plot time values, datetick

Hi,

I have been struggeling with this all night.

i have two columns of data: column 1 (HH:MM:SS), column 2 (random values) below an example

colmn 1:

09:04:21

09:04:36

09:04:51

09:05:06

09:05:51

09:06:06

colmn 2:

1

2

3

4

5

6

now i want to plot these two colmns as they are, so time on the x-as and the other values on the y-as. i was able to do this with datetick the only problem is that i have a large amount of data measuring for 3 days and so "the time will repeat" and data will stack on eachother so to say the x-as will only go from 00:00:00 to 00:00:00 (only 24 hours)

could it be possible that the time will just go on?

hope someone can help many thanks!

gr Paul

Answer 1

[dpb]

Add an arbitrary start yr/mo/day field for input to datenum. Or, if the timestamps are all precisely 15 second intervals you can simply create a 3-day long datenum as

>> dn=datenum(2015,1,1,9,4,21+[0:15:3*24*60*60].'); 
>> datestr(dn(1))
ans =
01-Jan-2015 09:04:21
>> datestr(dn(end))
ans =
04-Jan-2015 09:04:21
>>

[paul kaam]

Thanks you for your answer!

unfortunately it is not precisely 15 seconds, i found out. So i have to use the as you state arbitrary start yr/mo/dy. I tried this today and i couldn't get it to work

could you maybe help me out as to how i should do it?

i suppose it would be something like this: dn = datenum(2015,1,25,[column1]);

many thanks

gr Paul

[dpb]

Bummer... :) That does make it a little more complex but it's still not too bad.

What you have to do here then is multi-step process...

1. convert the timestamps to date numbers so are sequential,
2. compute the difference to find the roll-over locations that indicate the beginning of next day,
3. add one for each day between each section.

As example, from the list as I created above--

>> dn=datenum(2015,1,1,9,4,21+[0:15:3*24*60*60].');
>> dn=dn-fix(dn);  % convert to the time fraction only 
>> newDayIdx=find(diff(dn)<0)+1  % find the day rollover points
newDayIdx =
      3584
      9344
     15104
>> fprintf('%5d %f\n',[(3582:3586).' dn(3582:3586)].')
3582 0.999722
3583 0.999896
3584 0.000069
3585 0.000243
3586 0.000417
>>

So, you need to set the day 0 between 1:newDayIdx(1)-1, day 1 from newDayIdx(1):newDayIdx(2)-1, etc., ... then convert that adjusted vector.

NB: The "+1" in the index is owing to the fact the diff() vector of time differences is one shorter than the original and is missing the first location as dt(1)<--t(2)-t(1)