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# Creating List of Dates, Stepping by a Month6

Posted by Loren Shure,

Ever need to create a vector of dates using some sort of pattern? Perhaps these will be used as the edges argument for a histogram, with each a month.

What's the best way to create a datetime array where each element is the first of the month? And what does "best" even mean? - fewest keystrokes, fewest function calls, most readable, most flexible, most maintainable,etc. Suppose I want to produce something like this:

   2018-Jan-01, 2018-Feb-01, 2018-Mar-01, ...

In fact, there are lots of suitable ways. Here are a few.

### Contents

#### Simply Use datetime

Just use datetime and specify the month vectors. This works

mydates1 = datetime(2018,1:12,1)

mydates1 =
1×12 datetime array
Columns 1 through 5
01-Jan-2018   01-Feb-2018   01-Mar-2018   01-Apr-2018   01-May-2018
Columns 6 through 10
01-Jun-2018   01-Jul-2018   01-Aug-2018   01-Sep-2018   01-Oct-2018
Columns 11 through 12
01-Nov-2018   01-Dec-2018


While I'm at it, I can find the number of days in each month I have, using eomday.

numDays = eomday(year(mydates1),month(mydates1));
bar(mydates1,numDays)


#### What if Months Span Years?

If the months in question span years, you can do this fairly compactly using calmonths.

mydates2 = datetime(2017,1,1):calmonths(1):datetime(2018,7,1)

mydates2 =
1×19 datetime array
Columns 1 through 5
01-Jan-2017   01-Feb-2017   01-Mar-2017   01-Apr-2017   01-May-2017
Columns 6 through 10
01-Jun-2017   01-Jul-2017   01-Aug-2017   01-Sep-2017   01-Oct-2017
Columns 11 through 15
01-Nov-2017   01-Dec-2017   01-Jan-2018   01-Feb-2018   01-Mar-2018
Columns 16 through 19
01-Apr-2018   01-May-2018   01-Jun-2018   01-Jul-2018


In addition, you can go beyond 12 months and datetime still works as expected, without needing calmonths.

mydates3 = datetime(2018,1:24,1);


or

mydates4 = datetime(2017,7:18,1)

mydates4 =
1×12 datetime array
Columns 1 through 5
01-Jul-2017   01-Aug-2017   01-Sep-2017   01-Oct-2017   01-Nov-2017
Columns 6 through 10
01-Dec-2017   01-Jan-2018   01-Feb-2018   01-Mar-2018   01-Apr-2018
Columns 11 through 12
01-May-2018   01-Jun-2018


However keeping track of the relative month shifts to start has its own mental overhead for me.

#### Another Way Using calmonths

Try this instead. Find the right start date, and then add on the correct number of calmonths from there.

mydates5 = datetime(2017,7,1) + calmonths(0:11)

mydates5 =
1×12 datetime array
Columns 1 through 5
01-Jul-2017   01-Aug-2017   01-Sep-2017   01-Oct-2017   01-Nov-2017
Columns 6 through 10
01-Dec-2017   01-Jan-2018   01-Feb-2018   01-Mar-2018   01-Apr-2018
Columns 11 through 12
01-May-2018   01-Jun-2018


Do you have another way you like to produce lists of dates? What's your preference for these sorts of situations? Let me know here.

Get the MATLAB code

Published with MATLAB® R2018b

Gareth Thomas replied on : 1 of 6
Great post Loren, it is wonderful to see the flexibility and the power of datetime and the other functions around it. Personally I tend to use the mydates5 = datetime(2017,7,1) + calmonths(0:11). The eomday function was new to me and I had to look it up to see what the eomday stands for... End Of Month DAY. Knowing that makes it easier to remember, at least for me:)
Rob replied on : 2 of 6
I often work in day-of-year (how my source data appears) and often use a similar trick, i.e. day-of-year 50 is Feb 19th, but to put it in datenum or datetime: datenum(2018,1,50) or datetime(2018,1,50). I chose day-of-year 50 as it's before any leap-day issues, just for clarity, but this works for leap years as well, i.e. datetime(year,1,DayOfYear) As per your example, this also works with ranges of days, say I want every 6th day from day-of-year 50 to day-of-year 121.
datenum(2018,1,50:6:121)   or datetime(2018,1,50:121)

Great. Now we hit a difference. Say I want the same date range, but every hour. an hour is 1/24th of a day, so in datenum I can do datenum(2018,1,50:(1/24):121). This works. However, this is NOT allowed in datetime, since all input values must be integers (which is not the case with datenum), so the datetime solution would be: datetime(2018,1,1,(50*24):1:(121*24),0,0). You can do similar tricks with minutes and seconds too.
Personally - I still use datenum for nearly everything time related. I have to deeply care about leap-seconds for my work, and although Matlab has had a valiant attempt at dealing with leap seconds, I don't have an easy way to know how up to date it's leap-second list is which ultimately makes it useless to me. Since I share code with others, I don't know what version of Matlab they are using, so it's difficult to code in a way to check how many leap seconds their version of Matlab is aware of. Mathworks - this needs to be addressed if you want use to trust Matlab time with leap second correction.
However, one big advantage for me of datetime is converting UTC strings to datetime (and then datenum). My UTC strings are often of the form 2019-050T00:00:00:00.000 (day-of year format) or 2019-02-19T00:00:00.000 (year-month-day format). Converting that to a datenum value was computationally expensive when I've thousands of them in a char array (n x 21 for the day-of-year former version). e.g.
%TIME = char array of n by 21 for UTC time in day of year (ddd)
TIME(:,[5 9 12 15])=' '; % Remove the - T : from yyyy-dddTHH:MM:SS.sss
TIME = str2num(TIME);
T = datenum(TIME(:,1)-1,12,31+TIME(:,2),TIME(:,3),TIME(:,4),TIME(:,5)); % note that the 6th argument (Seconds) can be decimal to allow for milliseconds.

However, with datetime it's much quicker! MUCH MUCH quicker. Weirdly converting a UTC time string to a datenum was my biggest time-sink in my code. Anyway - my quick way was:
T = datenum(datetime(TIME,'Format','uuuu-DDD''T''HH:mm:ss.SSS'));
e.g.
T = datenum(datetime('2019-034T01:23:45.678','Format','uuuu-DDD''T''HH:mm:ss.SSS'));

If there's a quicker way to do this I'd love to know it!
My two time questions for the room: 1) Is there an easy (1-line) way to find out the last leap-second that Matlab version datetime is aware of? 2) Is there a way to find out what time-zone the machine is set to without using datetime?
For the latter, I can use datetime to figure this out, but if someone I've shared my code with has an older version of Matlab that does not have datetime yet (and there are a few who can't afford to upgrade), is there anyway I can figure out their timezone from Matlab using datenum or now or such?
Loren Shure replied on : 3 of 6
@Gareth- Good to know this helped. @Rob- I don't know about leap seconds in detail. Will try to find you the answer. --loren
Paul Shoemaker replied on : 4 of 6
Rob, For your #2 question, there is a FEX submission by Erwin Mayer that makes use of Java instead of datetime that could be helpful: https://www.mathworks.com/matlabcentral/fileexchange/27953-convert-between-world-time-zones-with-daylight-saving-times Using the above as a resource, take a look at the code snippet below and see if it gets you what you need:
import java.lang.String
import java.util.* java.awt.*
import java.util.Enumeration

myTime = java.util.GregorianCalendar();
get(myTime.getTimeZone())
Cheers, Paul MatlabInvesting.com
Peter Perkins replied on : 5 of 6
Rob, I don't know what to say to get you more comfortable with using datetime over datenum, other than to say that datetime addresses many of the shortcomings of datenum, notably no support for time zones, lower precision, and round-off when representing most points in time. If leap seconds are important to you, I don't understand how using datenum can suit your needs unless you have code that you have written that deals with them. We make sure every release is up to date with leap seconds as of its release date, so if you are on the current version, you should be ok. A sort of round-about way to confirm that is to subtract 1-Jan-1970 from the current date using the UTCLeapSeconds time zone. But I hear what you are saying about older versions and being able to get an explicit list. That's something we are actively working on. To create the sequence you want, do this:
   datetime(2018,1,50):hours(1):datetime(2018,1,122)

Fractional days have no meaning in most time zones. Perhaps you don't care about time zones, but that's why the datetime function will not accept fractional days.
Paul Shoemaker replied on : 6 of 6
Rob, for your #2 question, try the below code and see if it gets you what you need. It's inspired by a FEX submission by Erwin Mayer
import java.lang.String
import java.util.* java.awt.*
import java.util.Enumeration
myTime = java.util.GregorianCalendar();
get(myTime.getTimeZone());