Jun-2014 23:00:00
16-Jun-2014 00:00:00
16-Jun-2014 01:00:00
16-Jun-2014 02:00:00
16-Jun-2014 03:00:00
16-Jun-2014 04:00:00
>>

instead (add the integer hours, not the fractional days) to avoid rounding errors.

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Datestr problem with Hour

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