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

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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.
  1 Comment
Siddharth Bhutiya
Siddharth Bhutiya on 30 Dec 2021
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.

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Accepted Answer

Stephen23
Stephen23 on 18 Dec 2021
Edited: Stephen23 on 18 Dec 2021
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
  2 Comments

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More Answers (2)

madhan ravi
madhan ravi on 17 Dec 2021
doc ymd
doc hms

Steven Lord
Steven Lord on 17 Dec 2021
% 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
  2 Comments
Stephen23
Stephen23 on 18 Dec 2021
Edited: Stephen23 on 18 Dec 2021
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.

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