Calculating the Day of the Year

2012-04-19

In summary:

Given a date, the day of the year can be calculated:

var day = Math.floor((367 * month - 362) / 12) + date -
	((month > 2) ? (2 - isLeapYear) : 0);

Where month is 1 - January, 2 - February, etc and day_of_year is 1 - 1st Jan, 2 - 2nd Jan, etc

Introduction

If you are given any date, the number of days since the beginning of the year can be calculated. This value will range between 1 and 366:

Date	Day of Year
1 January	1
2 January	2
31 January	31
1 February	32
2 February	33
28 February	59
29 February	60 (if day exists)
1 March	60 (or 61 for leap years)
31 December	365 (or 366 for leap years)

Linear Equation Approximation

First, we will pretend that February has 30 days, instead of 28 or 29 days. Later we will remove these extra one or two days, if necessary. Now, if we calculate the number of days that have passed in all previous months, which we will call 'cumulative days':

Month Name	Month index (x)	Days in Month	Cumulative Days (y)
January	0	31	0
February	1	30 (for now!)	31
March	2	31	61
April	3	30	92
May	4	31	122
June	5	30	153
July	6	31	183
August	7	31	214
September	8	30	245
October	9	31	275
November	10	30	306
December	11	31	336

Graphing the cumulative days (y) vs. the month index (x):

This curve is pretty linear (a straight line), and the slope of the line can be approximated as the total number of days in the year (367, not 365 because we are pretending that February has 30 days) divided by the number of months in the year (12).

The difference between the actual cumulative days (y) and the approximation of 30.58*x can be shown as:

Month Index (x)	Cumulative Days (y)	30.58 * x	y - (30.58 * x)
0	0	0	+0.00
1	31	30.58	+0.42
2	61	61.16	−0.16
3	92	91.74	+0.26
4	122	122.32	−0.32
5	153	152.90	+0.10
6	183	183.48	−0.48
7	214	214.06	−0.06
8	245	244.64	+0.36
9	275	275.22	−0.22
10	306	305.80	+0.20
11	336	336.38	−0.38

If we round the approximate equation to the nearest integer, the approximate equation then exactly calculates the cumulative days:

var y = Math.round(30.58 * x);

It will be useful later on to use a floor() function instead of a round() function:

var y = Math.floor(30.58 * x + 0.5);

In this equation, x is the Month Index, which has the values (0, 1, 2, …, 11). If we prefer to use conventional month numbers (1, 2, 3, …, 12), we need to shift x by 1:

var y = Math.floor(30.58 * (month - 1) + 0.5);

It is interested to observe that 30.58 is not the only value that will work in this equation. In fact, using any number between 30.5625 and 30.5833 (between 489/16 and 367/12) will result in the correct value.

Adding the Days for the Current Month

Now we add the number of days in the current month:

var day = y + date;

Compensating for February

Finally, if the month is March or later we need to remove our extra one or two days to account for February having fewer days:

if (month > 2)
{
	if (isLeapYear) {
		day = day - 1;
	} else {
		day = day - 2;
}

This can also be written in ternary form:

day = day - ((month > 2) ? (2 - isLeapYear) : 0);

Putting It Together

Combined together the expression for the day of the year is:

day = Math.floor(30.58 * (month - 1) + 0.5) + date -
	((month > 2) ? (2 - isLeapYear) : 0);

Using Integer Mathematics

Unlike JavaScript, some programming languages have different data types for integers and floating point variables. It is useful to write the algorithm so that it can work with integer variables and not require floating point calculations:

var day = Math.floor((367 * month - 362) / 12) + date -
	((month > 2) ? (2 - isLeapYear) : 0);

Translating Into Other Languages

It is now trivial to convert this expression into other programming languages. In Perl or PHP, the expression becomes:

$day = (367 * $month - 362) / 12 + $date -
	(($month > 2) ? (2 - $isLeapYear) : 0);

In the C, C++ or Java programming languages, the expression can be written:

int day = (367 * month - 362) / 12 + date -
	((month > 2) ? (2 - isLeapYear) : 0);

Demo