I'm trying to figure out the most efficient way to find the closest payday from a given date.
Let's say we know payday are every 14 days and that January 6, 2022 was one of them. What is the most efficient way to find the next payday for a given date (the date is the beginning of a week actually) from a known payday?
Here is a current working ugly implementation
def payday_first_date
# TODO: Very reliable, but very inefficient
#payday_first_date ||= begin
date = known_pay_day
date -= 14.days until date < given_date
date += 14.days until date > given_date
date
end
end
def known_pay_day
# Thursday January 6, 2022 was a payday
#known_pay_day ||= Date.new(2022, 2, 17)
end
The given_date could be in the future and also in the past. It can be a date from another year aswell.
Thanks in advance!
First define the known pay date and two helper methods.
require 'date'
def date_str_to_date_obj(date_str)
DateTime.strptime(date_str, "%B %d, %Y").to_date
end
known_date = date_str_to_date_obj("January 6, 2022")
#=> #<Date: 2022-01-06 ((2459586j,0s,0n),+0s,2299161j)>
For presentation only:
def date_obj_to_date_str(date_obj)
date_obj.strftime("%B %d, %Y, %A")
end
date_obj_to_date_str(known_date)
#=> "January 06, 2022, Thursday"
The main method follows.
def closest_pay_day(known_date, target_date)
nbr_pay_dates = (target_date - known_date)/14.0
last_pay_date = known_date + nbr_pay_dates.floor * 14
next_pay_date = last_pay_date + 14
(next_pay_date - target_date < target_date - last_pay_date) ?
next_pay_date : last_pay_date
end
Suppose the target date is after the known pay date.
target_date_str = "March 26, 2022"
Then
target_date = date_str_to_date_obj(target_date_str)
#=> #<Date: 2022-03-26 ((2459665j,0s,0n),+0s,2299161j)>
date_obj_to_date_str(target_date)
#=> "March 26, 2022, Saturday"
The closest pay period to the target date is therefore
closest = closest_pay_day(known_date, target_date)
#=> #<Date: 2022-03-31 ((2459670j,0s,0n),+0s,2299161j)>
date_obj_to_date_str(closest)
#=> "March 31, 2022, Thursday"
which is
(target_date- closest).to_i
#=> -5
days before the target date, meaning 5 days after the target date.
Now suppose the target date is before the known pay date.
target_date_str = "October 19, 2021"
Then
target_date = date_str_to_date_obj(target_date_str)
#=> #<Date: 2021-10-19 ((2459507j,0s,0n),+0s,2299161j)>
date_obj_to_date_str(target_date)
#=> "October 19, 2021, Tuesday"
date_obj_to_date_str(closest_pay_day(known_date, target_date))
#=> "October 14, 2021, Thursday"
Related
How can I format this datetime attribute 2017-10-15 or how can I get the 15th of the current month?
#today = Time.now
#mid = time.strftime("%Y-%m-15")
into October 15, 2017? I tried using to_formatted_s(:long), but it gives an error of undefined method.
In Rails 4 or above
> Date.today.beginning_of_month + 14
#=> Sun, 15 Oct 2017
# formatted as per your requirement
> (Date.today.beginning_of_month + 14).strftime("%B %d, %Y")
#=> "October 15, 2017"
Date#beginning_of_month it will return you beginning date of month of specified date (which will be always 1st, add 14 days) so you will get 15th of that month
Mydate = "2017-10-15"
Mydate.to_date.strftime("%B %d, %Y")
=> "October 15, 2017"
Using #GaganGami solution, you can create a Date class method middle_of_current_month
class Date
def self.middle_of_current_month
(today.beginning_of_month + 14).strftime("%B %d, %Y")
end
end
Date.middle_of_current_month
#=> "October 15, 2017"
How can I get let's say All the dates for Saturday and Sunday from X year to Y year and store them as array? Pseudo code would be
(year_today..next_year).get_all_dates_for_saturday_and_sunday
Or perhaps there are gems that cater to this already?
Try this:
(Date.today..Date.today.next_year).select { |date|
date.sunday? or date.saturday?
}
#=> [Sat, 03 Sep 2016,Sun, 04 Sep 2016,Sat, 10 Sep 2016,Sun, 11 Sep 2016...
(Date.today..(Date.today + 1.year)).select do |date|
date.saturday? || date.sunday?
end # => [Sat, 03 Sep 2016, Sun, 04 Sep 2016, Sat, 10 Sep 2016, ...
This will then give you an array of 104 elements containing every date which is a saturday or a sunday between today and today in a year.
The following approach emphasizes efficiency over brevity, by avoiding the need to determine if every day in a range is a given day (or one of two given days) of the week.
Code
require 'date'
def dates_by_years_and_wday(start_year, end_year, wday)
(first_date_by_year_and_wday(start_year, wday)...
first_date_by_year_and_wday(end_year+1, wday)).step(7).to_a
end
def first_date_by_year_and_wday(year, wday)
d = Date.new(year)
d + (wday >= d.wday ? wday - d.wday : 7 + wday - d.wday)
end
Notice that the range is defined with three dots, meaning the first date in end_year is excluded.
Example
SATURDAY = 6
SUNDAY = 0
start_year, end_year = 2015, 2017
dates_by_years_and_wday(start_year, end_year, SATURDAY)
#=> [#<Date: 2015-01-03 ((2457026j,0s,0n),+0s,2299161j)>,
# #<Date: 2015-01-10 ((2457033j,0s,0n),+0s,2299161j)>,
# ...
# #<Date: 2017-12-30 ((2458118j,0s,0n),+0s,2299161j)>]
dates_by_years_and_wday(start_year, end_year, SATURDAY).size
#=> 157
dates_by_years_and_wday(start_year, end_year, SUNDAY)
#=> [#<Date: 2015-01-04 ((2457027j,0s,0n),+0s,2299161j)>,
# #<Date: 2015-01-11 ((2457034j,0s,0n),+0s,2299161j)>,
# ...
# #<Date: 2017-12-31 ((2458119j,0s,0n),+0s,2299161j)>]
dates_by_years_and_wday(start_year, end_year, SUNDAY).size
#=> 157
I am new with ruby and I want to get the day of the last Friday of each month.
For example, the last Friday of March is 29, the Last Friday of April 26.
So, how can I get a solution?
I'm using the rails framework.
The method .cweek returns the week of the year, but does not return the week of the current month.
#!/usr/bin/env ruby
require 'date'
(1..12).each do |month|
d = Date.new(2013, month, -1)
d -= (d.wday - 5) % 7
puts d
end
Source (second/third Google result..)
I would go with Lee's answer, I'm only posting this one because (I thought) it's pretty cool.
Using gem Chronic (https://github.com/mojombo/chronic):
#Last Friday of this coming December
require 'chronic'
last_friday = Chronic.parse("First Friday of next January") - 1.week
Retrieving the last friday of a month can be made in a single line:
def last_fridays_for_every_month_of_year(year)
(1..12).map do |month|
Date.new(year, month, -1).downto(0).find(&:friday?)
end
end
You may use it like this:
last_fridays_for_every_month_of_year 2013
#=> [#<Date: 2013-01-25 ((2456318j,0s,0n),+0s,2299161j)>,
#<Date: 2013-02-22 ((2456346j,0s,0n),+0s,2299161j)>,
#<Date: 2013-03-29 ((2456381j,0s,0n),+0s,2299161j)>,
#<Date: 2013-04-26 ((2456409j,0s,0n),+0s,2299161j)>,
#<Date: 2013-05-31 ((2456444j,0s,0n),+0s,2299161j)>,
#<Date: 2013-06-28 ((2456472j,0s,0n),+0s,2299161j)>,
#<Date: 2013-07-26 ((2456500j,0s,0n),+0s,2299161j)>,
#<Date: 2013-08-30 ((2456535j,0s,0n),+0s,2299161j)>,
#<Date: 2013-09-27 ((2456563j,0s,0n),+0s,2299161j)>,
#<Date: 2013-10-25 ((2456591j,0s,0n),+0s,2299161j)>,
#<Date: 2013-11-29 ((2456626j,0s,0n),+0s,2299161j)>,
#<Date: 2013-12-27 ((2456654j,0s,0n),+0s,2299161j)>]
require "active_support/core_ext"
end_of_month = Date.today.end_of_month
if end_of_month - end_of_month.beginning_of_week >= 4
end_of_month+5.days
else
end_of_month-2.days
end
# => Fri, 29 Mar 2013
For example today is 28/07/2011
How do i get the weeks first day the monday which is 25/07/2011 in ruby
>> Date.today.beginning_of_week.strftime('%d/%m/%Y')
#=> 25/07/2011
See the Time and Date classes under Rails for more info, and strftime for information on the formatting options.
Without Rails/ActiveSupport:
phrogz$ irb
> require 'date'
> now = Date.today
#=> #<Date: 2011-07-28 (4911541/2,0,2299161)>
> sunday = now - now.wday
#=> #<Date: 2011-07-24 (4911533/2,0,2299161)>
> monday = now - (now.wday - 1) % 7
#=> #<Date: 2011-07-25 (4911535/2,0,2299161)>
> monday.iso8601
#=> "2011-07-25"
> monday.strftime('%d/%m/%Y')
#=> "25/07/2011"
For more, see the Date class in the Standard Library.
Wrapped up as a method:
require 'date'
# For weekdays to start on Monday use 1 for the offset; for Tuesday use 2, etc.
def week_start( date, offset_from_sunday=0 )
date - (date.wday - offset_from_sunday)%7
end
sun = Date.parse '2011-07-24'
week_start(sun,0).strftime('%a, %b-%d') #=> "Sun, Jul-24"
week_start(sun,1).strftime('%a, %b-%d') #=> "Mon, Jul-18"
Given a Ruby date, does a one liner exist for calculating the next anniversary of that date?
For example, if the date is May 01, 2011 the next anniversary would be May 01, 2012, however if it is December 01, 2011, the next anniversary is December 01, 2011 (as that date hasn't yet arrived).
If you date variable is an instance of Date then you can use >>:
Return a new Date object that is n months later than the current one.
So you could do this:
one_year_later = date >> 12
The same approach applies to DateTime. If all you have is a string, then you can use the parse method:
next_year = Date.parse('May 01, 2011') >> 12
next_year_string = (Date.parse('May 01, 2011') >> 12).to_s
IMHO you're better off using the date libraries (Date and DateTime) as much as possible but you can use the Rails extensions (such as 1.year) if you know that Rails will always be around or you don't mind manually pulling in active_support as needed.
An excellent gem exists for doing this called recurrence. You can checkout the source code or some samples:
https://github.com/fnando/recurrence
http://blog.plataformatec.com.br/tag/recurrence/
For example, if you have a date set you could try:
date = ...
recurrence = Recurrence.new(every: :year, on: [date.month, date.day])
puts recurrence.next
You can do it using Ruby's Date class:
the_date = Date.parse('jan 1, 2011')
(the_date < Date.today) ? the_date + 365 : the_date # => Sun, 01 Jan 2012
the_date = Date.parse('dec 31, 2011')
(the_date < Date.today) ? the_date.next_year : the_date # => Sat, 31 Dec 2011
Or, for convenience use ActiveSupport's Date class extensions:
require 'active_support/core_ext/date/calculations'
the_date = Date.parse('jan 1, 2011')
(the_date < Date.today) ? the_date.next_year : the_date # => Sun, 01 Jan 2012
the_date = Date.parse('dec 31, 2011')
(the_date < Date.today) ? the_date.next_year : the_date # => Sat, 31 Dec 2011
Try this:
def next_anniversary(d)
Date.today > d ? 1.year.from_now(d) : d
end
Pulling in a gem just to do this is overkill.
your_date > Date.today ? your_date : your_date >> 12