binary vector that consists of a zero for normal years and a 1 for a leap year

I need to create a vector that consists of a binary vector that consists of a zero for normal years and a 1 for a leap year, with some data i already have. How would i do this?

[Image Analyst]

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[Dyuman Joshi]

I'm sure you can find the logic to check for a leap year online easily, or better if you know it already.

Try to apply that the same in MATLAB. If you need help in doing that, as Image Analyst has mentioned, take the free training course.

If after attempting to write the logic, you are stuck, mention a specific problem/error you are facing.

[Jakub]

ive already completed that 2 hour training. i know what vectors are and how to create them but i dont know how to implement it into the data set of 9000 days.

[John D'Errico]

@Jakub Reemmber that we don't know at all what skills you have. But when you ask a question that seems simple, we must assume you know nothing about MATLAB.

Do you know how to compute the year, for each date?

Do you know the formula to identify a leap year, given the numbered year? How would you use that with the mod function, for example, to identify leap years?

Answer 1

[Star Strider]

Use the eomday function —

YearVct = 1950:2050;
Lv = eomday(YearVct,2) == 29                                    % Create Logical Vector

Lv = 1×101 logical array
   0   0   1   0   0   0   1   0   0   0   1   0   0   0   1   0   0   0   1   0   0   0   1   0   0   0   1   0   0   0   1   0   0   0   1   0   0   0   1   0   0   0   1   0   0   0   1   0   0   0   1   0   0   0   1   0   0   0   1   0   0   0   1   0   0   0   1   0   0   0   1   0   0   0   1   0   0   0   1   0   0   0   1   0   0   0   1   0   0   0   1   0   0   0   1   0   0   0   1   0   0

LeapYears = YearVct(Lv)                                         % Leap Years

LeapYears = 1×25

1952 1956 1960 1964 1968 1972 1976 1980 1984 1988 1992 1996 2000 2004 2008 2012 2016 2020 2024 2028 2032 2036 2040 2044 2048

LeapYearsMtx = buffer(LeapYears,5)

LeapYearsMtx = 5×5

1952 1972 1992 2012 2032 1956 1976 1996 2016 2036 1960 1980 2000 2020 2040 1964 1984 2004 2024 2044 1968 1988 2008 2028 2048

EDIT — (1 Jan 2023 at 17:15)

This is based on the eomday Find Leap Years documentation section.

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