Sum date object and time object on RoR - ruby-on-rails

sorry for my english...
I have two object, one Date object and one Time object, I want sum them to get a DateTime object, this is I am trying...
[35] pry(main)> dateobj = Date.today
=> Sun, 06 Oct 2013
[36] pry(main)> timeobj = Time.parse("02:00:00")
=> 2013-10-06 02:00:00 -0600
[37] pry(main)> datetimeobj = dateobj + timeobj
TypeError: expected numeric
from /home/elquick/www/rails/vivsan/http/vendor/bundle/ruby/2.0.0/gems/activesupport-3.2.14/lib/active_support/core_ext/date/calculations.rb:90:in `+'
[38] pry(main)>
Some help?
Thanks!

Try to use
Date.today.to_datetime + Time.parse('02:00:00').seconds_since_midnight.seconds
Here is my result:
.0.0-p195 :019 > Date.today.to_datetime
=> Mon, 07 Oct 2013 00:00:00 +0000
2.0.0-p195 :020 > Time.parse('02:00:00').seconds_since_midnight.seconds
=> 7200.0 seconds
2.0.0-p195 :021 > Date.today.to_datetime + Time.parse('02:00:00').seconds_since_midnight.seconds
=> Mon, 07 Oct 2013 02:00:00 +0000

I found this...
dateobj = Date.new(2013, 10, 6)
timeobj = Time.parse("02:00:00")
datetime = DateTime.new(dateobj.year, dateobj.month, dateobj.day, timeobj.hour, timeobj.min, timeobj.sec)

Related

Is it possible in Delphi to union enumerations into a larger enumeration?

Delphi can have enumerated types, e.g.:
type
TDay = (Mon, Tue, Wed, Thu, Fri, Sat, Sun); // Enumeration values
Is it possible to union enumerated types:
type
TWeekDay = (Mon, Tue, Wed, Thu, Fri);
TWeekendDay = (Sat, Sun);
TDay = (TWeekday, TWeekendDay); //hypothetical syntax
In reality, i need to decompose a large list into the disjoint items they actually are, without breaking source-code compatibility:
type
TWeekDay = (Mon, Tue, Wed, Thu, Fri);
TWeekendDay = (Sat, Sun);
TDay = (Mon, Tue, Wed, Thu, Fri, Sat, Sun); //identifier redeclared syntax error
And then change some variables:
Day: TWeekday;TDay;
Day: TWeekendDay;TDay
It's sort of the moral equivalent of strict typing. 🕗
The answer is "No".
But a workaround available to you, if subrages are contiguous, is to use subranges:
TDay = (Mon, Tue, Wed, Thu, Fri, Sat, Sun);
TWeekDay2 = Mon..Fri;
TWeekday = type TWeekDay2;
TWeekendDay2 = Sat..Sun;
TWeekendDay = type TWeekendDay;

How to convert a localized string date to Ruby Date object?

I have a date which works fine when using it as English
> Date.strptime("Aug 02, 2015", "%b %d, %Y")
Sun, 02 Aug 2015
However when I use it another language es, that doesnt work. E.g.
> Date.strptime("ago 02, 2015", "%b %d, %Y")
ArgumentError: invalid date
The text strings ago 02, 2015 itself comes from another service, and I need to standardize it to a particular format such as
> I18n.localize(Date.strptime("Aug 02, 2015", "%b %d, %Y"), format: "%m/%d/%Y")
"08/02/2015"
Is there a way to do this in Ruby, so that
> I18n.localize(Date.strptime("ago 02, 2015", "%b %d, %Y"), format: "%m/%d/%Y")
"08/02/2015"
I'm assuming you've already tried the following function which handles most scenarios for turning something into a DateTime object.
#date = Date.parse("ago 02, 2015")
Other than that you can try appending to the date constants so that it properly picks them up. Personally haven't tried this approach but might work for you?
require 'date'
Date::MONTHNAMES = [nil] + %w( Enero Febrero Marzo Abril Mayo Junio Julio Agosto Septiembre Octubre Noviembre Diciembre )
Date::DAYNAMES = %w( Lunes Martes Miércoles Jueves Viernes Sábado Domingo )
Date::ABBR_MONTHNAMES = [nil] + %w( Ene Feb Mar Abr May Jun Jul Ago Sep Oct Nov Dic )
Date::ABBR_DAYNAMES = %w( Lun Mar Mié Jue Vie Sáb Dom )
Lastly, have you considered using the Chronic Gem for date parsing? I believe it should handle cross language cases.
You can substitute the Spanish words with English first. You can do it with #gsub, I18n yaml lookup, or a dictionary method, e.g.
dictionary = { "ago" => "Aug" }
date = "ago 02, 2015"
words = date.split
words[0] = dictionary[words[0]]
date = words.join(" ") # "Aug 02, 2015"
Refactor it to take advantage of the power of OOP:
class DateDictionary
LOOKUP = { "ene" => "Jan", "ago" => "Aug" }
def translate(date)
words = date.split
words[0] = LOOKUP[words[0]]
date = words.join(" ")
end
end
date = DateDictionary.new.translate("ago 02, 2015") # "Aug 02, 2015"

Sum the next value of hash in an array of hashes

I have an array of hashes like this:
[{Mon, 09 May 2016 14:49:17 UTC +00:00=>12},
{Sun, 17 Apr 2016 14:08:40 UTC +00:00=>30},
{Sun, 16 Apr 2016 14:08:40 UTC +00:00=>18},
{Sun, 15 Apr 2016 14:03:33 UTC +00:00=>21}]
How can I sum the previous value from the oldest date to the current date, my expected output will be:
[{Mon, 09 May 2016 14:49:17 UTC +00:00=>81},
{Sun, 17 Apr 2016 14:08:40 UTC +00:00=>69},
{Sun, 16 Apr 2016 14:08:40 UTC +00:00=>39},
{Sun, 15 Apr 2016 14:03:33 UTC +00:00=>21}]
Thanks!
Assuming that the key of every hash in your array is a DateTime object, you can get what your want with this:
balance = [
{DateTime.parse('Mon, 09 May 2016 14:49:17 UTC +00:00')=>12},
{DateTime.parse('Sun, 17 Apr 2016 14:08:40 UTC +00:00')=>30},
{DateTime.parse('Sun, 16 Apr 2016 14:08:40 UTC +00:00')=>18},
{DateTime.parse('Sun, 15 Apr 2016 14:03:33 UTC +00:00')=>21}
] # => your original array
# Get expected array.
balance.map{ |h|
{
h.keys.first => balance.select{ |e|
e.keys.first <= h.keys.first }.map{ |s|
s[s.keys.first] }.reduce(:+)
}
}
I split the code in lines in order to improve readability.
Another approach would be to sort the array first and then use the map function to keep a running total to collect the required data.
# sort the balances by date
balance = balance.sort {|a, b| a.keys.first <=> b.keys.first }
# get running total and collect for each date
total = 0
balance.map do |entry|
date, value = entry.first
total += value
{date => total}
end
I assume your array is in lastest-to-earliest date order and looks something like arr below:
a = [{ "Mon, 09 May 2016 14:49:17 UTC +00:00"=>12 },
{ "Sun, 17 Apr 2016 14:08:40 UTC +00:00"=>30 },
{ "Sun, 16 Apr 2016 14:08:40 UTC +00:00"=>18 },
{ "Sun, 15 Apr 2016 14:03:33 UTC +00:00"=>21 }]
require 'date'
arr = a.map do |h|
(d, v) = h.to_a.first
{ DateTime.parse(d) => v }
end
#=> [{#<DateTime: 2016-05-09T14:49:17+00:00 ((2457518j,53357s,0n),+0s,2299161j)>=>12},
# {#<DateTime: 2016-04-17T14:08:40+00:00 ((2457496j,50920s,0n),+0s,2299161j)>=>30},
# {#<DateTime: 2016-04-16T14:08:40+00:00 ((2457495j,50920s,0n),+0s,2299161j)>=>18},
# {#<DateTime: 2016-04-15T14:03:33+00:00 ((2457494j,50613s,0n),+0s,2299161j)>=>21}]
We can then compute the required array as follows.
cumv = 0
arr.reverse.
map { |h| h.to_a.first }.
each_with_object([]) do |(d,v),a|
cumv += v
a << { d => cumv }
end.
reverse
#=> [{#<DateTime: 2016-05-09T14:49:17+00:00 ((2457518j,53357s,0n),+0s,2299161j)>=>81},
# {#<DateTime: 2016-04-17T14:08:40+00:00 ((2457496j,50920s,0n),+0s,2299161j)>=>69},
# {#<DateTime: 2016-04-16T14:08:40+00:00 ((2457495j,50920s,0n),+0s,2299161j)>=>39},
# {#<DateTime: 2016-04-15T14:03:33+00:00 ((2457494j,50613s,0n),+0s,2299161j)>=>21}]

How do I get the first Thursday of the month in Ruby/Rails?

The following Ruby code gets me the first day of each month :
require 'active_support/all'
# get the date at the beginning of this month
date = Date.today.beginning_of_month
# get the first day of the next 5 months
5.times do |num|
date = date.next_month
p date
end
Which gives :
=> Fri, 01 Aug 2014
=> Mon, 01 Sep 2014
=> Wed, 01 Oct 2014
=> Sat, 01 Nov 2014
=> Mon, 01 Dec 2014
But how do I get the first Thursday of each month? i.e.
=> Thu, 07 Aug 2014
=> Thu, 04 Sep 2014
=> Thu, 02 Oct 2014
=> Thu, 06 Nov 2014
=> Thu, 04 Dec 2014
There's no need for iterations or conditions just get the so called delta of days till next thursday:
#4 is thursday because wday starts at 0 (sunday)
date = Date.today.beginning_of_month
date += (4 - date.wday) % 7
p date
=> Thu, 03 Jul 2014
That my opinion:
date_begin = Date.today.beginning_of_month
date_end = date_begin + 5.month
[*date_begin..date_end].select(&:thursday?).uniq(&:month)
=> [Thu, 03 Jul 2014, Thu, 07 Aug 2014, Thu, 04 Sep 2014, Thu, 02 Oct 2014, Thu, 06 Nov 2014]
Just for fun
class Date
def skip_to_thursday
# given current weekday, how many days we need to add for it to become thursday
# for example, for monday (weekday 1) it's 3 days
offset = lambda {|x| (4-x) % 7 }
self + offset[wday]
end
end
# get the date at the beginning of this month
date = Date.today.beginning_of_month
date.skip_to_thursday # => Thu, 03 Jul 2014
Here is my way :
def first_thursday
date = Date.today.beginning_of_month
date += 1 until date.wday == 4
date
end
first_thursday # => Thu, 03 Jul 2014
you can use something like this:
def first_thursday(months_ahead)
start_of_month = months_ahead.months.from_now.beginning_of_month.to_date
start_of_month += (4 - start_of_month.cwday) % 7
end
first_thursday 1
=> Thu, 07 Aug 2014
first_thursday 2
=> Thu, 04 Sep 2014
I ran into this problem for a recurring_events feature that I needed to build. I changed some of the variables to find the first Thursday but it also shows how you could evolve the answer to find the 2nd or 3rd Thursday (or any day of the week for that matter) if you had a week and day of the week count.
def find_thursday
start_of_month = DateTime.now.beginning_of_month
month_day = nil
loop do
month_day = start_of_month += 1.day
break if month_day.wday == find_weekday("Thu")
end
return month_day
end
def find_weekday
d = default_weekdays.find { |d| d[:day] == start_date.strftime("%a") }
d[:count]
end
def default_weekdays
return [
{ day: 'Sun', count: 0 },
{ day: 'Mon', count: 1 },
{ day: 'Tue', count: 2 },
{ day: 'Wed', count: 3 },
{ day: 'Thu', count: 4 },
{ day: 'Fri', count: 5 },
{ day: 'Sat', count: 6 },
]
end

Is the Time object suitable to create a calendar?

I want to make a database-backed calendar. Will the Time object make my life easier? It hasn't so far...
The .end_of_year method gives me some strange information. If it's contemporary time it works flawlessly:
date = '2012-3-2'.to_time(:utc) #=> 2012-03-02 00:00:00 UTC
date.end_of_year #=> 2012-12-31 23:59:59 UTC
However, if you go back in time things get strange.
date = '1399-3-2'.to_time(:utc) #=> 1399-03-02 00:00:00 UTC
date.end_of_year #=> 1399-12-23 23:59:59 UTC
23rd of December? Shouldn't that be 31st?
It's not even consistent:
date = '0000-3-2'.to_time(:utc) #=> 0000-03-02 00:00:00 UTC
date.end_of_year #=> 0001-01-02 23:59:59 UTC
Um, the 2nd of January? OF THE NEXT YEAR? What is going on?
Also, are leap years taken into account by the object?
You could use DateTime instead:
date = '2012-3-2'.to_datetime #=> Fri, 02 Mar 2012 00:00:00 +0000
date.end_of_year #=> Mon, 31 Dec 2012 23:59:59 +0000
date = '1399-3-2'.to_datetime #=> Sun, 02 Mar 1399 00:00:00 +0000
date.end_of_year #=> Wed, 31 Dec 1399 23:59:59 +0000
date = '0000-3-2'.to_datetime #=> Tue, 02 Mar 0000 00:00:00 +0000
date.end_of_year #=> Fri, 31 Dec 0000 23:59:59 +0000
It's mora accurate, and you can format the output
I've did some digging. Here's what I found.
Let's begin with end_of_year:
def end_of_year
change(:month => 12).end_of_month
end
Which relies on change and end_of_month:
def end_of_month
last_day = ::Time.days_in_month(month, year)
last_hour{ days_since(last_day - day) }
end
The most interesting part is happening inside of days_since:
def days_since(days)
advance(:days => days)
end
The advance method is a bit more complex:
def advance(options)
unless options[:weeks].nil?
options[:weeks], partial_weeks = options[:weeks].divmod(1)
options[:days] = options.fetch(:days, 0) + 7 * partial_weeks
end
unless options[:days].nil?
options[:days], partial_days = options[:days].divmod(1)
options[:hours] = options.fetch(:hours, 0) + 24 * partial_days
end
d = to_date.advance(options)
time_advanced_by_date = change(:year => d.year, :month => d.month, :day => d.day)
seconds_to_advance = options.fetch(:seconds, 0) +
options.fetch(:minutes, 0) * 60 +
options.fetch(:hours, 0) * 3600
if seconds_to_advance.zero?
time_advanced_by_date
else
time_advanced_by_date.since(seconds_to_advance)
end
end
And he is the guy we're looking for :
# in rails console
time = '0000-01-01'.to_time(:utc) #=> 0000-01-01 00:00:00 UTC
time.advance(days: 1) #=> 0000-01-04 00:00:00 UTC
time.advance(days: 2) #=> 0000-01-05 00:00:00 UTC
time.advance(days: 3) #=> 0000-01-06 00:00:00 UTC
That's all for now. I will continue to dig.

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