Convert time vector of Year, Month, Day, Hours, Minute to Decimal format

Hi everyone;

Suppose I have the following time vector:

....start from January

2019 10 02 12:00:00

2019 10 03 12:00:00

2019 10 04 12:00:00

2019 10 05 12:00:00

2019 10 06 12:00:00

2019 10 07 12:00:00

2019 10 08 12:00:00

2019 10 09 12:00:00

2019 10 10 12:00:00

2019 10 11 12:00:00

2019 10 12 12:00:00

2019 10 13 12:00:00

.....

I want to convert this time vector to decimal format, so it become:

2019,xxxx

2019,xxxx

2019,xxxx

etc

Any idea to do it, really appreciate it.

[Siddharth Bhutiya]

May I ask what you are planning to do with the (2019.xxxx) once you have calculated that? Because when you have a datetime object, it knows what operations can and cannot be done on a datetime. When you convert that into a double 2019.xxxx, that knowledge goes away. This could cause unexpected result if you are not careful later on in your code. For example, if you accidentally add 1 to it it changes the year from 2019 to 2020. You could multiply it with anything because its a double now, however, multiplication would not make sense for datetimes. So it would be helpful to know what you intend to do with those values.

Answer 1

[Stephen23]

Simpler:

format long G
D = datetime(2019,10,(2:13).',12,0,0)

D = 12×1 datetime array
   02-Oct-2019 12:00:00
   03-Oct-2019 12:00:00
   04-Oct-2019 12:00:00
   05-Oct-2019 12:00:00
   06-Oct-2019 12:00:00
   07-Oct-2019 12:00:00
   08-Oct-2019 12:00:00
   09-Oct-2019 12:00:00
   10-Oct-2019 12:00:00
   11-Oct-2019 12:00:00
   12-Oct-2019 12:00:00
   13-Oct-2019 12:00:00

B = dateshift(D, 'start', 'year'); % midnight at start of the year
E = B + calyears(1); % midnight at the end of the year (do not use DATESHIFT)
Y = year(D);
Y = Y + (D-B)./(E-B)

Y = 12×1

1.0e+00 * 2019.75205479452 2019.75479452055 2019.75753424658 2019.7602739726 2019.76301369863 2019.76575342466 2019.76849315068 2019.77123287671 2019.77397260274 2019.77671232877

Lets check the first fraction by hand:

between(B(1),D(1),'Time') % hours from start of year to midday 2nd Oct.

ans = calendarDuration
   6588h 0m 0s

between(B(1),E(1),'Time') % total hours in the year

ans = calendarDuration
   8760h 0m 0s

6588/8760

ans = 
         0.752054794520548

[Stephen23]

Another approach is to download DATESTR8601 (its output also confirms the values shown in my answer):

format long G
D = datetime(2019,10,(2:13).',12,0,0);
S = arrayfun(@(d)datestr8601(d,'y11'),D,'uni',0)

S = 12×1 cell array
    {'2019.75205479452'}
    {'2019.75479452054'}
    {'2019.75753424657'}
    {'2019.76027397260'}
    {'2019.76301369863'}
    {'2019.76575342465'}
    {'2019.76849315068'}
    {'2019.77123287671'}
    {'2019.77397260274'}
    {'2019.77671232876'}
    {'2019.77945205479'}
    {'2019.78219178082'}

Note that it returns character vectors, not numerics.

PS: in case anyone was curious, the dates I gave in my example here are actually:

datestr8601([2020,12,31,12,0,0],'y11')

ans = '2020.99863387978'

datestr8601([2021,1,1,6,11,34],'y11')

ans = '2021.00070693810'

Note that this function is now quite old and relies internally on datevectors and datenumbers, so is less precise than calculating with datetime objects.

[Marzuki Marzuki]

Thank you @Stephen it works

Answer 2

[madhan ravi]

 doc ymd
 doc hms

Answer 3

[Steven Lord]

% Sample data
d = datetime(2019, 10, (2:6).', 12, 0, 0)

d = 5×1 datetime array
   02-Oct-2019 12:00:00
   03-Oct-2019 12:00:00
   04-Oct-2019 12:00:00
   05-Oct-2019 12:00:00
   06-Oct-2019 12:00:00

startOfYear = dateshift(d, 'start', 'year')

startOfYear = 5×1 datetime array
   01-Jan-2019
   01-Jan-2019
   01-Jan-2019
   01-Jan-2019
   01-Jan-2019

p = years(d-startOfYear)

p = 5×1

0.7516 0.7543 0.7570 0.7598 0.7625

% Use that year fraction data
corresponding = datetime(2021, 1, 1) + years(p)

corresponding = 5×1 datetime array
   02-Oct-2021 12:00:00
   03-Oct-2021 12:00:00
   04-Oct-2021 12:00:00
   05-Oct-2021 12:00:00
   06-Oct-2021 12:00:00

[Marzuki Marzuki]

Thank you @Steven Lord but the desired result 2019.xx etc..

[Stephen23]

Note that by using YEARS this approach incorrectly assumes that a year has exactly

format long G
365.2425 * 24 % hours

ans = 
                   8765.82

whereas in fact the length of a calendar year is a) never equal to this value and b) not constant. The bug in this approach can be clearly demonstrated at the end of a leap year, where this calculation returns midday on december the 31st as a fraction greater than one:

d = datetime(2020, 12, 31, 12, 0, 0) % leap year!

d = datetime
   31-Dec-2020 12:00:00

startOfYear = dateshift(d, 'start', 'year')

startOfYear = datetime
   01-Jan-2020

d-startOfYear % correct

ans = duration
   8772:00:00

p = years(d-startOfYear) % flawed

p = 
           1.0007050110543

Adding this fraction to a year would give an ambiguous numeric value which can represent multiple dates, e.g. it would be impossible to tell the difference between 2020-12-31T12:00:00 and 2021-01-01T06:11:34.