Mac dayofweek issue - macos

Would anyone know why the following code works correctly on Windows and not on Mac??
Today (24/11/2010) should return 47 not 48 as per MacOS
def fm_date = '24/11/2010'
import java.text.SimpleDateFormat
def lPad = {it ->
st = '00' + it.toString()
return st.substring(st.length()-2, st.length())
}
dfm = new SimpleDateFormat("dd/MM/yyyy")
cal=Calendar.getInstance()
cal.setTime( dfm.parse(fm_date) )
now = cal.get(Calendar.WEEK_OF_YEAR)
cal.add(Calendar.DAY_OF_MONTH,-7)
prev = cal.get(Calendar.WEEK_OF_YEAR)
cal.add(Calendar.DAY_OF_MONTH,14)
next = cal.get(Calendar.WEEK_OF_YEAR)
prev = 'diary' + lPad(prev) + '.shtml'
next = 'diary' + lPad(next) + '.shtml'
return 'diary' + lPad(now) + '.shtml'

I believe it's an ISO week number issue...
If I use this code adapted (and groovyfied) from yours:
import java.text.SimpleDateFormat
def fm_date = '24/11/2010'
Calendar.getInstance().with { cal ->
// We want ISO Week numbers
cal.firstDayOfWeek = MONDAY
cal.minimalDaysInFirstWeek = 4
setTime new SimpleDateFormat( 'dd/MM/yyyy' ).parse( fm_date )
now = cal[ WEEK_OF_YEAR ]
}
"diary${"$now".padLeft( 2, '0' )}.shtml"
I get diary47.shtml returned
As the documentation for GregorianCalendar explains, if you want ISO Month numbers:
Values calculated for the WEEK_OF_YEAR
field range from 1 to 53. Week 1 for a
year is the earliest seven day period
starting on getFirstDayOfWeek() that
contains at least
getMinimalDaysInFirstWeek() days from
that year. It thus depends on the
values of getMinimalDaysInFirstWeek(),
getFirstDayOfWeek(), and the day of
the week of January 1. Weeks between
week 1 of one year and week 1 of the
following year are numbered
sequentially from 2 to 52 or 53 (as
needed).
For example, January 1, 1998 was a
Thursday. If getFirstDayOfWeek() is
MONDAY and getMinimalDaysInFirstWeek()
is 4 (these are the values reflecting
ISO 8601 and many national standards),
then week 1 of 1998 starts on December
29, 1997, and ends on January 4, 1998.
If, however, getFirstDayOfWeek() is
SUNDAY, then week 1 of 1998 starts on
January 4, 1998, and ends on January
10, 1998; the first three days of 1998
then are part of week 53 of 1997.
Edit
Even Groovier (from John's comment)
def fm_date = '24/11/2010'
Calendar.getInstance().with { cal ->
// We want ISO Week numbers
cal.firstDayOfWeek = MONDAY
cal.minimalDaysInFirstWeek = 4
cal.time = Date.parse( 'dd/MM/yyyy', fm_date )
now = cal[ WEEK_OF_YEAR ]
}
"diary${"$now".padLeft( 2, '0' )}.shtml"
Edit2
Just ran this on Windows using VirtualBox, and got the same result

Related

Swift 2.0+ finding the day of the week a month starts

I am using Swift 2.0+ and need it to be compatible with iOS 8.0+. I want to find out what day of the week a month starts. For example, November 15 would return Sunday, December 15 would return Tuesday, and January 16 would return Friday. I think the best approach would be to get the current day of the week and day of the month then count backwards until you reach 1 and get that day. Currently I have
// Returns the day of the week as an integer 1 being Sunday and 7 being Saturday
func dayOfWeek() -> Int {
let components = calender.component(NSCalendarUnit.Weekday, fromDate: t)
return components
}
Which calculates the current day of the week, after that I hit a roadblock.
I found the answer to be
// This function should return the value I need 0 being Sunday and 6 being Saturday
func dayMonthStarted() -> Int {
// 0 -> Sunday 6 -> Saturday
var component = calender.component(NSCalendarUnit.Day, fromDate: date)
var day = dayOfWeek()
while component > 1 {
component--
if day == 0 {
day = 6
} else {
day--
}
}
return day - 1
}

How to find out the date of second Monday of each month of given year?

My customer has an event each second Monday of each month.
I need to mark them with red in calendar.
How do i "cleanly" find out the date of that Mondays?
Here's my version.
If the eighth of the month is a Monday, then it is the second Monday. If it is not a Monday, then how many days until the next Monday?
oct_2012 = Date.new 2012, 10, 8
oct_2012.wday # => 1, We're done!
nov_2012 = Date.new 2012, 11, 8
nov_2012.wday # => 4
nov_2012 + (8 - nov_2012.wday) # => 2012-11-12
Does that help?
Edit
Easier version: Just add and be done. This algorithm works even if the month starts on a Monday.
oct_2012 = Date.new 2012, 10, 1
oct_2012 + (8 - oct_2012.wday) # => 2012-10-08
nov_2012 = Date.new 2012, 11, 1
nov_2012 + (8 - nov_2012.wday) # => 2012-11-12
One rule and done!
You second Monday will always fall within the 8th and 14th of each month.

Linq Query - get current month plus previous months

I need to build a Linq query that will show the results as follow:
Data:
Sales Month
----------------------
10 January
20 February
30 March
40 April
50 May
60 June
70 July
80 August
90 September
100 October
110 November
120 December
I need to get the results based on this scenario:
month x = month x + previous month
that will result in:
Sales Month
--------------------
10 January
30 February (30 = February 20 + January 10)
60 March (60 = March 30 + February 30)
100 April (100 = April 40 + March 60)
.........
Any help how to build this query ?
Thanks a lot!
Since you wanted it in LINQ...
void Main()
{
List<SaleCount> sales = new List<SaleCount>() {
new SaleCount() { Sales = 10, Month = 1 },
new SaleCount() { Sales = 20, Month = 2 },
new SaleCount() { Sales = 30, Month = 3 },
new SaleCount() { Sales = 40, Month = 4 },
...
};
var query = sales.Select ((s, i) => new
{
CurrentMonth = s.Month,
CurrentAndPreviousSales = s.Sales + sales.Take(i).Sum(sa => sa.Sales)
});
}
public class SaleCount
{
public int Sales { get; set; }
public int Month { get; set; }
}
...but in my opinion, this is a case where coming up with some fancy LINQ isn't going to be as clear as just writing out the code that the LINQ query is going to generate. This also doesn't scale. For example, including multiple years worth of data gets even more hairy when it wouldn't have to if it was just written out the "old fashioned way".
If you don't want add up all of the previous sales for each month, you will have to keep track of the total sales somehow. The Aggregate function works okay for this because we can build a list and use its last element as the current total for calculating the next element.
var sales = Enumerable.Range(1,12).Select(x => x * 10).ToList();
var sums = sales.Aggregate(new List<int>(), (list, sale) => list.Concat(new List<int>{list.LastOrDefault() + sale});

algorithm for calculating a week # from a date with custom start of week? (for iCal)

I can only find algorithm for getting ISO 8601 week (week starts on a Monday).
However, the iCal spec says
A week is defined as a seven day period, starting on the day of the
week defined to be the week start (see WKST). Week number one of the
calendar year is the first week that contains at least four (4) days
in that calendar year.
Therefore, it is more complex than ISO 8601 since the start of week can be any day of the week.
Is there an algorithm to determine what is the week number of a date, with a custom start day of week?
or... is there a function in iCal4j that does this? Determine a weekno from a date?
Thanks!
p.s. Limitation: I'm using a JVM language that cannot extend a Java class, but I can invoke Java methods or instantiate Java classes.
if (input_date < firstDateOfTheYear(WKST, year))
{
return ((isLeapYear(year-1))?53:52);
}
else
{
return ((dayOfYear(input_date) - firstDateOfTheYear(WKST, year).day)/7 + 1);
}
firstDateOfTheYear returns the first calendar date given a start of week(WKST) and the year, e.g. if WKST = Thursday, year = 2012, then it returns Jan 5th.
dayOfYear returns sequencial numerical day of the year, e.g. Feb 1st = 32
Example #1: Jan 18th, 2012, start of week is Monday
dayOfYear(Jan 18th, 2012) = 18
firstDateOfTheYear(Monday, 2012) = Jan 2nd, 2012
(18 - 2)/7 + 1 = 3
Answer Week no. 3
Example #2: Jan 18th, 2012, start of week is Thursday
dayOfYear(Jan 18th, 2012) = 18
firstDateOfTheYear(Thursday, 2012) = Jan 5th, 2012
(18 - 5)/7 + 1 = 2
Answer Week no. 2
Example #3: Jan 1st, 2012, start of week is Monday
firstDateOfTheYear(Monday, 2012) = Jan 2nd, 2012
IsLeapYear(2012-1) = false
Jan 1st, 2012 < Jan 2nd, 2012
Answer Week no. 52
Let daysInFirstWeek be the number of days on the first week of the year that are in January. Week starts on a WKST day. (e.g. if Jan 1st is a WKST day, return 7)
Set dayOfYear to the n-th days of the input date's year (e.g. Feb 1st = 32)
If dayOfYear is less than or equal to daysInFirstWeek
3.1. if daysInFirstWeek is greater than or equal to 4, weekNo is 1, skip to step 5.
3.2. Let daysInFirstWeekOfLastYear be the number of days on the first week of the previous year that are in January. Week starts on a WKST day.
3.3. if daysInFirstWeekOfLastYear is 4 or last year is Leap year and daysInFirstWeekOfLastYear is 5, weekNo is 53, otherwise weekNo is 52, skip to step 5.
Set weekNo to ceiling((dayOfYear - daysInFirstWeek) / 7)
4.1. if daysInFirstWeek greater than or equal to 4, increment weekNo by 1
4.2. if daysInFirstWeek equal 53 and count of days on the first week (starting from WKST) of January in the year of inputDate's year + 1 is greater than or equal to 4, set weekNo to 1
return weekNo

Using the Ruby Date class for Astronomical data

~ Approximate Solar Noon
lw = 88.743 # my longitude
jdate = Date.ordinal_to_jd(Time.now.year, Time.now.yday)
n = (jdate - 2451545 - 0.0009 - lw / 360).round # lw is users longitude west of 0.
j_noon = 2451545 + 0.0009 + lw / 360 + n
puts j_noon
=> 2455616.24740833
As an update, part of the confusion would be that solar noon is where all
calculations started since January 1, 4713 BC Greenwich noon.
The correct use of Date.ordinal_to_jd has not compensated for this fact. So by
adding or subtracting 12 hours like this:
jdn = Date.ordinal_to_jd(Time.now.year, Time.now.yday) - 0.5
we should get less errors. Just which do we use though since our calculations
start with yesterdays noon?
The code is derived from the two equations from this page Sunrise_equation.
The first answer I got from a user here was that we don't understand the use of
0.0009 and lw / 360. lw / 360 would appear to be a fractional day of arc from the
prime meridian. As for the 0.0009, it must be a small amount of variance in
seconds since January 1, 4713 BC Greenwich noon. see IAU standards for more info
I calculate it to be 0.007776 seconds according to this page.
I have a little bit of info from Date class not including method details.
=begin
--------------------------------------------------------------------- Class: Date
Class representing a date.
See the documentation to the file date.rb for an overview.
Internally, the date is represented as an Astronomical Julian Day Number, ajd.
The Day of Calendar Reform, sg, is also stored, for conversions to other date formats.
(There is also an of field for a time zone offset,
but this is only for the use of the DateTime subclass.)
A new Date object is created using one of the object creation class methods named
after the corresponding date format, and the arguments appropriate to that date
format; for instance, Date::civil()
(aliased to Date::new()) with year, month, and day-of-month, or Date::ordinal() with
year and day-of-year.
All of these object creation class methods also take the Day of Calendar Reform as an
optional argument.
Date objects are immutable once created.
Once a Date has been created, date values can be retrieved for the different date
formats supported using instance methods. For instance, #mon() gives the Civil month,
#cwday() gives the Commercial day of the week, and #yday() gives the Ordinal day of
the year. Date values can be retrieved in any format, regardless of what format was
used to create the Date instance.
The Date class includes the Comparable module, allowing date objects to be compared
and sorted, ranges of dates to be created, and so forth.
---------------------------------------------------------------------------------
Includes:
Comparable(<, <=, ==, >, >=, between?)
Constants:
MONTHNAMES: [nil] + %w(January February March April May June July August
September October November December)
DAYNAMES: %w(Sunday Monday Tuesday Wednesday Thursday Friday Saturday)
ABBR_MONTHNAMES: [nil] + %w(Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec)
ABBR_DAYNAMES: %w(Sun Mon Tue Wed Thu Fri Sat)
ITALY: 2299161
ENGLAND: 2361222
JULIAN: Infinity.new
GREGORIAN: -Infinity.new
Class methods:
_load, _parse, _strptime, ajd_to_amjd, ajd_to_jd, amjd_to_ajd, civil, civil_to_jd,
commercial, commercial_to_jd, day_fraction_to_time, gregorian?, gregorian_leap?, jd,
jd_to_ajd, jd_to_civil, jd_to_commercial, jd_to_ld, jd_to_mjd, jd_to_ordinal,
jd_to_wday, julian?, julian_leap?, ld_to_jd, mjd_to_jd, new, now, ordinal,
ordinal_to_jd, parse, s3e, strptime, time_to_day_fraction, today, valid_civil?,
valid_commercial?, valid_jd?, valid_ordinal?, valid_time?
Instance methods:
+, -, <<, <=>, ===, >>, _dump, ajd, amjd, asctime, civil, commercial, ctime, cwday,
cweek, cwyear, day, day_fraction, downto, england, eql?, gregorian, gregorian?, hash,
hour, inspect, italy, jd, julian, julian?, ld, leap?, mday, min, mjd, mon, month,
new_offset, new_start, next, next_day, offset, ordinal, sec, sec_fraction, start,
step, strftime, succ, time, to_s, to_yaml, upto, wday, weeknum0, weeknum1, wnum0,
wnum1, yday, year, zone
=end
As a side note, it's great that Ruby has a way to calculate the julian-date.
I'm looking into the Javascript code from NOAA.
Here is a class that I was inspired to write by the link.
class JulianDayNumber
def initialize(year = 2000, month = 1, day = 1) #defaults to Jan. 01, 2000
#year = year
#month = month
#day = day
end
def calcJDN
if (#month <= 2) then
#year -= 1
#month += 12
end
varA = (#year/100).floor
varB = 2 - varA + (varA/4).floor
jdn = (365.25*(#year + 4716)).floor \
+ (30.6001*(#month+1)).floor \
+ #day + varB - 1524.5
return jdn
end
end
jd = JulianDayNumber.new(2011, 3, 2)
julianday = jd.calcJDN
puts julianday
=> 2455622.5
Now this gets me there but I'm still researching for the way back for a number such
as the one calculated by the top most equation. Trying this we can see that we do
get a 0.5 in the JDN. Who is right? Ruby or NOAA?
NOAA uses the January 1st 2000 value of 2451545.0 that is subtracted from the jd to get time
in fractional century like this
def calcTimeJulianCent(j)
t = (j - 2451545.0)/36525.0
return t
end
Ruby has a number of ways of calculating Julian Day and you need to pick the right one. NOAA is calculating the JD since January 1, 4713 BC Greenwich noon as you know. It always ends in .5 because they are leaving out the fractional days.
Ruby's Julian Day is weird:
For scientific purposes, it is
convenient to refer to a date simply
as a day count, counting from an
arbitrary initial day. The date first
chosen for this was January 1, 4713
BCE. A count of days from this date is
the Julian Day Number or Julian Date,
which is abbreviated as jd in the Date
class. This is in local time, and
counts from midnight on the initial
day.
Which makes no sense for astronomical use. but wait..
The stricter usage is in UTC, and
counts from midday on the initial day.
This is referred to in the Date class
as the Astronomical Julian Day Number,
and abbreviated as ajd. In the Date
class, the Astronomical Julian Day
Number includes fractional days.
(rubydoc)
This is what you are looking for, ajd. Just get it without the fractional days:
julianday = Date.civil(#year, #month, #day).ajd
puts julianday
=> 2455622.5
No need to port 9 lines of JavaScript from NOAA. Ruby's got your back! ;)
Well thanks everybody, I guess that I can answer my own question now. I overlooked a simple method in the Date class. It is Date.day_fraction_to_time(day fractional). As I have a working program now I would like to share it with eveyone.
include Math
to_r = PI / 180.0
to_d = 180.0 / PI
latitude = 41.9478 # my latitude
longitude = 88.74277 # my longitude
lw = longitude / 360
jdate = Date.civil(Time.now.year, Time.now.month, Time.now.day).ajd
jdate = (jdate * 2).to_i/2 + 1
n = (jdate - 2451545 - 0.0009 - lw).round
j_noon = 2451545 + 0.0009 + lw + n
mean_anomaly = (357.52911 + 0.98560028 * (jdate - 2451545)) % 360
center = 1.9148 * sin(mean_anomaly * to_r) + 0.0200 * sin(2 * mean_anomaly * to_r) + \
0.0003 * sin(3 * mean_anomaly * to_r)
lambda = (mean_anomaly + 102.9372 + center + 180) % 360
j_transit = j_noon + (0.0053 * sin(mean_anomaly * to_r)) - (0.0069 * sin(2 * lambda * \
to_r))
delta = asin(0.397753054 * sin(lambda * to_r)) * to_d
omega = acos(sin(-0.83 * to_r)/cos(latitude * to_r) * cos(delta * to_r) \
- tan(latitude * to_r) * tan(delta * to_r)) * to_d
j_set = 2451545 + 0.0009 + ((omega + longitude)/360 + n + 0.0053 * sin(mean_anomaly * \
to_r)) - 0.0069 * sin(2 * lambda * to_r)
j_rise = j_transit - (j_set - j_transit)
rise = Date.day_fraction_to_time(j_rise - jdate)# + 0.25 for + 6 hours
risehour = rise[0].to_s
risemin = rise[1].to_s
risetime = "#{risehour}:#{risemin}"
puts "Sun rise = #{risetime} UTC"
transit = Date.day_fraction_to_time(j_transit - jdate)# + 0.25
transithour = transit[0].to_s
transitmin = transit[1].to_s
transittime = "#{transithour}:#{transitmin}"
puts "Solar noon = #{transittime} UTC"
set = Date.day_fraction_to_time(j_set - jdate)# + 0.25
sethour = set[0].to_s
setmin = set[1].to_s
settime = "#{sethour}:#{setmin} UTC"
puts "Sun set = #{settime}"
The method ordinal_to_jd converts the day with index 0 of the year 2011 (Gregorian calendar) to the corresponding day in the Julian calendar, then you are using the magical value of 0.0009 for which i dont know any reason, then you are adding the ratio of your longitude (east or west?) of the whole 360* circle and then adding todays day-of-year (54 if you evaluated it today). The combination of Julian calendar and longitudinal ratio makes not much sense, but hey its a nice number since you mixed a 0.0009 in.

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